To find the average speed of an object, you need to divide the total distance traveled by the time it took to travel that distance. Since the distance traveled and time taken are given, the average speed can be easily calculated.
Given, Distance traveled in the first 2 seconds = 11 meters Distance traveled in the next 1 second = 16 meters Total distance traveled = Distance traveled in the first 2 seconds + Distance traveled in the next 1 second= 11 meters + 16 meters= 27 meters
Total time taken = Time taken in the first 2 seconds + Time taken in the next 1 second= 2 seconds + 1 second= 3 seconds Average speed = Total distance traveled ÷ Total time taken= 27 meters ÷ 3 seconds= 9 meters per second Hence, the average speed of the object is 9 meters per second.
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List three (3) reasons why a PT100 to 4-20mA transmitter would
be used for plant temperature monitoring rather than just a 4-wire
PT100 RTD sensor with very long leads?
There are several reasons why a PT100 to 4-20mA transmitter may be used for plant temperature monitoring instead of just a 4-wire PT100 RTD sensor with very long leads:
Signal Integrity: Long leads in a temperature monitoring system can introduce additional resistance, which can affect the accuracy of temperature measurements.
By using a PT100 to 4-20mA transmitter, the resistance of the long leads is not directly connected to the measurement circuit, minimizing the impact on signal integrity.
Noise Immunity: Long leads are susceptible to picking up electrical noise and interference from other equipment or electromagnetic sources. A PT100 to 4-20mA transmitter can convert the resistance measurement to a current signal, which is less susceptible to noise and can travel longer distances without significant degradation.
Compatibility and Standardization: 4-20mA current loop signals are widely used in industrial automation systems,and many data acquisition and control devices are designed to accept and process this type of signal.
By using a PT100 to 4-20mA transmitter, the temperature measurement can be easily integrated into existing systems without the need for additional signal conditioning or conversion.
Overall, using a PT100 to 4-20mA transmitter provides advantages in terms of signal integrity, noise immunity, and compatibility, making it a preferred choice for plant temperature monitoring applications where long leads are involved.
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Suppose a point charge creates a 11500 N/C electric field at a distance of 0.75 m. A 50% Part (a) What is the magnitude of the point charge in coulombs? ∣Q∣= Hints: 090 deduction per hint. Hints remaining: 1 Feedback: deduction per feedback A 50% Part (b) What is the strength of the field, in newtons per coulomb, at a distance of 10 m ?
(a) the magnitude of the point charge in coulombs is 9.68 × 10^-6 C.
(b) the strength of the field, in newtons per coulomb, at a distance of 10 m is 9.68 N/C.
Part(a)
∣Q∣=The formula for the electric field is given by, E = kQ/r where, k = Coulomb's constant, Q = Charge of the point charge.
Solving for the unknown charge Q, we get
Q = Er/k.
On substituting the values of E, r and k, we get Q = (11500 N/C × 0.75 m) / (9 × 10^9 Nm^2/C^2)
Q = 9.68 × 10^-6 C Coulombs (C).
Therefore, the magnitude of the point charge in coulombs is 9.68 × 10^-6 C.
Part (b)
The electric field formula can also be expressed as
E = kQ/r².
On substituting the known values, we get E = (9 × 10^9 Nm²/C² × 9.68 × 10^-6 C) / (10 m)²
E = 9.68 N/C (Approx).
Therefore, the strength of the field, in newtons per coulomb, at a distance of 10 m is 9.68 N/C.
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A. The magnitude of the point charge in coulombs is 1.2 × 10^-6 C.
B. The strength of the electric field, in newtons per coulomb is 1.08 × 10^-7 N/C.
(a) The magnitude of the point charge in coulombs is given by Coulomb's law, which is written as:
k = 9 × 10^9 N m^2 C^-2
This equation relates the electrostatic force between two point charges to their quantities and the distance between them. Hence, the magnitude of the point charge in coulombs is:
|Q| = k × E × r²
|Q| = (9 × 10^9 N m^2 C^-2) × (11500 N/C) × (0.75 m)²
|Q| = 1.2 × 10^-6 C
(b) The strength of the electric field, in newtons per coulomb, at a distance of 10 m is given by Coulomb's law, which is written as:
k = 9 × 10^9 N m^2 C^-2
This equation relates the electrostatic force between two point charges to their quantities and the distance between them. Hence, the strength of the electric field, in newtons per coulomb, at a distance of 10 m is:
E = k × Q / r²
E = (9 × 10^9 N m^2 C^-2) × (1.2 × 10^-6 C) / (10 m)²
E = 1.08 × 10^-7 N/C
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t takes 2.0 μJ of work to move a 11 nC charge from point A to B. It takes -6.0 μJ of work to move the charge from C to B. What is the potential difference VC−VA ?
