An object of mass m1=4.00 kg is tied to an object of mass m2=2.50 kg with 5 string 1 of length f=0.500m. Tension in String 1 should be greater than the force of gravity to keep m1. String 2 is more likely to break first.
(a) To find the tension in String 1 at the top of its motion, we need to consider the forces acting on object m1.
At the top of the motion, the tension in String 1 provides the centripetal force to keep m1 moving in a circular path. Additionally, we have the force of gravity acting on m1.
Let's analyze the forces:
Tension in String 1 (T1): This force provides the centripetal force.
Force of gravity (m1 * g): This force acts downward
Since the object is at the top of its motion, the tension in String 1 should be greater than the force of gravity to keep m1 moving in a circular path.
Therefore, T1 > m1 * g.
(b) To find the tension in String 2 at the top of its motion, we need to consider the forces acting on object m2.
At the top of the motion, the tension in String 2 provides the centripetal force to keep m2 moving in a circular path. Additionally, we have the force of gravity acting on m2.
Let's analyze the forces:
Tension in String 2 (T2): This force provides the centripetal force.
Force of gravity (m2 * g): This force acts downward.
Since the object is at the top of its motion, the tension in String 2 should be greater than the force of gravity to keep m2 moving in a circular path
Therefore, T2 > m2 * g.
(c) The string that will break first if the combination is rotated faster and faster depends on the tension each string can withstand. The tension in String 1 is generally greater than the tension in String 2 because m1 has a greater mass than m2. Therefore, if the combination is rotated faster and faster, String 2 is more likely to break first because it experiences lower tension compared to String 1.
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guitar string is 90.0 cm long and has a mass of 3.28 g Part A From the bridge to the support post (=ℓ) is 60.0 cm and the string is under a tension of 506 N What are the frequencies of the fundamental and first two overtones? Enter your answers numerically separated by commas.
The frequencies of the fundamental and the first two overtones are 250.31 Hz, 750.93 Hz and 1251.55 Hz respectively. ,Guitar string length = ℓ = 90.0 cm = 0.9 m Mass of the guitar string = m = 3.28 g = 0.00328 kg.
Tension in the guitar string = F = 506 N (newton).
Fundamental frequency of a guitar string is given by formula,f1=1/2L√(F/m)Where,L = Length of string F = Tension in stringm = Mass of string.
Now,Substitute the given values,f1=1/2 × 0.9m√(506 N/0.00328 kg)= 250.31 Hz.
The fundamental frequency of the guitar string is 250.31 Hz.
The first overtone frequency is given by,f2=3f1= 3 × 250.31 Hz= 750.93 Hz.
The second overtone frequency is given by,f3=5f1= 5 × 250.31 Hz= 1251.55 Hz.
Hence, the frequencies of the fundamental and the first two overtones are 250.31 Hz, 750.93 Hz and 1251.55 Hz respectively.
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Answer the following question. Show your work. Two capacitors have the same plate separation, but one has square plates and the other has circular plates. The square plates are a length L on each side, and the diameter of the circular plate is L. The capacitors have the same capacitance because they contain different dielectric materials. The dielectric constant of the material between the circular plates has a value of κcircle =3.82. What is the dielectric constant κsquare of the material between the square plates?
The dielectric constant κsquare of the material between the square plates is approximately 2.78.
The capacitance of a parallel plate capacitor is given by the formula C = κε₀A/d, where C is the capacitance, κ is the dielectric constant, ε₀ is the vacuum permittivity, A is the area of the plates, and d is the plate separation.
Since the capacitors have the same capacitance, we can equate their respective capacitance formulas:
κcircleε₀A/d = κsquareε₀A/d
Canceling out ε₀, A, and d from both sides of the equation, we are left with:
κcircle = κsquare
Substituting the given value of κcircle = 3.82, we find:
3.82 = κsquare
Therefore, the dielectric constant κsquare of the material between the square plates is approximately 3.82.
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You throw a stone down from a 100 m cliff at 3 m/s. A bird is on a ledge 47 meters below the top of the cliff. How long does it take the stone to hit the ledge and scare the bird? g=-9.8 m/s2. Round to one decimal place.
The time it takes for the stone to hit the ledge and scare the bird is approximately 5.4 seconds.
To solve the problem, we can use the kinematic equation for vertical motion:
[tex]h = h_0 + v_0t + \frac{1}{2}gt^2[/tex]
where:
h is the final height of the stone (47 m),
h0 is the initial height of the stone (100 m),
v0 is the initial velocity of the stone (3 m/s),
g is the acceleration due to gravity (-9.8 m/s²),
and t is the time.
Since the stone is thrown downward, the initial velocity (v0) will have a negative sign (-3 m/s).
Plugging in the values into the equation, we have:
[tex]47 = 100 - 3t + \frac{1}{2}(-9.8)t^2[/tex]
Rearranging the equation, we get:
[tex]0 = \frac{1}{2}(-9.8)t^2 - 3t + (100 - 47)[/tex]
Simplifying further:
0 = -4.9t^2 - 3t + 53
Now, we can solve this quadratic equation for time (t). We can either factor it or use the quadratic formula. Let's use the quadratic formula:
[tex]t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
where a = -4.9, b = -3, and c = 53.
Plugging in the values, we get:
[tex]t = \frac{3 \pm \sqrt{(-3)^2 - 4(-4.9)(53)}}{2(-4.9)}[/tex]
Simplifying the expression under the square root:
[tex]\frac{3 \pm \sqrt{9 + 1024.4}}{-9.8}[/tex]
[tex]t = \frac{3 \pm \sqrt{1033.4}}{-9.8}[/tex]
Now, let's calculate the two possible solutions:
[tex]t_1=\frac{3 + \sqrt{1033.4}}{-9.8}[/tex]
[tex]t_2=\frac{3 - \sqrt{1033.4}}{-9.8}[/tex]
Calculating t1 and t2 using a calculator, we find:
t1 ≈ -0.6 s (ignoring the negative solution as time cannot be negative in this context)
t2 ≈ 5.4 s
Therefore, it takes approximately 5.4 seconds for the stone to hit the ledge and scare the bird.
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Two charged particles of charges Q
1
=3nC and Q
2
=.9nC are separated by a distance of 20 cm. What is the magnitude of the electric fieid (in N/C) due to these charges at a point midway between them?
The magnitude of the electric field is **13.7 N/C**. The magnitude of the electric field due to two charged particles at a point midway between them can be calculated using the following equation:
E = k |Q1 + Q2| / r^2
where:
* E is the electric field
* k is the Coulomb constant
* |Q1 + Q2| is the sum of the magnitudes of the charges
* r is the distance between the charges
In this case, the charges are Q1 = 3 nC and Q2 = 0.9 nC. The distance between the charges is r = 20 cm = 0.2 m. The Coulomb constant is k = 8.988 * 10^9 N m^2/C^2.
So, the electric field is:
E = k |3 + 0.9| / 0.2^2 = 13.7 N/C
The electric field is a vector field, so it has both magnitude and direction. The direction of the electric field is from positive charges to negative charges. The electric field due to two charged particles at a point midway between them is the vector sum of the electric fields due to each individual charge. The electric field due to each individual charge can be calculated using the Coulomb's law. In this case, the charges are positive, so the electric field due to each charge is directed away from the charge. The electric field due to the two charges add together, so the net electric field is directed away from both charges.
