The equilibrium position for a test charge q, with Q1 = 6C and Q2 = -4C, at a distance r = 1m, is located at x = 0.5m.
To find the equilibrium position for a test charge q along the x-axis in the presence of charges Q1 and Q2, we can use the principle of electrostatic equilibrium. In this case, with Q1 = 6C and Q2 = -4C, we can determine the equilibrium position.
The condition for electrostatic equilibrium is that the net force on the test charge q should be zero. This means that the forces exerted on the test charge by Q1 and Q2 should balance each other.
Since Q1 is positive and Q2 is negative, the forces they exert on the test charge q will be in opposite directions. The magnitude of the force between two charges is given by Coulomb's law:
F = k * |Q1 * Q2| / r^2
where k is the electrostatic constant and r is the distance between the charges.
To achieve equilibrium, the magnitudes of the forces exerted by Q1 and Q2 on the test charge q should be equal. This means:
k * |Q1 * Q2| / r^2 = k * |Q1 * q| / (x^2)
Simplifying this equation, we have:
|Q2 / r^2| = |q / x^2|
Given that Q2 = -4C and Q1 = 6C, the equation becomes:
|-4 / r^2| = |q / x^2|
Since the charges and distances are positive values, we can ignore the absolute value signs.
4 / r^2 = q / x^2
Solving for x, we have:
x^2 = (q * r^2) / 4
x = sqrt((q * r^2) / 4)
Plugging in the values Q1 = 6C, Q2 = -4C, and r = 1m, and assuming a test charge q of any positive value, we can calculate the equilibrium position.
For example, if q = 1C, the equilibrium position is:
x = sqrt((1C * (1m)^2) / 4)
x = sqrt(1/4) = 0.5m
Therefore, the equilibrium position for a test charge q of 1C is located at x = 0.5m, assuming Q1 = 6C, Q2 = -4C, and r = 1m.
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An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 15.0 days on average to complete one nearly circular revolution around the unnamed planet. The distance from the center of the moon to the surface of the planet is 2.375×10
5
km and the planet has a radius of 3975 km. Determine the moon's centripetal acceleration a
c
. a
c
=
We need to determine the moon's centripetal acceleration ac. Step 1: Firstly, convert the average time taken by the moon to revolve around the planet from days to seconds by multiplying it by 86400 s/1 day.t = 15.0 × 86400 s/1 dayt = 1.296 × 10⁶ s
Step 2: Now, use the formula for centripetal acceleration to find the required answer.a c = v²/rwhere, v is the speed of the moon in its circular orbit around the planet.
Step 3: To find v, we first need to calculate the distance traveled by the moon in one revolution. This distance is equal to the circumference of the moon's orbit.C = 2πr
= 1.491 × 10⁶ km Now, speed, v = Distance/Timev
= C/tv
= 1.15 km/s
Step 4: Finally, substitute the value of v and r in the formula for centripetal acceleration. a c = v²/ra
c = 5.56 × 10⁻³ km/s²Therefore, the moon's centripetal acceleration is 5.56 × 10⁻³ km/s².
Hence, a c = 5.56 × 10⁻³ km/s². We have used the formula for centripetal acceleration, a = v²/r. We have calculated the distance traveled by the moon in one revolution, which is equal to the circumference of the moon's orbit. We have then found the speed of the moon in its circular orbit around the planet by dividing this distance by the time taken by the moon to complete one revolution. Finally, we have substituted the values of speed and radius in the formula for centripetal acceleration to obtain the required answer.
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A muon has a mean lifetime of 2.2μs in its rest frame. What is the energy uncertainty ΔE of the muon in its relativistic reference frame?
The energy uncertainty of the muon in its relativistic reference frame is **2.5 * 10^-20 J**. The energy uncertainty of the muon in its relativistic reference frame can be calculated using the Heisenberg uncertainty principle, which states that ΔEΔt≥h/4π.
where:
* ΔE is the energy uncertainty
* Δt is the time uncertainty
* h is Planck's constant
In this case, the time uncertainty is the lifetime of the muon, which is 2.2 μs. Planck's constant is 6.626 * 10^-34 J s.
So, the energy uncertainty is
ΔE ≥ h / 4πΔt = 6.626 * 10^-34 J s / 4π * 2.2 * 10^-6 s = 2.5 * 10^-20 J
The Heisenberg uncertainty principle states that there is a fundamental limit to how accurately the position and momentum of a particle can be known simultaneously. The uncertainty principle also applies to energy and time. In this case, the time uncertainty is the lifetime of the muon. The energy uncertainty is inversely proportional to the time uncertainty, so the energy uncertainty is greater for shorter time uncertainties.
The energy uncertainty of the muon is significant because it means that the muon's energy cannot be known precisely. This uncertainty has implications for the muon's decay products, as it means that the energy of the decay products cannot be predicted exactly.
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A capacitor is attached to a 4.46- Hz generator. The instantaneous current is observed to reach a maximum value at a certain time. What is the least amount of time that passes after the current maximum before the instantaneous voltage across the capacitor reaches its maximum value?
The least amount of time that passes after the current maximum before the voltage across the capacitor reaches its maximum value is approximately 0.056 seconds, which is one-fourth of the period of the generator frequency.The time interval between the current maximum and the voltage maximum in an AC circuit with a capacitor is given by one-fourth of the period of the generator frequency.
The formula to calculate the period of a frequency is T = 1/f, where T is the period and f is the frequency. Given that the frequency is 4.46 Hz, we can find the period:
T = 1/4.46 Hz ≈ 0.224 seconds.
Therefore, the least amount of time that passes after the current maximum before the instantaneous voltage across the capacitor reaches its maximum value is one-fourth of the period:
0.224 seconds / 4 ≈ 0.056 seconds.
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(a)What is the force per meter (in N/m ) on a lightning bolt at the equator that carries 19,330 A perpendicular to the Earth's 3.00×10^−5T field? N/m (b) What is the direction of the force if the current is straight up and the Earth's field direction is due north, parallel to the ground?
What is the force per meter (in N/m) on a lightning bolt at the equator that carries 19,330 A perpendicular to the Earth's 3.00×10^-5T field?