Express your answer in volts.
2. A point charge with charge q1 = 2.20 μC is held stationary at the origin. A second point charge with charge q2 = -4.60 μC moves from the point (0.135 mm , 0) to the point (0.230 mm , 0.280 mm ). How much work is done by the electrostatic force on the moving point charge?
Express your answer in joules.
1. The potential difference VC−VA = (-4.0 * 10^-6 J) / q
2. The work done by the electrostatic force on the moving point charge is approximately 2.95 * 10^-12 Joules.
1. To calculate the potential difference (VC−VA) between points C and A, we can use the formula for potential difference:
Potential Difference (V) = Work (W) / Charge (q)
Work from A to B (WA→B) = 2.0 μJ = 2.0 * 10^-6 J
Work from C to B (WC→B) = -6.0 μJ = -6.0 * 10^-6 J
We need to find the potential difference VC−VA.
Using the formula, we have:
VC−VA = (WC→B - WA→B) / q
Since the charge (q) is not given, we need to find it first. The total work done on the charge from A to C can be found by adding the work done from A to B and from C to B:
Total Work done (WTOTAL) = WA→B + WC→B
Therefore,
Total Work (WTOTAL) = 2.0 * 10^-6 J + (-6.0 * 10^-6 J)
= -4.0 * 10^-6 J
Now, we can use the formula to find the charge (q):
WTOTAL = V * q
q = WTOTAL / V
Substituting the values:
q = (-4.0 * 10^-6 J) / (VC−VA)
Since we want to find VC−VA, we can rearrange the equation:
VC−VA = (-4.0 * 10^-6 J) / q
Therefore, we need the value of charge (q) to calculate the potential difference VC−VA.
2. To calculate the work done by the electrostatic force on the moving point charge, we can use the formula:
Work (W) = Electric Force (F) * Distance (d)
Given:
Charge q1 = 2.20 μC = 2.20 * 10^-6 C
Charge q2 = -4.60 μC = -4.60 * 10^-6 C
To find the work, we need to calculate the electric force and the distance between the two points.
The electric force (F) between the two charges can be calculated using Coulomb's Law:
F = (k * |q1 * q2|) / r²
Where:
k is the electrostatic constant = 8.99 * 10^9 N m²/C²
|q1 * q2| is the absolute value of the product of the charges
r is the distance between the charges
Given:
r = √((0.230 mm - 0.135 mm)² + (0.280 mm - 0)²) (distance between two points using the Pythagorean theorem)
Substituting the values and calculating r:
r = √((0.095 mm)² + (0.280 mm)²)
= √(0.009025 mm² + 0.0784 mm²)
= √(0.087425 mm²)
≈ 0.2956 mm
Converting r to meters:
r = 0.2956 mm * (1 m / 1000 mm)
= 0.0002956 m
Now, we can calculate the electric force:
F = (k * |q1 * q2|) / r²
= (8.99 * 10^9 N m²/C² * |2.20 * 10^-6 C * -4.60 * 10^-6 C|) / (0.0002956 m)²
Calculate the the magnitude of the product of the charges:
|q1 * q2| = |2.20 * 10^-6 C * -4.60 * 10^-6 C|
= |-1.012 * 10^-11 C²|
≈ 1.012 * 10^-11 C²
Substituting the values and calculating the electric force:
F = (8.99 * 10^9 N m²/C² * 1.012 * 10^-11 C²) / (0.0002956 m)²
Next, we multiply the electric force by the distance between the two points to find the work done:
W = F * d
= (8.99 * 10^9 N m²/C² * 1.012 * 10^-11 C²) / (0.0002956 m)² * 0.0002956 m
Calculate the value to find the work done:
W ≈ 2.95 * 10^-12 J
Therefore, the work done by the electrostatic force on the moving point charge is approximately 2.95 * 10^-12 Joules.