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1(a) What is the maximum energy in eV of photons produced in a CRT using a 25.0-kV accelerating potential, such as a color TV? (b) What is their frequency?
2 (a) A γ-ray photon has a momentum of 8.00×10−21kg⋅m/ss. What is its wavelength? (b) Calculate its energy in MeV.
3 (a) Find the momentum of a 100-keV x-ray photon. (b) Find the equivalent velocity of a neutron with the same momentum. (c) What is the neutron’s kinetic energy in keV?
1. a. To calculate the maximum energy of photons produced in a CRT using a 25.0 kV accelerating potential, we can use the equation E = qV,
where E is the energy,
q is the elementary charge (1.602 × 10⁻¹⁹ C),
and V is the accelerating potential in volts.
We need to convert 25.0 kV to volts by multiplying by 1000.
Thus, V = 25,000 V. Plugging in the values we get:E = qV = (1.602 × 10⁻¹⁹ C) × (25,000 V) = 4.01 × 10⁻¹⁵ J
To convert joules to electron volts (eV), we need to divide by the elementary charge. Thus, the maximum energy of the photons is:4.01 × 10⁻¹⁵ J / 1.602 × 10⁻¹⁹ J/eV = 2.50 × 10⁴ eV (or 25 keV)
b. To find the frequency of the photons, we can use the equation E = hf,
where h is Planck's constant (6.626 × 10⁻³⁴ J s),
and f is the frequency. Solving for f,
we get:
f = E/h = (2.50 × 10⁴ eV) × (1.602 × 10⁻¹⁹ J/eV) / (6.626 × 10⁻³⁴ J s) = 1.03 × 10¹⁷ Hz (or 103 THz)
2. a. The momentum of a photon is given by p = h/λ, where h is Planck's constant (6.626 × 10⁻³⁴ J s) and λ is the wavelength. Solving for λ, we get:λ = h/p = (6.626 × 10⁻³⁴ J s) / (8.00 × 10⁻²¹ kg⋅m/s) = 8.28 × 10⁻¹⁵ m
b. The energy of a photon is given by E = hf, where h is Planck's constant (6.626 × 10⁻³⁴ J s) and f is the frequency.
To find the frequency, we can use the fact that the speed of light
(c) is related to wavelength and frequency by c = λf. Solving for f, we get:
f = c/λ = (2.998 × 10⁸ m/s) / (8.28 × 10⁻¹⁵ m) = 3.62 × 10²² Hz
Then, we can find the energy of the photon
:E = hf = (6.626 × 10⁻³⁴ J s) × (3.62 × 10²² Hz) = 2.40 × 10⁻¹³ JTo convert joules to megaelectronvolts (MeV),
we can use the conversion factor 1 MeV = 1.602 × 10⁻¹³ J. Thus, the energy of the photon is
E = 2.40 × 10⁻¹³ J / (1.602 × 10⁻¹³ J/MeV) = 1.50 MeV
3. a. The momentum of a photon is given by p = h/λ,
where h is Planck's constant (6.626 × 10⁻³⁴ J s)
and λ is the wavelength.
To find the momentum of a 100 keV x-ray photon, we need to convert the energy to joules and then find the corresponding wavelength. To convert keV to joules, we can use the conversion factor 1 eV = 1.602 × 10⁻¹⁹ J. Thus, the energy of the photon is:E = 100 keV × (1.602 × 10⁻¹⁹ J/eV) = 1.60 × 10⁻¹⁵ J
To find the wavelength, we can use the equation
E = hc/λ,
where c is the speed of light (2.998 × 10⁸ m/s) and λ is the wavelength
Solving for λ, we get:λ = hc/E = (6.626 × 10⁻³⁴ J s) × (2.998 × 10⁸ m/s) / (1.60 × 10⁻¹⁵ J) = 1.24 × 10⁻¹⁰ m
Then, we can find the momentum:
p = h/λ = (6.626 × 10⁻³⁴ J s) / (1.24 × 10⁻¹⁰ m) = 5.35 × 10⁻²⁴ kg⋅m/s
b. To find the velocity of a neutron with the same momentum as the 100 keV x-ray photon, we can use the equation p = mv, where m is the mass of the neutron and v is its velocity. Solving for v, we get:v = p/m = (5.35 × 10⁻²⁴ kg⋅m/s) / (1.675 × 10⁻²⁷ kg) = 3.19 × 10³ m/s c. To find the kinetic energy of the neutron, we can use the equation KE = ½mv², where m is the mass of the neutron and v is its velocity. Solving for KE, we get:
KE = ½mv² = ½ (1.675 × 10⁻²⁷ kg) × (3.19 × 10³ m/s)² = 1.70 × 10⁻²¹ JTo convert joules to kiloelectronvolts (keV), we can use the conversion factor 1 keV = 1.602 × 10⁻¹⁶ J.
Thus, the kinetic energy of the neutron is:KE = 1.70 × 10⁻²¹ J / (1.602 × 10⁻¹⁶ J/keV) = 10.6 keV
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A microscope has an objective lens with a focal length of 16.22 mm and an eyepiece with a focal length of 9.20 mm. With the length of the barrel set at 30.0 cm, the diameter of a red blood cell's image subtends an angle of 1.43mrad with the eye. If the final image distance is 30.0 cm from the eyepiece, what is the actual diameter of the red blood cell? Hint: To solve this equation, qo back to basics and use the thin-lens equation. μm
1/f = 1/do + 1/di
To find the actual diameter of the red blood cell, we need to divide the angular subtense by the magnification:
Actual diameter = angle / M_total
Substituting the given angle and calculated M_total:
Actual diameter = (1.43 × 10^-3) / M_total
Now, we can plug in the values and calculate the actual diameter.
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What is the magnitude of the momentum of a 0.0063−kg marble whose speed is 0.55 m/s ? Express your answer to two significant figures and include appropriate units. Part B What is the speed of a 0.130−kg baseball whose momentum has a magnitude of 3.34 kg⋅m/s ? Express your answer to two significant figures and include appropriate units
The magnitude of the momentum of the 0.0063 kg marble is approximately 0.0035 kg·m/s (final answer).
The momentum of an object is calculated by multiplying its mass by its velocity. In this case, the mass of the marble is 0.0063 kg and its speed is 0.55 m/s. Therefore, the momentum is given by:
momentum = mass * velocity
momentum = 0.0063 kg * 0.55 m/s
Calculating the product:
momentum ≈ 0.0035 kg·m/s
Therefore, the magnitude of the momentum of the marble is approximately 0.0035 kg·m/s.
For the second part of the question, we are given the magnitude of the momentum of a 0.130 kg baseball, which is 3.34 kg·m/s. We need to find the speed of the baseball.
The momentum of the baseball is given by the equation
momentum = mass * velocity
To find the velocity, we rearrange the equation:
velocity = momentum / mass
Plugging in the given values:
velocity = 3.34 kg·m/s / 0.130 kg
Calculating the division:
velocity ≈ 25.69 m/s
Therefore, the speed of the baseball is approximately 25.69 m/s.
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Calculate or Find the mass (kg unit) of an object that has a gravitational force of 750 N acting upon it.
The mass of the object is approximately 76.37 kg.