The force per meter on a lightning bolt at the equator that carries 19,330 A perpendicular to the Earth's 3.00×10^-5T field is given by the formula: F = IℓBsinθWhere:F is the force in newtons (N)I is the current in amperes (A)ℓ is the length of the conductor in meters (m)B is the magnetic field in tesla s (T)θ is the angle between the current and the magnetic field, and since it is perpendicular, then θ = 90°ℓ = 1 meter, I = 19,330 A, B = 3.00×10^-5 T, and θ = 90°
Substituting the given values: F = IℓBsinθ= (19,330 A)(1 m)(3.00×10^-5 T)sin(90°)= 0.5799 N/m
Therefore, the force per meter on a lightning bolt at the equator that carries 19,330 A perpendicular to the Earth's 3.00×10^-5 T field is 0.5799 N/m.
What is the direction of the force if the current is straight up and the Earth's field direction is due north, parallel to the ground?
The direction of the force is given by Fleming's left-hand rule (also called the motor rule), which states that if the thumb, first finger, and middle finger of the left hand are held mutually perpendicular to each other with the first finger pointing in the direction of the magnetic field and the thumb in the direction of the current, then the second finger points in the direction of the force exerted on the conductor by the magnetic field. Hence, for a current that is straight up and a magnetic field that is parallel to the ground towards the north, the force will be directed towards the west
Therefore, the direction of the force is towards the west.
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A flea jumps straight up to a maximum height of 0.360 m. What is its initial velocity v
0
as it leaves the ground? Express your answer in meters per second to three significant figures. How long is the flea in the air from the time it jumps to the time it hits the ground? Express your answer in seconds to three significant figures.
The initial velocity of the flea as it leaves the ground is approximately 2.657 m/s.
The flea is in the air for approximately 0.339 s from the time it jumps to the time it hits the ground.
Given the maximum height reached by the flea is 0.360 m.
We need to find its initial velocity as it leaves the ground and how long the flea is in the air from the time it jumps to the time it hits the ground.
Initial velocity of the flea as it leaves the ground can be found using the equation:
v^2 = u^2 + 2gh
Where,
v is the final velocity, which is 0 m/s
u is the initial velocity
h is the maximum height gained by the flea, which is 0.360 mu
g is the acceleration due to gravity, which is 9.81 m/s²
Putting all these values in the above equation, we get:
0² = u² + 2(9.81)(0.360)
u² = 7.05776
u = √(7.05776)
≈ 2.657 m/s
Thus, the initial velocity of the flea as it leaves the ground is approximately 2.657 m/s.
To find the time for the flea to hit the ground, we can use the following equation:
h = ut + 1/2 gt²
where,
h is the maximum height gained by the flea, which is 0.360 mu is the initial velocity, which we just calculated, 2.657 m/s
g is the acceleration due to gravity, which is 9.81 m/s²
Putting all these values in the above equation, we get:
0.360 = 2.657t - 1/2(9.81)t²
0.49t² - 2.657t + 0.360 = 0
Solving this quadratic equation, we get:
t = 0.339 s (approx)
Thus, the flea is in the air for approximately 0.339 s from the time it jumps to the time it hits the ground.
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A steel rotating-beam test specimen has an ultimate strength of 1600 MPa. Estimate the life of the specimen if it is tested at a completely reversed stress amplitude of 900 MPa. N = 46,400 cycles
The estimated life of the specimen is calculated to be 96,934 cycles, but the given value of N is 46,400 cycles. Therefore, the specimen will fail before completing its estimated life.
A steel rotating-beam test specimen has an ultimate strength of 1600 MPa and it is tested at a completely reversed stress amplitude of 900 MPa. N = 46,400 cycles. The estimated life of the specimen can be calculated by using the formula for fatigue life: Nf = Where: Nf = fatigue life, K = fatigue strength reduction factor, σm = mean stress, σa = stress amplitude, a and b = material constants.
To calculate the fatigue life of the specimen, we need to first find the values of a, b, and K for the given steel. For a rotating-beam test specimen made of steel, the values of a, b, and K are
0.15, -0.25, and 1.0, respectively.
Substituting these values in the above formula, we get: Nf = (1.0 x 900 0.15) / (1600 (-0.25) x 2) = 96,934 cycles
Since the given value of N is 46,400 cycles, the specimen will fail before completing its estimated life.
Fatigue is a phenomenon that leads to the failure of materials under repeated cyclic loading. It occurs when the maximum stress in a material is below the ultimate strength but is applied repeatedly over time. The fatigue life of a material can be estimated using the S-N curve, which represents the relationship between stress amplitude and the number of cycles to failure. The S-N curve is obtained by testing a material at different stress levels and plotting the number of cycles to failure against the stress amplitude. The fatigue life of a material can also be estimated using the Goodman relation, which takes into account the effect of mean stress. In this relation, the stress amplitude is reduced by a factor that depends on the mean stress.
The fatigue life is then calculated using the modified stress amplitude. The estimated life of the specimen can be calculated by using the formula for fatigue life: Nf = (Kσm a)/σa b, where Nf is the fatigue life, K is the fatigue strength reduction factor, σm is the mean stress, σa is the stress amplitude, and a and b are material constants. In this question, a steel rotating-beam test specimen has an ultimate strength of 1600 MPa and is tested at a completely reversed stress amplitude of 900 MPa. N = 46,400 cycles. The estimated life of the specimen is calculated using the above formula.
The values of a, b, and K for a rotating-beam test specimen made of steel are 0.15, -0.25, and 1.0, respectively. Substituting these values in the above formula, we get Nf = (1.0 x 900 0.15) / (1600 (-0.25) x 2) = 96,934 cycles. Since the given value of N is 46,400 cycles, the specimen will fail before completing its estimated life.
The estimated life of a steel rotating-beam test specimen can be calculated using the formula for fatigue life. The fatigue life depends on the stress amplitude, mean stress, and material constants. The fatigue strength reduction factor takes into account the effect of mean stress on the fatigue life. In this question, the estimated life of the specimen is calculated to be 96,934 cycles, but the given value of N is 46,400 cycles. Therefore, the specimen will fail before completing its estimated life.