Thus,
1. The potential difference VC−VA = (-4.0 * 10^-6 J) / q
2. The work done by the electrostatic force on the moving point charge is approximately 2.95 * 10^-12 Joules.
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Suppose a horizontal block-spring system has a spring constant of 1240 N/m and a block of mass 15.0 kg. Calculate: - the angular frequency in rad/s: - the frequency in Hz : , and - the period in s:
The angular frequency is approximately 9.10 rad/s, the frequency is approximately 1.45 Hz, and the period is approximately 0.69 s.V
To calculate the angular frequency, frequency, and period of a block-spring system, we can use the formulas:
Angular frequency (ω) = √(k / m)
Frequency (f) = ω / (2π)
Period (T) = 1 / f
Spring constant (k) = 1240 N/m
Mass of the block (m) = 15.0 kg
1. Angular frequency (ω): ω = √(k / m)
= √(1240 N/m / 15.0 kg)
≈ √(82.67 rad/s²)
≈ 9.10 rad/s
2. Frequency (f): f = ω / (2π)
= 9.10 rad/s / (2π)
≈ 1.45 Hz
3. Period (T): T = 1 / f
= 1 / 1.45 Hz
≈ 0.69 s
Therefore, the angular frequency is approximately 9.10 rad/s, the frequency is approximately 1.45 Hz, and the period is approximately 0.69 s.
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A certain parallel plate capacitor consists of two plates, each with area of 200 cm
2, separated by a 0.40−cm air gap: a. Compute its capacitance b. If the capacitor is connected across a 500 V source, find the charge, the energy stored, and the strength of electric field between the plates. c. If a liquid with a dielectric constant of 2.6 is poured between the plates to fill the air gap, how much additional charge will flow on the capacitor from the 500 V source?
Compute its capacitance The formula for capacitance is given by; C=ε0A/d Where;ε0 is the electric constant A is the area of the plates d is the distance between the plates Substituting the given values in the equation gives,
C = ε0A/d
C = 8.85 x 10^-12 x 200 / 0.40
C = 4.43 x 10^-9 Fb.
If the capacitor is connected across a 500 V source, find the charge, the energy stored, and the strength of electric field between the plates.
Charge, Q = VC
Q = 500 V x 4.43 x 10^-9 F
Q = 2.215 x 10^-6 C
The energy stored in a capacitor is given by the equation;
E = 0.5QV
Substituting the given values in the equation gives,
E = 0.5 x 2.215 x 10^-6 C x 500 VE = 0.5537 J
The strength of electric field between the plates is given by the equation;
E = V/d
Substituting the given values in the equation gives,
E = 500 V / 0.40 cm
E = 1.25 x 10^3 V/cm (or N/C)c.
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The world's fastest pulsed laser outputs 4.352×1016 Hz photons at a rate of 2.7×1029 photons/sec. If the pulse lasts for 43asec (a is the symbol for the prefix "atto" which means 10−18 ), how much energy (in J) is contained in one pulse? Note that h=6.626×10−34 J⋅s
Energy contained in one pulse of the laser is 2.10 joules (J).
To calculate the energy contained in one pulse of the laser, we can use the formula:
Energy = Photon energy × Number of photons
The photon energy can be calculated using the formula:
Photon energy = Planck's constant (h) × Frequency
Given:
Frequency of photons (ν) = 4.352×10^16 Hz
Number of photons per second (n) = 2.7×10^29 photons/sec
Pulse duration (Δt) = 43 asec = 43×10^(-18) s
Planck's constant (h) = 6.626×10^(-34) J·s
First, let's calculate the photon energy:
Photon energy = h × ν
Photon energy = (6.626×10^(-34) J·s) × (4.352×10^16 Hz)
Now, let's calculate the energy in one pulse:
Energy = Photon energy × Number of photons × Pulse duration
Energy = (Photon energy) × (Number of photons per second) × (Pulse duration)
Substituting the values, we have:
Energy = (6.626×10^(-34) J·s) × (4.352×10^16 Hz) × (2.7×10^29 photons/sec) × (43×10^(-18) s)
Now we can calculate the energy:
Energy ≈ 2.10 J
Therefore, the energy contained in one pulse of the laser is approximately 2.10 joules (J).
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You (m
1
=90.0 kg) are standing in an elevator (m
2
=2200.kg) that is moving downward with a speed of 4.60 m/s. The elevator comes to a stop with a constant acceleration over a distance of 2.00 m.
The elevator comes to a stop with an acceleration of approximately 5.29 m/s² over a distance of 2.00 m, and it takes approximately 0.869 seconds to come to a stop.