To calculate the mass of an object given its gravitational force, use the formula below:F = ma, Where:F = gravitational force (in Newtons)m = mass (in kilograms)a = acceleration due to gravity (9.81 m/s²)Given:
F = 750 NUsing the formula:
F = ma750
= m(9.81). Solving for m, we have:
m = 750/9.81m
≈ 76.37 kg. Therefore, the mass of the object is approximately 76.37 kg.
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An object falls under the influence of gravity, starting from initial position y0 = 25.4 m and initial velocity vyo = −2.5 m/s. Find its y position after 2.0 seconds. Find the y velocity of the particle at the same moment
the answer should be 0.78m and -22.12m/s but I dont know how it got that
The y velocity of the particle at the same moment is -22.1 m/s (approx).
Given that,
Initial position of the object,
y0 = 25.4 m
Initial velocity of the object,
vyo = -2.5 m/s
Time after which we have to calculate the position and velocity of the particle,
t = 2.0 s
Using the formula of displacement,
we have:
y = y0 + vyo × t + (1/2) × a × t²
Here, a is the acceleration due to gravity which is equal to -9.8 m/s² (negative because the object is falling in the downward direction).
Now substituting the values,
we get:
y = 25.4 m + (-2.5 m/s) × (2.0 s) + (1/2) × (-9.8 m/s²) × (2.0 s)²y = 25.4 m - 5 m - 19.6 m = 0.8 m
Therefore, the y position after 2.0 seconds is 0.8 m.
To calculate the y velocity of the particle at the same moment, we can use the formula:
v = vyo + a × tv = -2.5 m/s + (-9.8 m/s²) × (2.0 s) = -2.5 m/s - 19.6 m/sv = -22.1 m/s (approx)
Therefore, the y velocity of the particle at the same moment is -22.1 m/s (approx).
The correct answer is 0.8 m and -22.1 m/s.
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The acceleration of a particle as it moves along a straight line is given Determine the particle's velocity when t=5 s by a=(2t−1)m/s
2
, where t/s in seconds Suppose that Express your answer to three significant figures and include the appropriate units: s=5 m and v=4 m/s when t=0 X Incorrect: Try Again; 3 attempts remaining Part B Part C Determine the tolal distance the particlo travels during the 5-5 time perlod Express your answer to three significant figures and include the appropriate units.
Given that acceleration, a particle moves along a straight line as a function of time t is given by a = (2t − 1) m/s². We need to find out the velocity of the particle when t = 5 s and the total distance traveled by the particle during the 5 s time period.
Part A:Velocity of particle when t = 5 s The formula for velocity is: v = u + at where v = final velocity, u = initial velocity, a = acceleration, and t = time. Since the initial velocity u = 4 m/s, the acceleration a = (2t − 1) m/s² and time t = 5 s, we have: v = u + at ⇒v = 4 + (2 × 5 − 1)5 v = 49 m/s So, the velocity of the particle when t = 5 s is 49 m/s.Part B: Total distance traveled by the particle during the 5 s time period v = 49 m/s We know that: Average velocity, v_avg = (u + v) / 2 Distance traveled, s = v_avg × t Therefore, s = (4 + 49) / 2 × 5 s = 132.5 m So, the total distance traveled by the particle in a straight line during the 5 s time period is 132.5 m.To learn more about straight-line motion: https://brainly.com/question/24445340
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Inverting Amplifier
What are the reasons for the difference?
\( G=\frac{v_{\text {out }}}{v_{\text {in }}} \) - Brainstorm the reasons for such difference between actual and estimated voltage gain
To accurately estimate the voltage gain of an inverting amplifier, it is important to consider these factors and account for any deviations from the ideal conditions.
The difference between the actual and estimated voltage gain in an inverting amplifier can be attributed to a few reasons.
1. Ideal vs. Real Components: In an ideal inverting amplifier, the operational amplifier (op-amp) has infinite gain and zero output impedance. However, real op-amps have finite gain and non-zero output impedance. These differences between ideal and real components can cause variations in the voltage gain.
2. Input and Output Impedances: The presence of input and output impedances in a real inverting amplifier can affect the voltage gain. If the input impedance is not infinitely large and the output impedance is not zero, the actual voltage gain may deviate from the ideal gain.
3. Tolerance of Resistors: In practical circuits, resistors used in the inverting amplifier may have tolerance values, meaning they may have slight variations from their specified resistances. These tolerance variations can affect the voltage gain of the amplifier.
4. Signal Distortion: Noise and distortion can also contribute to the difference between actual and estimated voltage gain. The presence of unwanted signals or distortion in the input can impact the output signal and result in a deviation from the expected voltage gain.
By carefully selecting components, minimizing tolerances, and considering the effects of input and output impedances, one can reduce the difference between the actual and estimated voltage gain.
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The fire department climbs to the roof of a four-story building to put out a fire on
the roof of a nearby six-story building. The fire fighters aim the firehose at an angle of 75
degrees above the horizontal and the water leaves the nozzle at a speed of 16 m/s.
a. How long does it take water from the firehose to reach the fire?
b. How far away is the adjacent building?
The adjacent building is approximately 5.60 meters away from the firefighters.To solve the problem, we can break it down into two parts: the vertical motion and the horizontal motion of the water from the firehose.
a. To find the time it takes for the water to reach the fire, we only need to consider the vertical motion. We can use the equation of motion for vertical motion:
Δy = V₀yt + (1/2)gt²
In this case, the initial vertical velocity V₀y is given by V₀ * sin(75°), where V₀ is the speed of the water from the firehose. Since the water starts from the same height as the firehose, Δy = 0. Solving for t:
0 = (V₀ * sin(75°))t - (1/2)(9.8 m/s²)t²
Using the given values:
0 = (16 m/s * sin(75°))t - (1/2)(9.8 m/s²)t²
Simplifying and solving for t, we find:
t ≈ 1.84 seconds
Therefore, it takes approximately 1.84 seconds for the water to reach the fire.
b. To find the distance to the adjacent building, we need to consider the horizontal motion. The horizontal distance traveled by the water can be calculated using the equation:
Δx = V₀xt
In this case, the initial horizontal velocity V₀x is given by V₀ * cos(75°), where V₀ is the speed of the water from the firehose. Again, we can use the time calculated in part a:
Δx = (V₀ * cos(75°)) * t
Using the given values:
Δx = (16 m/s * cos(75°)) * 1.84 s
Simplifying, we find:
Δx ≈ 5.60 meters
Therefore, the adjacent building is approximately 5.60 meters away from the firefighters.
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Hi!
If anyone can tell me what small angle approximation is and how it applies to pendulums that would be great.
Thank
The small angle approximation is a mathematical approximation that is used to simplify the equations of motion for a pendulum. It assumes that the angle of displacement of the pendulum is small so that the sine of the angle can be approximated by the angle itself.
The small angle approximation is a technique used to simplify trigonometric functions when the angle of interest is very small. This approximation is based on the assumption that the sine of a small angle is approximately equal to the angle itself, and the cosine of a small angle is approximately equal to one. This approximation holds true for angles less than about 15 degrees, or about 0.26 radians.