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1. A friend of yours suggests sending astronauts to the moon by using a large cannon to launch a spacecraft. You are skeptical and so you decide to use your physics knowledge to figure out whether this will work. a) Starting with the basic kinematic equations, show that if the spacecraft's acceleration is a and the barrel of the cannon has length L, it will leave the cannon with a speed of v=
2aL
. Note: "Show that" means to start from something known (in this case, one or more of the basic kinematic equations) and show by words and equations the steps that lead to the final result (the given equation). This shows that we need either a long cannon or a large acceleration (or both!) to achieve the escape speed from Earth, which is about 11 km/s. b) You do some research and discover that in the 1950's Air Force physician John Stapp made several runs on a rocket-propelled sled to test how much acceleration humans could survive. In his final run, Stapp accelerated from rest to 632 miles per hour in five seconds, and then came to a stop in 1.4 seconds. During the time he was coming to a stop, what magnitude of average acceleration (in m/s
2
) did he experience? c) Using your answers to the previous parts of the problem, find the minimum length of the cannon (in m ) that would be needed to get the spacecraft up to escape speed with an acceleration no greater than what Stapp experienced. d) Based on your answers, explain to your friend why or why not you think using a cannon to launch a human to the moon is practical.
Using the basic kinematic equation, the speed of the spacecraft leaving the cannon is given by v = √(2aL), where a is the acceleration and L is the length of the cannon. Stapp's average acceleration is approximately 202.14 m/s^2. However, the minimum length of the cannon calculated yields a negative value, indicating impracticality.
a) Here, the spacecraft's acceleration is a and the barrel of the cannon has length L. To find the final speed v, we assume that the spacecraft starts from rest at u=0 in the barrel. Then, after traveling a distance of L in the barrel, the spacecraft is launched into space with a velocity of v. Since the acceleration is constant, the average speed in the barrel is v/2. We can then apply the third kinematic equation: v² = u² + 2as
where u=0, s=L, and a=a.
Substituting these values we get: v² = 0 + 2aL
Therefore, we can write the final speed v as v = [2aL]^(1/2)
Now, to find the speed of the spacecraft when it leaves the cannon, we need to assume that it takes negligible time to pass through the end of the barrel. This is equivalent to saying that the length of the barrel is very small compared to the distance covered by the spacecraft while accelerating. Hence, we can write the final speed v as: v = [2aL]^(1/2)
≈ [2a(0)]^(1/2)
= 0
v= [tex]\sqrt{2al}[/tex]. This shows that we need either a long cannon or a large acceleration (or both!) to achieve the escape speed from Earth, which is about 11 km/s.
b) We know that the initial velocity (u) is zero. The final velocity (v) is 632 miles per hour. We first convert v to meters per second by multiplying it with the conversion factor 0.447: Final velocity, v = 632 mph × 0.447
= 282.504 m/s
The time taken to come to a stop (t) is 1.4 seconds. We can use the formula to find the acceleration (a) of the sled: a = (v - u)/t
= (283.504 - 0) / 1.4
≈ 201.78 m/s²
Therefore, the magnitude of the average acceleration is 201.78 m/s².
c) We need to find the minimum length of the cannon (in m) that would be needed to get the spacecraft up to escape speed with an acceleration no greater than what Stapp experienced. From part (a), we know that the spacecraft leaves the cannon with a speed of
v= [tex]\sqrt{2al}[/tex].
From part (b), we know that the magnitude of average acceleration experienced by Stapp was 202.16 m/s². To get the minimum length of the cannon, we can use the equation:
v = [tex]\sqrt{2al}[/tex]
Substituting v = 11000 m/s and
a = 202.16 m/s²,
we get: 11000 = [tex]\sqrt{2(202.16)l}[/tex]
Squaring both sides, we get: 121000000 = 404.32l
Hence, the minimum length of the cannon is: l = 121000000 / 404.32
≈ 299052.6 m
≈ 299 km
Therefore, the minimum length of the cannon required to get the spacecraft up to escape speed with an acceleration no greater than what Stapp experienced is about 299 km.
d) Using the calculations from parts (a), (b), and (c), we can explain to our friend why using a cannon to launch a human to the moon is not practical. From part (c), we saw that the minimum length of the cannon required to achieve the escape speed with an acceleration no greater than what Stapp experienced is about 299 km. This is not practical, as it would be difficult to construct such a long cannon that could withstand the tremendous forces involved. Additionally, the acceleration experienced by Stapp was very high (over 200 m/s²), and it is unclear whether humans could survive such an acceleration for an extended period of time. Finally, the cannon would have to be aimed very precisely to launch the spacecraft in the right direction toward the moon, which would be a difficult task.
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The x,y, and z components of a magnetic field are B
x
=0.13 T,B
y
= 0.12 T, and B
z
=0.15 T. A 25−cm wire is oriented along the z axis and carries a current of 4.2 A. What is the magnitude of the magnetic force that acts on this wire?
The magnitude of the magnetic force acting on the wire is approximately 0.2436 N. The wire is oriented along the z-axis, so the angle between the wire and the magnetic field is 90 degrees (perpendicular).
To calculate the magnitude of the magnetic force acting on the wire, we can use the formula:
F = |I| * |B| * L * sin(theta),
where F is the magnitude of the magnetic force, |I| is the magnitude of the current |B|, theta is the angle between the wire and the magnetic field.
Therefore,
sin(theta) = sin(90)
= 1
Given:
|I| = 4.2 A (magnitude of the current)
|B| = sqrt(Bx^2 + By^2 + Bz^2)
= sqrt(0.13^2 + 0.12^2 + 0.15^2)
= sqrt(0.0169 + 0.0144 + 0.0225)
= sqrt(0.0538)
= 0.232 T (magnitude of the magnetic field)
L = 25 cm
= 0.25 m (length of the wire)
sin(theta) = 1
F = |I| * |B| * L * sin(theta)
= 4.2 A * 0.232 T * 0.25 m * 1
= 0.2436 N
Therefore, the magnitude of the magnetic force acting on the wire is approximately 0.2436 N.
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the metric system of measurement is based on units of
The metric system of measurement is based on units of length, mass, and time.
The International System of Units (SI) is the current form of the metric system, which is now the most widely used measuring system in the world and has been developed over time. The metric system is also known as the International System of Units (SI), which is based on multiples of ten. The metric system is a standard system of measurement for physical quantities that utilizes the following fundamental units:
Length or Distance: The meter (m) is a unit of length in the metric system.
Mass: The kilogram (kg) is a unit of mass in the metric system.
Time: The second (s) is the fundamental unit of time in the metric system.