To solve this problem, we can apply the principles of Newton's second law and use the equations of motion to determine the acceleration and the time taken for the elevator to come to a stop.
Given:
Mass of the person (m₁) = 90.0 kg
Mass of the elevator (m₂) = 2200 kg
Initial velocity of the elevator (v) = -4.60 m/s (negative sign indicates downward direction)
Distance over which the elevator comes to a stop (s) = 2.00 m
Step 1: Calculate the net force acting on the system.
The net force is given by the equation:
F_net = m_total * a
where m_total is the sum of the masses (m₁ + m₂).
Substituting the known values:
F_net = (90.0 kg + 2200 kg) * a
F_net = 2290.0 kg * a
Step 2: Calculate the acceleration (a).
To find the acceleration, we can use the equation:
a = (v² - u²) / (2s)
where u is the initial velocity and v is the final velocity.
Substituting the known values:
a = (-4.60 m/s)² / (2 * 2.00 m)
a = 21.16 m²/s² / 4.00 m
a ≈ 5.29 m/s² (approx.)
Step 3: Calculate the time taken to come to a stop.
We can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Substituting the known values:
0 = -4.60 m/s + 5.29 m/s² * t
Solving for t:
4.60 m/s = 5.29 m/s² * t
t = 4.60 m/s / 5.29 m/s²
t ≈ 0.869 s (approx.)
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You come across an empty room that is 5 m high, 20 m wide, and 30 m long. Being a physics student who is interested in sound waves you remember that acoustic resonances (standing waves) will occur in the room at certain frequencies. a. Calculate the three fundamental frequencies of the resonances that the engineer should expect. b. How do these frequencies change if one of the walls that is 30 m long is removed
The fundamental frequencies of the resonances that the engineer should expect are f1 = v/2L = v/2 × 5 = 1/10vf2 = v/2L = v/2 × 20 = 1/2vf3 = v/2L = v/2 × 30 = 3/10v, the frequency of the first resonance would be 1/10v, and the frequencies of the second and third resonances would no longer exist since the length of the room is zero.
a. To calculate the fundamental frequencies of the resonances, we use the formula:f = v/2LIn this formula, f represents frequency, v represents the speed of sound, and L represents the length of the room. We can use the same formula for all three fundamental frequencies. We only need to adjust L to account for the other two dimensions of the room. Therefore, the fundamental frequencies of the resonances that the engineer should expect are f1 = v/2L = v/2 × 5 = 1/10vf2 = v/2L = v/2 × 20 = 1/2vf3 = v/2L = v/2 × 30 = 3/10v
b. If one of the walls that is 30 m long is removed, we have a room that is 5 m high, 20 m wide, and 0 m long. The frequency of the first resonance is unaffected by the change because it depends only on the height of the room. However, the second and third resonant frequencies are determined by the length of the room. Since the length of the room is now zero, there will be no resonances at those frequencies. Therefore, the frequency of the first resonance would be 1/10v, and the frequencies of the second and third resonances would no longer exist since the length of the room is zero.
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When a fighter pilot makes a quick turn he experiences a centripetal acceleration. When this acceleration is greater than 5.7 x g.
When a fighter pilot makes a quick turn, he experiences a centripetal acceleration. When this acceleration is greater than 5.7 times g, the pilot may black out. This is because of the excessive force or acceleration acting on the pilot.
Centripetal acceleration is defined as the acceleration of a body directed towards the center of a circular path. The formula to calculate centripetal acceleration is given as: a_c = v^2/r, where v is the velocity of the body and r is the radius of the circular path.
On the other hand, g is the acceleration due to gravity. It is equal to 9.8 m/s². This means that when a fighter pilot experiences an acceleration greater than 5.7 times the acceleration due to gravity, the centripetal force is so strong that it may cause a pilot to black out.
This is because the pilot's body cannot withstand such a strong force or acceleration.
Thus, it is important for fighter pilots to undergo training to withstand these forces.
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A truck is traveling at a certain speed when it decelerates at -2 m/52 to a speed of 8
m/s in a distance of 176 m. Find the initial speed.
The initial speed of the truck can be calculated using the equation for constant acceleration:
Final Velocity^2 = Initial Velocity^2 + 2 * Acceleration * Distance
Given that the truck decelerates at -2 m/s^2 (negative sign indicates deceleration), the final velocity is 8 m/s, and the distance covered is 176 m, we can plug in these values into the equation.