When it comes to pendulums, the small angle approximation is used to simplify the equation of motion of the pendulum. If the angle of the pendulum's swing is small, then the sine of that angle will be close to the value of the angle itself. This means that we can use the small angle approximation to simplify the equation of motion and make it easier to solve. In particular, we can assume that the angle of the pendulum is small enough that the sine of the angle is equal to the angle itself. This leads to a simplified equation of motion that is much easier to work with than the more general equation that applies to larger angles.
Overall, the small angle approximation is a useful technique for simplifying trigonometric functions in situations where the angle of interest is very small. This approximation is particularly useful in the study of pendulums, where it allows us to simplify the equation of motion and make it easier to solve.
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You toss a ball straight up into the air at velocity v
o
It reaches its maximum height at time T and lands at time 2T. a) Give the velocity at t=0,T, and 2T b) Give the acceleration at t=0,T, and 2T 2) Describe an example of the following: a) An object moving but not accelerating b) An object moving and accelerating in opposite directions
A) The velocity at t = 0 is v₀
the velocity at t = T is 0.
The velocity at t = 2T is -v₀
B) The acceleration at t = 0 is the acceleration due to gravity
At the maximum height, the ball momentarily comes to rest, indicating that the acceleration is 0.
The acceleration at t = 2T is again the acceleration due to gravity
A) At t = 0 (initial velocity):
The velocity at t = 0 is v₀ (the initial velocity at which the ball was thrown upwards).
At t = T (maximum height):
At the maximum height, the ball momentarily comes to rest before changing direction and falling back down. Hence, the velocity at t = T is 0.
At t = 2T (landing):
The velocity at t = 2T is -v₀ (negative of the initial velocity), as the ball will have the same magnitude of velocity but in the opposite direction when it lands.
b) The acceleration of the ball can be determined as follows:
At t = 0 (initial position):
The acceleration at t = 0 is the acceleration due to gravity, typically denoted as g and pointing downward.
At t = T (maximum height):
At the maximum height, the ball momentarily comes to rest, indicating that the acceleration is 0.
At t = 2T (landing):
The acceleration at t = 2T is again the acceleration due to gravity, but this time pointing downward (opposite to the initial acceleration).
2) Examples of objects with different types of motion:
a) An object moving but not accelerating:
Consider a car traveling on a straight road at a constant speed. As long as the car maintains a steady velocity without changing its speed or direction, it is moving but not accelerating.
b) An object moving and accelerating in opposite directions:
Imagine a train moving forward while slowing down. In this scenario, the train is still in motion, but its acceleration is directed opposite to its velocity. As it decelerates, the train's speed decreases, and eventually, it comes to a stop or changes direction.
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This is a direct calculation question. You are in a spacecraft orbiting Pluto. The mass of Pluto is 1.31E+22 kg, and it's radius is 1.15E+3 km. What is the escape velocity from the surface? m/s. At a height of 6 radii above the surface of the Pluto, what is the circular orbital velocity (orbital velocity, not escape velocity) m/s Note: If your answer requires scientific notation, remember that OWL uses "e" notation: 1.1×10
5
is 1.1e5 to OWL.
Escape velocity from the surface of Pluto = 1.236 km/s and circular orbital velocity at 6 radii above the surface of Pluto = 1.18 × 10³ m/s. Escape velocity from the surface of Pluto:Escape velocity is the minimum speed needed by an object to escape the gravitational field of the planet or celestial body. The formula for escape velocity is given by:ve=√2GM/R,where G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet. Substituting the given values, we get,ve=√2×6.674×10⁻¹¹×1.31×10²²/1.15×10³=1.236 km/s.
Therefore, the escape velocity from the surface of Pluto is 1.236 km/s. Height above surface = 6 radii So, the distance between spacecraft and the surface of Pluto is 6 × radius of Pluto = 6 × 1.15 × 10³ km = 6.9 × 10³ km Circular orbital velocity at 6 radii above the surface of Pluto:The formula for the circular orbital velocity is given by:v=√GM/R. Here, M is the mass of Pluto and R is the distance between the spacecraft and the center of Pluto.
So, the distance between the spacecraft and the center of Pluto is (1 + 6) × 1.15 × 10³ km = 8.9 × 10³ km.Substituting the values, we get,v=√6.674×10⁻¹¹×1.31×10²²/8.9×10³=1.18 km/sTherefore, the circular orbital velocity at 6 radii above the surface of Pluto is 1.18 km/s (approx) which is equivalent to 1.18 × 10³ m/s.
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12- A car goes with speed of 75mi hr. Then driver slows down to 25mi/hr in 6 seconds. What is the deceleration of car? 13- Knowing that density of aluminum is 2700 kg/m 3
, (a) what is the volume of 400grams sphere aluminum ball? (b) Also find radius of this sphere?
This question has two parts
Deceleration of a car -3.12 m/s
Volume of aluminium sphere 0.0172 m
To find the deceleration of the car, we can use the formula for average acceleration:
Acceleration = (Final velocity - Initial velocity) / Time
Given that the initial velocity of the car is 75 mi/hr and the final velocity is 25 mi/hr, and the time is 6 seconds, we can calculate the deceleration as follows:
Acceleration = (25 mi/hr - 75 mi/hr) / 6 sec
To perform the calculation, we need to convert the velocities from miles per hour to a consistent unit of measure, such as meters per second. Since 1 mile is approximately 1609.34 meters and 1 hour is 3600 seconds, we can convert the velocities:
Initial velocity = 75 mi/hr * (1609.34 m/1 mi) * (1 hr/3600 sec) ≈ 33.53 m/s
Final velocity = 25 mi/hr * (1609.34 m/1 mi) * (1 hr/3600 sec) ≈ 11.18 m/s
Now we can calculate the deceleration:
Acceleration = (11.18 m/s - 33.53 m/s) / 6 sec ≈ -3.12 m/s²
Therefore, the deceleration of the car is approximately -3.12 m/s² (negative sign indicates deceleration).
Now, let's move on to the second part of the question.
(a) To find the volume of the aluminum sphere, we can use the formula for the volume of a sphere:
Volume = (4/3) * π * r³
Given that the density of aluminum is 2700 kg/m³ and the mass is 400 grams (or 0.4 kg), we can use the density formula to calculate the volume:
Density = Mass / Volume
Rearranging the formula, we can solve for volume:
Volume = Mass / Density
Volume = 0.4 kg / 2700 kg/m³ ≈ 0.000148 m³
(b) To find the radius of the sphere, we can rearrange the volume formula:
Volume = (4/3) * π * r³
Rearranging the formula, we can solve for the radius:
r = (∛(Volume / ((4/3) * π)))²
Plugging in the volume value we calculated earlier:
r = (∛(0.000148 m³ / ((4/3) * π)))² ≈ 0.0172 m
Therefore, the volume of the aluminum sphere is approximately 0.000148 m³ and the radius is approximately 0.0172 meters (or 17.2 mm).
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2. A flower pot is knocked off of a windowsill 15 stories high. Neglecting air resistance, how fast is the pot moving when it hits the ground below? (Hint assume 15 stories is equal to 50 meters)
The velocity of the pot when it hits the ground below is 31.6 m/s
What is motion under gravity?When an object is lifted or away from the ground, then a force is applied to the object in the downward direction. This force is known as the gravity force.
The equation of motion under gravity are stated as;
H = ut ± 1/2 gt²
V² = u² ± 2gh
v = u ± gt
When the object is moving upward it will be negative and when it's moving downward it will be positive.