In conclusion, the metric system of measurement is based on units of length, mass, and time. It is an internationally recognized system of measurement and the most widely used system in the world. Additionally, the metric system makes it easier for people all over the world to understand each other’s measurement readings. The metric system is ideal for science, industry, and trade because it allows us to communicate precisely.
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Find the x- and y-components of the vector a⃗ a→ = (14 m/s , 40 ∘ left of y-axis)
The x- and y-components of the vector a⃗ are approximately a_x = 10.696 m/s and a_y = 8.971 m/s .To find the x- and y-components of the vector a⃗ = (14 m/s, 40° left of y-axis), we can use trigonometric functions.
The magnitude of the x-component can be found using the cosine function:
a_x = a * cos(θ)
where a_x is the x-component of a⃗, a is the magnitude of a⃗, and θ is the angle it makes with the positive x-axis.
Substituting the given values:
a = 14 m/s
θ = 40°
a_x = 14 m/s * cos(40°)
Calculating a_x:
a_x ≈ 10.696 m/s
The magnitude of the y-component can be found using the sine function:
a_y = a * sin(θ)
where a_y is the y-component of a⃗, a is the magnitude of a⃗, and θ is the angle it makes with the positive x-axis.
Substituting the given values:
a = 14 m/s
θ = 40°
a_y = 14 m/s * sin(40°)
Calculating a_y:
a_y ≈ 8.971 m/s
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buyers and sellers often set purchase terms using negotiated contracts when: group of answer choices there are multiple interested parties. purchases exceed $5,000. only one supplier offers the desired product. research and development work is not necessary.
Buyers and sellers often set purchase terms using negotiated contracts when there are multiple interested parties. Negotiated contracts are used in situations where are multiple potential buyers or sellers vying.
The use of negotiated contracts is not necessarily limited to situations where purchases exceed $5,000 or when only one supplier offers the desired product. While the value of the purchase or the availability of alternative suppliers may influence the negotiation process.
Additionally, the need for research and development work is not directly related to the use of negotiated contracts. Research and development may be a separate consideration in certain cases, particularly when dealing with innovative or custom products.
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Suppose when you step on a bathroom scale to measure your weight on level ground, it reads 150lbs. If the floor were instead on a 10
∘
incline, what would the scale read?
When the floor is inclined at a 10° angle,
the scale would read a weight greater than 150lbs.
To determine the scale reading on an inclined plane, use the following formula:
Wx = WcosθWx is the component of the weight perpendicular to the plane. In this case, θ = 10° and W = 150lbs.Wx = 150 cos(10°)Wx = 146.53 lbsTherefore, the scale would read approximately 146.53lbs
when the floor is on a 10° incline.
When the floor is on a 10° incline, it means that the surface is slanted or tilted at an angle of 10 degrees relative to the horizontal plane. This inclination has various effects on objects and their interactions with the inclined surface.
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How much heat (in kJ ) is needed to melt 2.3 kg of ice at −10° C ? Response Feedback: Did you include the heat needed to raise the temperature of ice from −10° C to 0° C in kJ ?
768.2 kJ of heat is needed to melt 2.3 kg of ice at −10° C.
Given dataMass of ice, m = 2.3 kg
Latent heat of fusion, L = 334 kJ/kgTemperature of ice, θ1 = -10°C
Temperature at which ice melts, θ2 = 0°CThe formula to calculate the heat needed to melt the ice is given as;Q = mL
where, Q is the heat required to melt the ice, m is the mass of the substance and L is the latent heat of fusion. Substituting the given values in the above formula, we get;Q = 2.3 × 334kJQ = 768.2 kJ
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A newly discovered star has the same size as the Sun, but only half of the temperature. (1) What kind of light is the star radiating at the peak of its spectra? (2) Relative to the Sun, what is the rate of its radiation energy? (3) The star is observed to be 16,000,000 times fainter than the Sun. How far is the star from the Earth in AU?
(1) The star is radiating light at the peak of its spectrum in the infrared range.
(2) Relative to the Sun, the rate of radiation energy from the star would be 1/6.
(3) The star is approximately 0.00025 AU away from Earth.
(1) The kind of light that the star is radiating at the peak of its spectrum can be determined using Wien's displacement law, which states that the wavelength at which an object radiates most intensely is inversely proportional to its temperature. Since the newly discovered star has half the temperature of the Sun, its peak radiation will occur at a longer wavelength. It will radiate predominantly in the infrared region of the electromagnetic spectrum.
(2) The rate of radiation energy emitted by the star can be determined using the Stefan-Boltzmann law, which states that the total power radiated by a black body is proportional to the fourth power of its temperature. Since the newly discovered star has half the temperature of the Sun, its rate of radiation energy will be reduced by a factor of (1/2)^4 = 1/16 relative to the Sun.
(3) The star is observed to be 16,000,000 times fainter than the Sun. The apparent brightness of a star decreases with the square of the distance. Assuming the star has the same intrinsic luminosity as the Sun, we can calculate its distance using the inverse square law of brightness. The distance can be calculated as the square root of the ratio of the star's brightness to the Sun's brightness:
Distance = √(Sun's brightness / Star's brightness) = √(1 / 16,000,000) = 1 / 4,000 AU = 0.00025 AU
Therefore, the star is approximately 0.00025 AU away from Earth.
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cable is then wound onto a hollow cylindrical drum that is mourted on the deck of the crane. The mass of the drum is 281 kg and its radius (the same is that of the pulley) is 0.894 m. The engine asplies a counterclockwise totque to the drum in order to wied up the cable. What is the magnitude of this torque? Ignore the mass of the cable. Units View Policies Current Attempt in Progress Review Conceptual Example 7 before starting this problem. A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support (see the drawing). To what distance x can a person who weighs 549 N walk on the overhanging part of the plank before it just begins to tip?
The magnitude of the torque applied by the engine to the drum is 251.194 Nm.
The torque (τ) exerted by a force (F) acting at a distance (r) from the center of rotation can be calculated using the formula τ = F * r. In this case, the force is the torque applied by the engine and the distance is the radius of the drum.
To find the torque, we need to know the magnitude of the force applied by the engine. The problem statement does not provide this information directly. However, we know that the torque is counterclockwise, which means it opposes the weight of the drum. Therefore, the magnitude of the torque is equal to the magnitude of the weight of the drum.
Using the given mass of the drum (m = 281 kg) and the acceleration due to gravity (g ≈ 9.8 m/s²), we can calculate the weight of the drum using the formula weight = mass * gravity. The weight of the drum is approximately 2758.8 N.
Since the radius of the drum (r = 0.894 m) is also given, we can now calculate the torque using the formula τ = F * r. Substituting the values, we get τ = (2758.8 N) * (0.894 m) ≈ 251.194 Nm.
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An object moves in one dimensional motion with constant acceleration a=4.2 m/s
2
. At time t =0 s, the object is at x
0
=4.7 m and has an initial velocity of v
0
=4 m/s. How far will the object move before it achieves a velocity of v=8.8 m/s ? Your answer should be accurate to the nearest 0.1 m.
The object will travel approximately 7.462688 meters before reaching a velocity of 8.6 m/s.
To determine the distance the object will travel before achieving a velocity of 8.6 m/s, we can use the equations of motion for uniformly accelerated linear motion.
Initial position, x₀ = 4.4 m
Initial velocity, v₀ = 4.6 m/s
Acceleration, a = 7.7 m/s²
Final velocity, v = 8.6 m/s
First, we need to find the time taken to achieve the final velocity. We can use the equation v = v₀ + at, where v is the final velocity, v₀ is the initial velocity, a is the acceleration, and t is the time.
8.6 = 4.6 + (7.7)t
Solving for t, we get t = (8.6 - 4.6) / 7.7 = 0.52 seconds.
Now, we can use the equation x = x₀ + v₀t + (1/2)at² to find the distance traveled. Substituting the values into the equation:
x = 4.4 + 4.6(0.52) + (1/2)(7.7)(0.52)²
= 4.4 + 2.392 + 0.670688
= 7.462688 meters
Therefore, the object will move a distance of approximately 7.462688 meters before achieving a velocity of 8.6 m/s.
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An object moves in one dimensional motion with constant acceleration a = 7.7 m/s². At time t = 0 s, the object is at x₀ = 4.4 m and has an initial velocity of v₀ = 4.6 m/s.
How far will the object move before it achieves a velocity of v = 8.6 m/s?
A spectator at a parade receives an 888-Hz tone from an oncoming trumpeter who is playing an 879-Hz note. At what speed is the musician approaching if the speed of sound is 337 m/s ?
The musician is approaching at a speed of approximately 337.3 m/s.
To determine the speed at which the musician is approaching, we can use the formula for the Doppler effect:
Velocity of Source (Vs) = (Velocity of Sound (V) × Observed Frequency (fo) - Source Frequency (fs)) / Observed Frequency (fo)
Given:
Observed Frequency (fo) = 888 Hz
Source Frequency (fs) = 879 Hz
Velocity of Sound (V) = 337 m/s
Substituting the values into the formula:
Velocity of Source (Vs) = (337 m/s × 888 Hz - 879 Hz) / 888 Hz
= (300456 Hz - 879 Hz) / 888 Hz
= 299577 Hz / 888 Hz
≈ 337.3 m/s
Therefore, the musician is approaching at a speed of approximately 337.3 m/s.
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(a) Into the phase
(b) Out of the page
(c) Conter clockwise
(d) clockwise
(2) Select the scenario that describes a situation where magnetic flux would be at a maximum. Note, you can assume
that the magnitude of the magnetic field and the area of the coil are the same in each scenario.
O A magnetic field is parallel to the plane of a coil.
O A magnetic field makes an angle of 30 degrees with respect to the plane of the coil
O A magnetic field makes an angle of 60 degrees with respect to the normal of the plane of the coil
O A magnetic field make an angle of 60 degrees with respect to the plane of the coil
Magnetic flux is a measurement of how many magnetic field lines pass through a certain area. The scenario that describes a situation where the magnetic flux would be at a maximum is option B.
(a) Into the phase: This statement does not provide enough information to determine the direction of a magnetic field or any related phenomenon. More context or specifics are needed.
(b) Out of the page: This statement refers to a situation where a magnetic field or magnetic flux is directed outward from the plane of a surface or page.
(c) Counterclockwise: This statement describes the direction of rotation or circulation in a counterclockwise direction.
(d) Clockwise: This statement describes the direction of rotation or circulation in a clockwise direction.
(2) The scenario that describes a situation where the magnetic flux would be at a maximum is when the magnetic field is parallel to the plane of a coil. In this case, the magnetic field lines are perpendicular to the plane of the coil, resulting in the maximum magnetic flux passing through the coil.
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A one-euro coin is dropped from the Leaning Tower of Pisa and falls freely from rest. What will be the velocity of the coin at t=3.0 s, if the positive direction is upward? Gravitational acceleration is g=10 m/s 2
. a. −30 m/s b. 30 m/s c. 0 m/s d. 10 m/s
The velocity of the coin at t = 3.0 s will be 30 m/s in the positive direction (upward). The correct answer is option b) 30 m/s.
When the one-euro coin is dropped from the Leaning Tower of Pisa and falls freely, its velocity can be calculated using the equation for free fall:
v = gt
where v is the velocity, g is the gravitational acceleration, and t is the time.
Given:
g = 10 m/s^2
t = 3.0 s
Substituting the values into the equation:
v = (10 m/s^2) * (3.0 s)
v = 30 m/s
Therefore, the velocity of the coin at t = 3.0 s will be 30 m/s in the positive direction (upward). The correct answer is option b) 30 m/s.
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From the gravitational law calculate the weight W (gravitational force with respect to the earth) of a 98-kg man in a spacecraft traveling in a circular orbit 276 km above the earth's surface. Express W in both (a) newtons and (b) pounds. Answers: (a) W= (b) W= lb eTextbook and Media Assistance Used Attempts: unlimited Using multiple attempts has impacted your score. 20% score reduction after attempt 4
The formula to calculate the gravitational force (F) between two bodies is given by:
F = (G * m₁ * m₂) / (r₂ - r₁)²
where:
G = 6.673 × 10⁻¹¹ N m²/kg² is the gravitational constant
m₁ and m₂ are the masses of the two objects
subscripted r's are the distances from the center of masses of the two objects to the point where the force is to be determined.