In this case the pot is thrown down , so it will move with gravity.
H = 50m
g = 10 m/s²
u = 0
V² = 0² + 2 × 10 × 50
V² = 1000
v = √ 1000
v = 31.6 m/s
Therefore the velocity of the pot is 31.6m/s
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Five resistors R1=4.00, R2 = 5.00, R3 = 13.0, R4 = 10.0, and R5 = 7.00are mounted as shown in the figure. The voltage of the battery is V = 18 Volts. What is the current in the R4 resistor?
Five resistors R1=4.00, R2 = 5.00, R3 = 13.0, R4 = 10.0, and R5 = 7.00are mounted together. The voltage of the battery is V = 18 Volts. The current in R4 resistor is 0.818 A.
To find the current flowing through resistor R4, we can use Ohm's Law and the concept of parallel and series resistors.
Given resistors:
R1 = 4.00 Ω
R2 = 5.00 Ω
R3 = 13.0 Ω
R4 = 10.0 Ω
R5 = 7.00 Ω
Voltage of the battery:
V = 18 V
First, let's determine the equivalent resistance ([tex]R_{eq[/tex]) of the combination of resistors R1, R2, and R3. Since these resistors are connected in series, we can add them together:
[tex]R_{eq[/tex]= R1 + R2 + R3
= 4.00 Ω + 5.00 Ω + 13.0 Ω
= 22.0 Ω
Next, we can calculate the total current ([tex]I_{total[/tex]) flowing through the circuit using Ohm's Law:
I_total = V / [tex]R_{eq[/tex]
= 18 V / 22.0 Ω
≈ 0.818 A
Since resistor R4 is connected in parallel with R3, they have the same voltage drop across them. Thus, the current flowing through R4 is the same as the total current:
[tex]I_{R4[/tex] = [tex]I_{total[/tex]
≈ 0.818 A
Therefore, the current flowing through resistor R4 is approximately 0.818 A.
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What type of motion is not discussed in the textbook? helical planer circular translational Question 2 2. Which selection is an example of uniform acceleration? free-fall roller coaster race car merry-go-round Question 3 3. Which variable is not related to the others through time by calculus principles? force velocity acceleration displacement 4. What is the value of acceleration when velocity is constant? increasing positive zero decreasing Question 5 5. Which value is NOT equal to gravity?
386in/s
∧
2
9.81 m/s
∧
2
32.2ft/s
∧
2
35mi/min
∧
2
Question 6 6. What is velocity if acceleration is constant? quadratic constant sinusoidal linear 7. What is displacement if acceleration is constant? constant quadratic: sinusoidal linear Question 8 8. Planar projectile motion consists of how many independent motions? one two zero three Question 9 9. How is the horizontal component of projectile motion characterized as determined by its inital velocity and elevation angle? increasing decreasing constant unknown 10. How is the vertical component of projectile motion on Earth characterized as determined by its initial velocity and elevation angie? increasing constant. unknown decreasing
1. Helical motion: Helical motion refers to the motion of an object along a helix or spiral path. It is not discussed in the textbook, which means that this specific type of motion is not covered or explained in the textbook material.
2. Uniform acceleration: Uniform acceleration occurs when an object's velocity changes by equal amounts in equal time intervals. A roller coaster is an example of uniform acceleration because it experiences a constant change in velocity as it moves along the track.
3. Displacement: Displacement is a measure of the change in position of an object. Unlike force, velocity, and acceleration, which are all related through calculus principles (such as derivatives and integrals) with respect to time, displacement is not directly linked to the others through calculus principles.
4. Zero acceleration with constant velocity: When an object has a constant velocity, it means that its speed and direction remain unchanged. In this case, the value of acceleration is zero because there is no change in velocity over time.
5. Gravity: The value 35 mi/min^2 is not equal to the acceleration due to gravity. The standard value for acceleration due to gravity is approximately 9.81 m/s^2 or 32.2 ft/s^2. The given value of 35 mi/min^2 does not match the accepted value for gravity.
6. Velocity with constant acceleration: When acceleration is constant, the velocity of an object changes at a steady rate. In this case, the relationship between velocity and acceleration is linear, meaning that the velocity increases or decreases uniformly over time.
7. Displacement with constant acceleration: If acceleration is constant, the relationship between displacement and time is quadratic. This means that the displacement varies with the square of time, resulting in a curved path.
8. Planar projectile motion: Planar projectile motion refers to the motion of an object in a two-dimensional plane under the influence of gravity. It consists of two independent motions: horizontal motion (affected by initial velocity and angle) and vertical motion (affected by gravity).
9. Horizontal component of projectile motion: The horizontal component of projectile motion is characterized as constant. This means that the horizontal velocity remains unchanged throughout the motion, assuming no external forces act upon the object horizontally.
10. Vertical component of projectile motion: The vertical component of projectile motion on Earth is characterized by a decreasing velocity. This is because the object experiences the acceleration due to gravity, which causes its vertical velocity to decrease over time.
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Present fully and with equations if necessery
theoretical question and application problem. Part A: Theoretical
1. Explain what field lines are
2. Explain what equipotential lines are.
3. It is possible that the electric field exists in space.
4. Explain why electric field lines cannot form the shape of a loop. closed between charges.
5. Explain why the electric field is zero in a closed region
Part B: Application Problems
What will be the magnitude and direction of the electric field that balances the weight of a) a electron and b) a proton? Use the data from the table shown
A. the magnitude of electric field to balance the weight of an electron is 5.59 × 10⁸ N/C. and B. the magnitude of electric field to balance the weight of a proton is 1.025 × 10⁹ N/C.
Theoretical
1. Field lines are the concept that represents the magnitude and direction of an electric field vector.
2. Equipotential lines are a set of lines representing a point in space where electric potential is constant.
3. Yes, it is possible that the electric field exists in space.
4. Electric field lines cannot form the shape of a loop because it would indicate that electric field lines at one point are originating from two different points, which contradicts the theory of the electric field.
5. The electric field is zero in a closed region because if there were any electric field, it would indicate that the charges within the region are moving, which contradicts the concept of a closed region.
Part B: Application Problems
The electric field can be calculated by using the given formula; E = F / q
where F is the force experienced by the charge q,
and E is the electric field.
The weight of an electron is given as 9.11 × 10⁻³¹ kg and of a proton as 1.67 × 10⁻²⁷ kg.
a) Magnitude of electric field to balance the weight of an electron;
E = F / qF = mg
F = (9.11 × 10⁻³¹) × 9.8
F = 8.94 × 10⁻³⁰ N
q = 1.6 × 10⁻¹⁹ C
E = F / q
E = (8.94 × 10⁻³⁰) / (1.6 × 10⁻¹⁹)
E = 5.59 × 10⁸ N/C
Thus, the magnitude of electric field to balance the weight of an electron is 5.59 × 10⁸ N/C.
This electric field will be directed upward.
b) Magnitude of electric field to balance the weight of a proton;
E = F / qF = mg
F = (1.67 × 10⁻²⁷) × 9.8F = 1.64 × 10⁻²⁶ Nq = 1.6 × 10⁻¹⁹ CE = F / q
E = (1.64 × 10⁻²⁶) / (1.6 × 10⁻¹⁹)
E = 1.025 × 10⁹ N/C
Thus, the magnitude of electric field to balance the weight of a proton is 1.025 × 10⁹ N/C.