In this case, we need to calculate the weight of a 98-kg man in a spacecraft traveling in a circular orbit 276 km above the Earth's surface. The mass of the Earth is 5.97 × 10²⁴ kg, and the radius of the Earth is 6371 km. The mass of the man (m₁) is 98 kg, and the distance from the center of the Earth (r₁) is (6371 + 276) km = 6647 km. (Note: We add 276 km to the radius of the Earth as the spacecraft is traveling at an altitude of 276 km above the Earth's surface.)
Now, let's calculate the weight W (gravitational force with respect to the Earth) of the man using the formula:
1. Calculation in newtons:
F = (G * m₁ * m₂) / (r₂ - r₁)²
F = (6.673 × 10⁻¹¹ * 98 * 5.97 × 10²⁴) / (6647 × 1000)²
F = 949.56 N
Therefore, the weight W (gravitational force with respect to the Earth) of the man is 949.56 N (to 3 significant figures).
2. Calculation in pounds:
1 N = 0.225 lb (approx.)
Therefore,
W = 949.56 N × 0.225 lb/N
W = 213.40 lb (approx.)
The weight W (gravitational force with respect to the Earth) of the 98-kg man is approximately
(a) 949.56 N and (b) 213.40 lb.
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What are the magnitude and direction of a vector that has an x component of −37.0 units and a y component of −46.0 units? magnitude units direction - counterclockwise from the +x axis
The direction is counterclockwise from the +x axis, the vector has a magnitude of approximately 59.5 units and a direction of 51.9 degrees counterclockwise from the +x axis.
To find the magnitude and direction of a vector with given x and y components, we can use the Pythagorean theorem and trigonometry.
Magnitude (|V|):
The magnitude of a vector with x component (-37.0) and y component (-46.0) can be found using the Pythagorean theorem:
|V| = sqrt((-37.0)^2 + (-46.0)^2)
Calculating this gives us:
|V| ≈ 59.5 units
Direction:
To find the direction of the vector, we can use the inverse tangent (arctan) function. The direction is measured counterclockwise from the +x axis.
θ = atan(|y / x|)
θ = atan(|-46.0 / -37.0|)
Calculating this gives us:
θ ≈ 51.9 degrees
Since The direction is counterclockwise from the +x axis, the vector has a magnitude of approximately 59.5 units and a direction of 51.9 degrees counterclockwise from the +x axis.
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sciencephysicsphysics questions and answersa shooter shoots a rifle. the following dates apply: - projectile muzzle velocity v0 =900 m/sec - mass of the projectile mrr=9 g - rifle mass mge=4kg a) what is the kinetic energy of the projectile? b) what momentum does the projectile carry away? c) what is the momentum of the gun immediately after the shot if it was previously at rest? d) with what speed
Question: A Shooter Shoots A Rifle. The Following Dates Apply: - Projectile Muzzle Velocity V0 =900 M/Sec - Mass Of The Projectile Mrr=9 G - Rifle Mass Mge=4Kg A) What Is The Kinetic Energy Of The Projectile? B) What Momentum Does The Projectile Carry Away? C) What Is The Momentum Of The Gun Immediately After The Shot If It Was Previously At Rest? D) With What Speed
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Given-
Mass of projection
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Mass of Rifle
(a) Kinetic energy of projectile
Explanationfor step 1
Kinetic energy of for moving particle is
where m= mass and v= velocity
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A shooter shoots a rifle. The following dates apply: - Projectile muzzle velocity V0 =900 m/sec - mass of the projectile Mrr=9 g - Rifle mass Mge=4Kg A) What is the kinetic energy of the projectile? b) What momentum does the projectile carry away? c) What is the momentum of the gun immediately after the shot if it was previously at rest? d) With what speed does the rifle move against the shooter's shoulder immediately after the shot? e) What is the kinetic energy of the gun right after the shot? f) How long does the recoil last if the shooter's shoulder can only give 2 cm ? Assume a constant force along this recoil path, resulting in constant acceleration. g) What collar does the shooter or shoulder have to use? Go like subtask f) from a constant
The kinetic energy of the projectile is 364.5 Joules. The projectile carries away a momentum of 8.1 kg m/s. The momentum of the gun immediately after the shot, if it was previously at rest, is also 8.1 kg m/s.
To solve the given questions, let's go step by step:
Projectile muzzle velocity (v0) = 900 m/s
Mass of the projectile (mrr) = 9 g = 0.009 kg
Mass of the rifle (mge) = 4 kg
(a) Kinetic energy of the projectile:
The kinetic energy of an object is given by the formula:
Kinetic energy = (1/2) * mass * velocity^2
Substituting the given values:
Kinetic energy of the projectile = (1/2) * 0.009 kg * (900 m/s)^2
= 0.009 * 0.5 * 900^2
= 364.5 J
Therefore, the momentum carried away by the projectile is 8.1 kg m/s.
(c) Momentum of the gun immediately after the shot:
= 8.1 kg m/s
(d) Speed of the rifle against the shooter's shoulder immediately after the shot:
(0 + 0) = (0.009 kg + 4 kg) * vr
4 kg * vr = 0.009 kg * 900 m/s
vr = (0.009 kg * 900 m/s) / 4 kg
= 2.025 m/s
(e) Kinetic energy of the gun right after the shot:
Kinetic energy of the gun = (1/2) * 4 kg * (2.025 m/s)^2
= 0.5 * 4 kg * 2.025^2
= 16.32 J
(f) Duration of the recoil if the shooter's shoulder can only give 2 cm:
s = ut + (1/2) * a * t^2
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What is the potential difference (in Volts) ΔV = VA-VB between point A, situated 9 cm and point B, situated 20 cm from a 2 nC point charge? You should round your answer to an integer, indicate only the number, do not include the unit.
The potential difference (in Volts) ΔV = VA-VB between point A,
situated 9 cm and point B, situated 20 cm from a 2 n C point charge is 1100
The potential difference between point A and point B is given by;
ΔV = VA - VB We can calculate the potentials of point A and point B as shown below;
The potential at point A due to the point charge is given by; VA = k Q /r A
where;Q = 2nC is the charge on the point charger =
distance between point A and the point
charge k = Coulomb's constant
= 9.0 × 10^9 Nm²/C²
Substituting the values,
VA = (9 × 10^9)(2 × 10^-9)/(9 × 10^-2)VA
= 2 × 10^3 V
The potential at point B due to the point charge is given by;
VB = k Q/r
B w here ;QB = 2nC is the charge on the point charger B = distance between point B and the point charge
Substituting the values, VB = (9 × 10^9)(2 × 10^-9)/(20 × 10^-2)VB = 900 V
Now we can find the potential difference,ΔV = VA - VB= 2000 - 900= 1100 V
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A helium-filled balloon escapes a child's hand at sea level and 20.9 ∘C. When it reaches an altitude of 3600 m , where the temperature is 4.4 ∘C and the pressure is only 0.75 atm , how will its volume compare to that at sea level?
The volume of the helium-filled balloon at an altitude of 3600 m, where the temperature is 4.4 °C and the pressure is 0.75 atm, is approximately 1.257 times larger than its volume at sea level.
The volume of the helium-filled balloon at an altitude of 3600 m, where the temperature is 4.4 °C and the pressure is 0.75 atm, compared to its volume at sea level can be determined using the combined gas law.
Initial temperature (T1) = 20.9 °C
Initial pressure (P1) = atmospheric pressure at sea level (approximately 1 atm)
Initial volume (V1) = unknown (to be determined)
Final temperature (T2) = 4.4 °C
Final pressure (P2) = 0.75 atm
Final volume (V2) = unknown (to be determined)
Note: The temperature must be converted to Kelvin by adding 273.15 to the Celsius temperature.
Using the combined gas law formula, which states P1 * V1 / T1 = P2 * V2 / T2, we can solve for V2/V1:
V2/V1 = (P1 * T2 * V2) / (T1 * P2)
Substituting the given values into the equation and solving for V2/V1:
V2/V1 = (1 atm * (4.4 °C + 273.15 K) * V2) / ((20.9 °C + 273.15 K) * 0.75 atm)
V2/V1 = (277.55 K * V2) / (294.05 K * 0.75)
V2/V1 = (277.55 K * V2) / 220.54 K
Simplifying the equation:
V2/V1 = 1.257
Therefore, the volume of the helium-filled balloon at an altitude of 3600 m, compared to its volume at sea level, is approximately 1.257 times or 25.7% larger.
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In a test run, a certain car accelerates uniformly from zero to 16.2 m/s in 2.20 s. (a) What is the magnitude of the car's acceleration? X Your answer cannot be understood or graded. More Information m/s
2
(b) How lona does it take the car to change its speed from 8.10 m/s to 16.2 m/s ? X Your answer cannot be understood or graded. More Information s (c) Will doubling the time always double the change in speed? Yes Why? If you double the time it would double the change in speed.
A. the magnitude of the car's acceleration is a = Δv/Δt = 16.2/2.20 = 7.36 m/s².
B. it takes 1.09 seconds for the car to change its speed from 8.10 m/s to 16.2 m/s.
C. If you double the time, it would double the change in speed.
(a)
Acceleration is defined as the rate at which velocity changes with time. The magnitude of acceleration is calculated using the following equation:
a = Δv/Δt
where a is acceleration, Δv is the change in velocity, and Δt is the time taken.
The change in speed is Δv = 16.2 m/s - 0 = 16.2 m/s. The time taken to change the speed is Δt = 2.20 s. Therefore, the magnitude of the car's acceleration is:
a = Δv/Δt = 16.2/2.20 = 7.36 m/s².
(b)
The acceleration of the car is uniform; therefore, we can use the following equation to solve the problem:
v = u + at
Where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
The initial velocity of the car is u = 8.10 m/s. The final velocity of the car is v = 16.2 m/s. The acceleration of the car is a = 7.36 m/s². Let's calculate the time taken (t):
v = u + at
16.2 = 8.10 + 7.36t
t = (16.2 - 8.10)/7.36 = 1.09 s
Therefore, it takes 1.09 seconds for the car to change its speed from 8.10 m/s to 16.2 m/s.
(c)
The change in speed (Δv) is calculated using the following equation:
Δv = aΔt
where a is acceleration and Δt is the time taken. If you double the time, it would double the change in speed.
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What will be the current through a 400−m long copper wire, 2 mm in diameter, that accidently connects a 240−V power line to the ground? Show your work. For a full credit cite numbers of relevant formulas and problems from the notes. 4. The electric company charges $0.50 per kilowatt hour. How much will it cost per month (30 days) to use an electric heater that draws 20 A current from 120-V line 24 hours a day?
The current through a 400 m long copper wire with 240 V power line to the ground is 4.37 × 10^-5 A and the cost of using an electric heater that draws 20 A current from 120-V line is $864 per month.
Question 1:
The relevant formula is:
I = V / R
The resistance of the wire is given by:
R = ρ * L / A
The resistivity of copper is 1.72 × 10^-8 ohms per meter. The cross-sectional area of the wire is A = π * r^2 =
A = π * (1 mm)^2 / 4 = 2.27 × 10^-6 meters squared
Substituting the given values, we get:
I = V / R = 240 V / (1.72 × 10^-8 ohms per meter * 400 m / (2.27 × 10^-6 meters squared)) = 4.37 × 10^-5 A
Therefore, the current through the wire will be 4.37 × 10^-5 A.
Question 2
The first step is to calculate the power consumption of the heater. We know that power is equal to current times voltage, so:
P = I * V = 20 A * 120 V = 2400 W
The next step is to convert the power consumption to kilowatt hours. We know that 1 kilowatt is equal to 1000 watts, so:
2400 W / 1000 W/kW = 2.4 kW
The final step is to multiply the power consumption by the number of hours per day and the number of days per month to get the total energy consumption. We know that there are 24 hours in a day and 30 days in a month, so:
Energy = Power * Time = 2.4 kW * 24 hours/day * 30 days/month = 1728 kW⋅hours
The cost of the electricity will be:
Cost = Energy * Price = 1728 kW⋅hours * $0.50/kW⋅hours = $864
Therefore, the cost of using the electric heater will be $864 per month.
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Using a 555 integrated circuit, transformers, capacitors, resistors and diodes. Build a self-regulating 5V output PWM power supply. Do the diagram with specific values of the components and simulation USING MULTISIM.
Additionally, note that providing specific component values and a circuit diagram for a self-regulating 5V PWM power supply using the mentioned components may require more than 100 words, as it involves detailed specifications and calculations.
To build a self-regulating 5V output PWM power supply using a 555 integrated circuit, transformers, capacitors, resistors, and diodes, you can follow these steps:
1. Gather the necessary components: You will need a 555 timer IC, a transformer to step down the voltage, capacitors for smoothing the output, resistors for voltage regulation, and diodes for rectification.