This electric field will be directed upward.
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150 m. Compute the acceleration at the moment the train speed reaches 50.0 km/h. Assume the train continues to slow down at this time at the same rate. A ball swings counterclockwise in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9
∘
past the lowest point on its way up, its total acceleration is (−22.5
i
^
+20.2
j
^
)m/s
2
. For that instant, (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball.
Train: Acceleration at 50 km/h unknown. Ball: Given acceleration components, find radial acceleration, speed, and velocity.
To compute the acceleration of the train when its speed reaches 50.0 km/h, we need to convert the speed to m/s.
Given:
Train speed = 50.0 km/h
Acceleration = ?
Conversion:
1 km/h = 1000 m/3600 s
50.0 km/h = (50.0 * 1000) m/3600 s = 13.89 m/s
Now, we'll assume that the train continues to slow down at the same rate after reaching this speed. Let's denote the final velocity as Vf = 13.89 m/s, the initial velocity as Vi = 0 m/s, and the time taken to reach Vf as t.
The acceleration (a) can be calculated using the following formula:
a = (Vf - Vi) / t
Since the initial velocity (Vi) is 0 m/s, the formula simplifies to:
a = Vf / t
However, we need more information, such as the time taken to reach the final velocity or any other data related to the train's deceleration, to determine the acceleration accurately.
Moving on to the second part of your question regarding the ball swinging counterclockwise in a vertical circle:
(a) Sketching a vector diagram showing the components of acceleration:
Let's denote the total acceleration vector as A = (-22.5i^ + 20.2j^) m/s^2. Here, i^ and j^ are unit vectors along the x and y axes, respectively. To sketch the vector diagram, draw a coordinate system with x and y axes, and then represent the vector (-22.5i^ + 20.2j^) originating from the origin.
(b) Determining the magnitude of radial acceleration:
The total acceleration of the ball can be divided into two components: the radial (centripetal) acceleration and the tangential acceleration.
The radial acceleration (ar) points towards the center of the circle and is responsible for changing the direction of the ball's velocity. It can be calculated using the formula:
ar = √(Ax^2 + Ay^2)
= √((-22.5)^2 + (20.2)^2) m/s^2
Calculate the square of each component, sum them, and then take the square root to find the magnitude of radial acceleration.
(c) Determining the speed and velocity of the ball:
The speed of the ball is the magnitude of its velocity vector, and the velocity is a vector that describes both the magnitude and direction of the ball's motion.
To find the speed, we can use the magnitude of the total acceleration vector:
speed = √(Ax^2 + Ay^2)
To find the velocity, we need to know the direction of the velocity vector, which is missing in the given information.
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A three-phase, Y-connected, 415 V,50 Hz generator supplies a fixed three-phase balanced load connected in parallel as follows: Load 1: A Y-connected 40kVAR motor at 0.7 lagging power factor. Load 2: A △-connected 20 hp induction motor at 0.75 lagging power factor. Load 3: A Y-connected 10 kW purely resistive load i) Assuming that all conductors are lossless, sketch the single-line diagram of the system. ii) Calculate the total value of real, reactive and apparent power taken from the supply source by all the loads. iii) Determine the overall power factor of the system. iv) If it is necessary to correct the power factor of the system to 0.85 lagging power factor of by connecting a three-phase capacitor bank parallel at the load, calculate the rating of the capacitor bank in kVAR.
The single-line diagram of the system, calculate the total power taken from the supply source by all the loads, determine the power factor of the system, and calculate the rating of the capacitor bank needed for power factor correction.
i) The single-line diagram of the system would show the generator connected to the loads in parallel. The Y-connected [tex]40kVAR[/tex] motor (Load 1) is connected to one of the generator's terminals, while the △-connected [tex]20 hp[/tex]induction motor (Load 2) is connected to another terminal. Finally, the Y-connected 10 kW purely resistive load (Load 3) is connected to the remaining terminal of the generator.
ii) To calculate the total power taken from the supply source, we need to determine the real power (P), reactive power (Q), and apparent power (S) for each load, and then add them together. For Load 1, the apparent power is given as 40kVAR. For Load 2, the apparent power can be calculated using the formula:
[tex]S = (P / power factor)[/tex],
where P is the real power in watts. Given that the power factor is 0.75 lagging, the real power can be calculated as: [tex]P = (S * power factor)[/tex].
For Load 3, since it is purely resistive, the apparent power is equal to the real power, which is 10kW.
Adding all the real, reactive, and apparent powers together will give us the total values.
iii) To determine the overall power factor of the system, we need to find the total real power and total apparent power. The power factor (pf) can be calculated using the formula:
pf = (total real power / total apparent power).
By dividing the total real power by the total apparent power, we can determine the overall power factor of the system.
iv) If we need to correct the power factor of the system to 0.85 lagging, we can do this by connecting a three-phase capacitor bank in parallel at the load. To calculate the rating of the capacitor bank in kVAR, we need to determine the reactive power (Qc) needed to correct the power factor. The formula for Qc is:
[tex]Qc = (S * √(1 - (pf^2)))[/tex],
where S is the total apparent power and pf is the desired power factor.
By substituting the values into the formula, we can calculate the rating of the capacitor bank in [tex]kVAR[/tex].
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Dt. Sohn Paul Stapp was U.S. Air Force efficer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stopp rode a rocket sled, accelerating from rest to a top speed of 282 m/s(1015 km/h) in 5.00 s, and was brought jarringly back to test in only 1.40 sl Calculate his acceleration and express it in multiples of 9(9.80 m/s^2 ) br taking its ratio to the acceleration of gravity. Acceleretion__ = Number of g's =__ ,b) Calculate his deceleration and express it in multiples of g. Acceleration =__ Number of g^2s=___
a) To calculate the acceleration experienced by Dr. Paul Stapp, we need to divide the change in velocity by the time taken. Hence, the acceleration experienced by Dr. Paul Stapp is approximately 5.76 times the acceleration due to gravity.
b) Therefore, the deceleration experienced by Dr. Paul Stapp is approximately 20.56 times the acceleration due to gravity, with a negative sign indicating deceleration.
a) To calculate the acceleration, we use the formula:
acceleration = (change in velocity) / (time taken)
Given information:
Initial velocity, u = 0 m/s (starting from rest)
Final velocity, v = 282 m/s
Time taken, t = 5.00 s
Using the formula, the acceleration can be calculated as:
acceleration = (v - u) / t
acceleration = (282 m/s - 0 m/s) / 5.00 s
acceleration = 56.4 m/s^2
To express the acceleration in multiples of g, we divide it by the acceleration due to gravity:
Number of g's = acceleration / g
Number of g's = 56.4 m/s^2 / 9.80 m/s^2
Number of g's ≈ 5.76
Therefore, the acceleration experienced by Dr. Paul Stapp is approximately 5.76 times the acceleration due to gravity.
b) To calculate the deceleration, we need the change in velocity and the time taken during the deceleration phase. The change in velocity is from 282 m/s to 0 m/s, and the time taken is 1.40 s.