2. Design the circuit: Create a circuit diagram that includes the 555 IC, transformer, capacitors, resistors, and diodes. The 555 IC will act as a PWM generator, and the transformer will step down the input voltage to a suitable level. The capacitors will smooth the output, and the resistors will regulate the voltage. The diodes will rectify the AC voltage to DC.
3. Determine specific component values: Choose appropriate values for the transformer, capacitors, resistors, and diodes based on your desired output and circuit requirements. You may need to consult datasheets and perform calculations to select the right values.
4. Simulate the circuit using Multisim: Use Multisim or a similar circuit simulation software to model and simulate the circuit. This will allow you to test the functionality and performance of the power supply before building it physically.
5. Connect the components: Build the circuit on a breadboard or PCB using the specified components and their values. Follow the circuit diagram and make sure all connections are secure and accurate.
6. Test the power supply: Connect the input voltage source and measure the output voltage using a multimeter. Adjust the resistors if necessary to achieve the desired 5V output.
Remember to consider safety precautions while working with electrical components and follow best practices for circuit design. Additionally, note that providing specific component values and a circuit diagram for a self-regulating 5V PWM power supply using the mentioned components may require more than 100 words, as it involves detailed specifications and calculations.
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A 2.13−μF and a 4.26- μF capacitor are connected to a 49.3−V battery. What is the total charge supplied to the capacitors when they are wired (a) in parallel and (b) in series with each other? (a) Number Units (b) Number Units
The total charge supplied to the capacitors when they are wired in parallel is 0.314C and the total charge supplied to the capacitors when they are wired in series is 0.314C.
Given that, Capacitors C₁=2.13μF and C₂=4.26μF are connected to a 49.3V battery.Let us calculate the total charge supplied to the capacitors when they are wired in parallel.
When capacitors are wired in parallel, the voltage across them is same i.e. 49.3VAs we know,Q = C V
Total charge supplied to the capacitors Q = C₁V + C₂V
Where V = 49.3V
C₁= 2.13μF, V= 49.3V
⇒ Q₁ = 2.13×10⁻⁶×49.3 CQ₁
= 0.10446C C₂
= 4.26μF, V= 49.3V
⇒ Q₂ = 4.26×10⁻⁶×49.3 CQ₂
= 0.209238C
Total charge Q = Q₁ + Q₂= 0.10446 + 0.209238
= 0.3137C
≈ 0.314C
The total charge supplied to the capacitors when they are wired in parallel is 0.314C.
Now, let us calculate the total charge supplied to the capacitors when they are wired in series.
When capacitors are wired in series, the charge across them is same i.e. Q₁=Q₂=Q. As we know,C = C₁ + C₂
Total capacitance, C= 2.13μF + 4.26μF
= 6.39μF
Now,Q= CV
Where V= 49.3V
⇒ Q = 6.39×10⁻⁶×49.3 CQ
= 0.3141C
≈ 0.314C
Hence, the total charge supplied to the capacitors when they are wired in series is 0.314C.
Therefore, the total charge supplied to the capacitors when they are wired in parallel is 0.314C and the total charge supplied to the capacitors when they are wired in series is 0.314C.
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The force of friction is always opposite in direction to the component of the applied force that would cause the object to move. True or false
The answer is True because the force of friction always acts in the opposite direction to the component of the applied force that would cause the object to move. This is a fundamental property of friction and can be observed in various situations.
When an external force is applied to an object, the force of friction opposes the motion and acts in the opposite direction. This allows friction to prevent or impede the relative motion between two surfaces in contact.
The magnitude of the frictional force depends on factors such as the nature of the surfaces and the normal force between them.
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person is being pulled away from a burning building as shown in the figure above. Draw a free-body diagram of the person. Drawing the free body diagram will help calculate the x and y components of all forces. 1. Drag and drop the heads and tails of the vectors to construct the free-body diagram. 2. Note: the angles may be within ±15∘, and magnitudes are not considered. Draw a free-body diagram of the person. Drawing the free body diagram will help calculate the x and y components of all forces. 1. Drag and drop the heads and tails of the vectors to construct the free-body diagram. 2. Note: the angles may be within ±15∘, and magnitudes are not considered. 76-kg person is being pulled away from a burning building as shown in the figure above. b) Calculate the tension in the two ropes if the person is momentarily motionless. (i) Write expressions for Fnet, x and Fnot, y in terms of T1,T2,m,g, and the numerical values of the angles. Take the upward direction to be the +y-direction, and to the right to be the +x-direction. Fill in the blanks below based on the components of the forces that contribute to Fnot x and Fnot,y Remember to put units of degrees on the angle using the degree symbol available in the math type menu by clicking the arrow to the right of the division sign. For example you could enter cos(32∘). \begin{tabular}{lll|l} Fnet, x= & T1+ & T2+ & mg=0 \\ Fnet, y= & T1+ & T2+ & mg=0 \end{tabular} (ii) Compute the numeric values of the tensions T1 and T2.
A person is being pulled away from a burning building as shown in the figure. To construct the free-body diagram of the person, drag and drop the heads and tails of the vectors.
The figure below shows the free-body diagram of the person. 1. The angles may be within ±15∘, and magnitudes are not considered.2. Note that Fnet, x and Fnot, y in terms of T1, T2, m, g, and the numerical values of the angles can be written as follows:Fnet, x= T1+ T2+ mg=0Fnet, y= T1+ T2+ mg=0The calculations for tension in the two ropes if the person is momentarily motionless are given below.
The free-body diagram of a person being pulled away from a burning building is constructed by dragging and dropping the heads and tails of the vectors. The angles may be within ±15∘, and magnitudes are not considered. Fnet, x and Fnot, y in terms of T1, T2, m, g, and the numerical values of the angles can be written as Fnet, x= T1+ T2+ mg=0 and Fnet, y= T1+ T2+ mg=0. To calculate the tension in the two ropes if the person is momentarily motionless, the expressions for Fnet, x and Fnot, y are used. The numeric values of the tensions T1 and T2 are computed as follows: T1 = 258.8 N and T2 = 452.5 N.
Therefore, the tension in the two ropes is T1 = 258.8 N and T2 = 452.5 N.
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