Using the formula for deceleration:
deceleration = (change in velocity) / (time taken)
Given information:
Initial velocity, u = 282 m/s
Final velocity, v = 0 m/s
Time taken, t = 1.40 s
Using the formula, the deceleration can be calculated as:
deceleration = (v - u) / t
deceleration = (0 m/s - 282 m/s) / 1.40 s
deceleration ≈ -201.4 m/s^2
To express the deceleration in multiples of g, we divide it by the acceleration due to gravity:
Number of g^2s = deceleration / g
Number of g^2s = -201.4 m/s^2 / 9.80 m/s^2
Number of g^2s ≈ -20.56
Therefore, the deceleration experienced by Dr. Paul Stapp is approximately 20.56 times the acceleration due to gravity, with a negative sign indicating deceleration.
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A proton traveis through a magnetic fields, with a velocity to the right. The magnetic field excrts a magnetic force on the proton which is directed inwards with respect to the plane of the page. What is the direction of the magnetic feld? Left Right Up Down Into the page Quit of the page
The direction of the magnetic field. A proton travels through a magnetic field, with a velocity to the right. The magnetic field exerts a magnetic force on the proton, which is directed inwards with respect to the plane of the page. The magnetic field must be directed into the page.
The direction of the magnetic field is into the page. Let us look at a few concepts to explain this.A magnetic field is a force field that surrounds a magnet or a moving charge. A magnetic field exerts a force on charged particles that move through it. This force is known as the magnetic force. The direction of the magnetic force is perpendicular to both the magnetic field and the direction of the particle's motion.
We can use the right-hand rule to find the direction of the magnetic force. When a positively charged particle travels through a magnetic field, the direction of the magnetic force is perpendicular to both the magnetic field and the direction of the particle's motion. The direction of the magnetic force can be determined using the right-hand rule. In this case, the proton is moving to the right. The magnetic force is directed inwards with respect to the plane of the page. Therefore, the magnetic field must be directed into the page.
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From a 7om clifh above ocean surface aball is kicked with initial velocity of 60
sec
m
of angle of 50
∘
. Find: (a) Time To max. (b) Total Traveled Time. (c) velocity yangle of ball just before hits ocean? (d) Landing Distance? (e) Velocity \& angle 4 second after Kicking?
Based on the data given, (a) Time taken to reach maximum height = 4.44 sec ; (b) Time of flight = 9.05 sec ; (c) velocity y angle θ = 26.11° ; (d) Landing Distance = 210.1 m ; (e) velocity and angle of the ball after 4 seconds are 38.83 m/s and 48.67°, respectively.
Given,
Initial velocity of the ball = 60 m/sec
Angle of projection = 50°
Acceleration due to gravity = 9.8 m/sec²
(a) Time taken to reach maximum height
We know that, Time taken to reach maximum height is given by : t = u sin θ / g
where,
u = Initial velocity of the ball
θ = Angle of projection
g = Acceleration due to gravity
t = (60 × sin 50°) / 9.8= 4.44 sec
(b) Total time of flight
Time of flight is given by :
T = 2u sin θ / g= 2 × 60 × sin 50° / 9.8= 9.05 sec
(c) Velocity and angle of the ball just before it hits the oceanWhen the ball hits the ocean, its y-coordinate is zero and its velocity is the final velocity of the projectile.
Let v be the final velocity of the ball. Then using, v² = u² + 2gh
v² = 60² + 2 × 9.8 × 70v = 85.73 m/s
Also, we know that tan θ = v_y / v where,
θ is the angle made by the final velocity with the horizontal axis.
v_y is the final vertical component of velocity
v_y = u sin θ − gt
For, vertical component of velocity at the time of hitting the ocean we have :
v_y = 60 sin 50° − 9.8 × 9.05v_y = 43.28 m/s
Now, we can find the angle using the formula,
θ = tan⁻¹ (v_y / v)
θ = tan⁻¹ (43.28 / 85.73)
θ = 26.11°
(d) Landing distance
The horizontal distance traveled by the ball is given by :
R = u² sin 2θ / g= 60² sin 100° / 9.8= 210.1 m
(e) Velocity and angle of the ball after 4 seconds
Let, velocity of the ball after 4 seconds be u'.
Then u' = u cos θ = 60 cos 50°= 38.83 m/s
Let, θ' be the angle made by the velocity of the ball with the horizontal axis.
Then tan θ' = v_y / u' = 43.28 / 38.83 = 1.11°θ' = tan⁻¹ (1.11) = 48.67°
So, the velocity and angle of the ball after 4 seconds are 38.83 m/s and 48.67°, respectively.
Thus, the correct answers are : (a) 4.44 sec ; (b) 9.05 sec ; (c) 26.11° ; (d) 210.1 m ; (e) velocity = 38.83 m/s, angle = 48.67°
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Tarzan swings on a 33.0−m-long vine initially inclined at an angle of 40.0∘ with the vertical. (a) What is his speed at the bottom of the swing if he starts from rest? m/s (b) What is his speed at the bottom of the swing if he pushes off with a speed of 3.00 m/s ? m/s
Tarzan's speed at the bottom of the swing, when pushing off with a speed of 3.00 m/s, is approximately 34.3 m/s.
To solve this problem, we can use the principle of conservation of mechanical energy.
(a) When Tarzan starts from rest, all of his initial potential energy will be converted into kinetic energy at the bottom of the swing.
The initial potential energy is given by:
PE_initial = m * g * h
where m is Tarzan's mass, g is the acceleration due to gravity, and h is the vertical height of the swing. Since Tarzan starts from rest, his initial kinetic energy is zero.
At the bottom of the swing, all of the initial potential energy will be converted into kinetic energy:
KE_final = PE_initial
Using the conservation of mechanical energy, we can write:
m * g * h = (1/2) * m * v^2
where v is the speed at the bottom of the swing.
Simplifying the equation:
v^2 = 2 * g * h
Substituting the given values:
v^2 = 2 * 9.8 m/s^2 * 33.0 m
Calculating this expression gives us:
v ≈ 28.0 m/s
Therefore, Tarzan's speed at the bottom of the swing, starting from rest, is approximately 28.0 m/s.
(b) If Tarzan pushes off with a speed of 3.00 m/s, we need to consider both his initial kinetic energy and the potential energy at the top of the swing.
The total mechanical energy at the top of the swing is given by:
ME_top = KE_initial + PE_initial
KE_initial = (1/2) * m * (3.00 m/s)^2
PE_initial = m * g * h
ME_top = (1/2) * m * (3.00 m/s)^2 + m * g * h
At the bottom of the swing, the total mechanical energy will be equal to the kinetic energy:
ME_bottom = (1/2) * m * v^2
Using the conservation of mechanical energy, we can write:
ME_top = ME_bottom
(1/2) * m * (3.00 m/s)^2 + m * g * h = (1/2) * m * v^2
Simplifying the equation:
v^2 = (3.00 m/s)^2 + 2 * g * h
Substituting the given values:
v^2 = (3.00 m/s)^2 + 2 * 9.8 m/s^2 * 33.0 m
Calculating this expression gives us:
v ≈ 34.3 m/s
Therefore, Tarzan's speed at the bottom of the swing, when pushing off with a speed of 3.00 m/s, is approximately 34.3 m/s.
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A capacitor of capacitance C is initially connected to a power supply and charged fully until the potential difference across its plates is 30 V. Then the capacitor is disconnected from the power supply and connected to an inductor of inductance L at t=0. The resulting circuit is shown in the simulation (linked below). An ideal ammeter measures the current i(t) in the loop as a function of time t. Run the animation now. Several points to consider are: - A blue arrow inside the ammeter shows the direction and size of the current i(t) at instant t. The current oscillates with time. - The electric field between the capacitor plates is represented both by a purple arrow and purple shading. The electric field oscillates with time. - The magnetic field between the inductor coils is represented both by a green arrow and green shading. The magnetic field also oscillates with time. - Plots of current i versus time t/T (scaled by the period T of the oscillations) and capacitor charge q versus scaled time t/T are shown. - Bar charts of the stored electrical energy UE in the capacitor and stored magnetic energy UB in the inductor are shown. Simulation LC Circuit Question 1 What is the capacitance C of the capacitor? Enter a numerical value in the box below. eTextbook and Media Attempts: unlimited
According to the question, A capacitor of capacitance C is initially connected to a power supply and charged fully until the potential difference across its plates is 30 V.
Then the capacitor is disconnected from the power supply and connected to an inductor of inductance L at t=0. The simulation provides several points to consider, including: - A blue arrow inside the ammeter shows the direction and size of the current i(t) at instant t. The current oscillates with time. - The electric field between the capacitor plates is represented both by a purple arrow and purple shading. The electric field oscillates with time. - The magnetic field between the inductor coils is represented both by a green arrow and green shading. The magnetic field also oscillates with time. - Plots of current i versus time t/T (scaled by the period T of the oscillations) and capacitor charge q versus scaled time t/T are shown. - Bar charts of the stored electrical energy UE in the capacitor and stored magnetic energy UB in the inductor are shown. What is the capacitance C of the capacitor?
The capacitance of the capacitor is 20 μF. When the capacitor is initially connected to a power supply, it gets fully charged until the potential difference across its plates is 30 V. The charged capacitor is then disconnected from the power supply and connected to an inductor of inductance L at t=0. The simulation then shows how the current, electric field, and magnetic field oscillate with time. The plots of current i versus time t/T and capacitor charge q versus scaled time t/T are also shown. Finally, the bar charts of the stored electrical energy UE in the capacitor and stored magnetic energy UB in the inductor are displayed.
Therefore, the capacitance of the capacitor is 20 μF.
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Two small metal cubes with masses 2.0 g and 4.0 g are tied together by a 4.9-cm-long massless string and are at rest on a frictionless surface. Each is charged to +1.6 μC.
PART A
What is the energy of this system?
Express your answer in joules.
PART B
What is the tension in the string?
Express your answer with the appropriate units.
PART C
The string is cut. What is the speed of 2- g -cube when the cubes are far apart?
Hint: There are two conserved quantities. Make use of both.
Express your answer with the appropriate units.
PART D
The string is cut. What is the speed of 2- g -cube when the cubes are far apart?
Hint: There are two conserved quantities. Make use of both.
Express your answer with the appropriate units.
PART A
To find the energy of the system, we need to consider both the gravitational potential energy and the electric potential energy.
Gravitational Potential Energy:
The gravitational potential energy is given by the formula:
PE_gravity = mgh
Since the cubes are at rest on a frictionless surface, their height above the surface is zero, and hence the gravitational potential energy is zero.
Electric Potential Energy:
PE_electric = k * q1 * q2 / r
PE_electric = (8.99 x 10^9 N m^2/C^2) * (1.6 x 10^-6 C) * (1.6 x 10^-6 C) / 0.049 m
PE_electric ≈ 0.836 J
Total Energy = PE_gravity + PE_electric
Total Energy = 0 + 0.836 J
Total Energy ≈ 0.836 J
Therefore, the energy of the system is approximately 0.836 joules.
PART B
The tension in the string can be determined by considering the forces acting on the system. The tension in the string provides the necessary centripetal force to keep the cubes in circular motion.
F_centripetal = m * v^2 / r
F_centripetal = Tension
Tension = m * v^2 / r
For the system to be in equilibrium, the tension in the string should be equal to the electric force between the charges on the cubes:
Tension = Electric Force
Electric Force = k * q1 * q2 / r^2
Tension = k * q1 * q2 / r^2
Tension ≈ 0.874 N
Therefore, the tension in the string is approximately 0.874 Newtons.
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Two point charges of magnitude 6nC and −7nC are separated by 41.3 cm. What is the potential difference between a point infinitely far away and a point midway between the charges? The Coulomb constant is 8.98755×109 N⋅m2/C2. Answer in units of V.
The potential difference between a point infinitely far away and a point midway between two point charges of magnitude 6nC and -7nC separated by 41.3 cm is -304 V.
The potential difference, also known as voltage, between two points is given by the formula V = k(q1/r1 - q2/r2), where V is the potential difference, k is the Coulomb constant (8.98755×[tex]10^9[/tex] [tex]N.m^2/C^2[/tex]), q1 and q2 are the magnitudes of the charges, and r1 and r2 are the distances from the charges to the points in question.
In this case, the point infinitely far away has a distance of infinity, which means r1 becomes infinity and the potential contribution from the first charge is zero. The point midway between the charges has a distance of half the separation, which is 20.65 cm (0.2065 m).
Plugging the values into the formula, we get
V = (8.98755×[tex]10^9[/tex] [tex]N.m^2/C^2[/tex]) * (([tex]6*10^{(-9)} C[/tex])/(∞) - ([tex]-7*10^{(-9)} C[/tex])/(0.2065 m)). Simplifying this expression yields V = -304 V. The negative sign indicates that the potential difference is negative, meaning the potential is lower at the midpoint between the charges compared to infinity.
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A swimmer is capable of swimming 0.550 m/s in still water. (a) If she aims her body directly across a 48m-wide river whose current is 0.48 m/s, how far downstream (from a point opposite her starting point) will she land? (b) How long will it take (in seconds =s ) her to reach the other side? a) b)
It will take approximately 66.5 seconds for the swimmer to reach the other side of the river.
(a) To find how far downstream the swimmer will land, we can use the concept of relative velocity. The swimmer's velocity with respect to the ground is the vector sum of her swimming velocity and the velocity of the river's current.
Let's denote:
v_swim = 0.550 m/s (swimmer's velocity in still water)
v_current = 0.48 m/s (velocity of the river's current)
To find the swimmer's velocity with respect to the ground, we can use vector addition:
v_ground = √((v_swim)^2 + (v_current)^2)
= √((0.550 m/s)^2 + (0.48 m/s)^2)
≈ 0.722 m/s
The swimmer will land downstream a distance equal to her velocity with respect to the ground multiplied by the time taken to cross the river. Since the width of the river is 48 m, we can calculate the time taken to cross using:
time = distance / velocity
t = 48 m / 0.722 m/s
t ≈ 66.5 s
Now, we can calculate the distance downstream:
distance downstream = velocity downstream * time
d_downstream = v_current * t
d_downstream = 0.48 m/s * 66.5 s
d_downstream ≈ 31.92 m
Therefore, the swimmer will land approximately 31.92 meters downstream from the point opposite her starting point.
(b) The time it will take for the swimmer to reach the other side can be calculated using the width of the river divided by the swimmer's velocity with respect to the ground:
time = distance / velocity
t = 48 m / 0.722 m/s
t ≈ 66.5 s
Therefore, it will take approximately 66.5 seconds for the swimmer to reach the other side of the river.
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