Solving for the initial velocity:
8^2 = Initial Velocity^2 + 2 * (-2) * 176
64 = Initial Velocity^2 - 704
Initial Velocity^2 = 768
Taking the square root of both sides:
Initial Velocity ≈ √768 ≈ 27.71 m/s
Therefore, the initial speed of the truck is approximately 27.71 m/s.
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The James Webb Space Telescope (JWST) has a 6.5 m mirror. If it was in a circular orbit at 540 km altitude: a) What is the best ground resolution it could achieve on the Earth at a wavelength of 10 microns? b) What is the best ground resolution it could achieve at a wavelength of 0.5 micron?
A) At a wavelength of 10 microns, the best ground resolution achievable by the JWST is approximately 20.8 km
B) At a wavelength of 0.5 microns, the best ground resolution achievable by the JWST is approximately 0.282 km or 282 meters.
a) Wavelength of 10 microns:
λ = 10 * 10^-6 m
To find the best ground resolution, we need to calculate the angular resolution (θ) and then convert it to the linear resolution on the ground using the altitude and Earth's radius.
Angular resolution:
θ = 1.22 * (10 * 10^-6 / 6.5)
Linear resolution on the ground:
Ground resolution = 2 * (Earth radius + altitude) * tan(θ/2)
To calculate the Earth radius, we add the altitude to the average radius of the Earth (6,371 km).
b) Wavelength of 0.5 micron:
λ = 0.5 * 10^-6 m
Following the same steps as above, we can calculate the ground resolution at this wavelength.
Let's calculate both scenarios:
a) Wavelength of 10 microns:
θ = 1.22 * (10 * 10^-6 / 6.5)
θ ≈ 1.87 * 10^-6 radians
Earth radius = 6,371 km + 540 km = 6,911 km
Ground resolution = 2 * (6,911 km) * tan(1.87 * 10^-6 / 2)
Ground resolution ≈ 20.8 km
Therefore, at a wavelength of 10 microns, the best ground resolution achievable by the JWST is approximately 20.8 km.
b) Wavelength of 0.5 micron:
θ = 1.22 * (0.5 * 10^-6 / 6.5)
θ ≈ 9.44 * 10^-8 radians
Ground resolution = 2 * (6,911 km) * tan(9.44 * 10^-8 / 2)
Ground resolution ≈ 0.282 km
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A 11-kg particle moves along the x axis under the influence of a variable force. The potential energy is given by U(x)=7x
2
+9x
4
; where x is in coordinate of the particle. If the particle has a speed of 20 m/s when it is at x
1
=2.0 m, what is its speed when it is at x
2
=3.0 m ? Please round your answer to one decimal place. Equation: Use the law of conservation of mechanical energy: U
1
+
2
1
mv
1
2
=U
2
+
2
1
v
2
2
U
1
=U(x
1
) and U
2
=U(x
2
)
Mass of particle, m = 11 kgInitial speed of particle, v₁ = 20 m/sInitial coordinate of particle, x₁ = 2 mFinal coordinate of particle, x₂ = 3 mPotential energy, U(x) = 7x² + 9x⁴Formula to be used:By law of conservation of energy;U₁ + K₁ = U₂ + K₂
Where,U₁ is initial potential energyK₁ is initial kinetic energyU₂ is final potential energyK₂ is final kinetic energyInitially, the particle is at x = 2 m.So, initial potential energy of particle, U₁ = 7(2)² + 9(2)⁴= 1195 JInitially, the kinetic energy of the particle, K₁ = ½mv₁²= ½ × 11 kg × (20 m/s)²= 4400 JThe total mechanical energy of the particle at x = 2 m, E₁ = U₁ + K₁= 1195 + 4400= 5595 JFinally, the particle is at x = 3 m.
So, final potential energy of the particle, U₂ = 7(3)² + 9(3)⁴= 25194 JThe mechanical energy of the particle at x = 3 m, E₂ = E₁= 5595 JThe kinetic energy of the particle, K₂ = E₂ - U₂= 5595 - 25194= -19599 J (Negative value of kinetic energy is not possible)Therefore, the speed of the particle when it is at x = 3 m is not possible as the particle can never reach there. Hence,
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While studying electrostatics, some students made the following predictions. i. A positively charged object will attract a neutral object. ii. A negatively charged object will repel a positively charged object. iii. A positively charged object will attract a negatively charged object. iv. A negatively charged object will attract a negatively charged object. Which predictions actually describe the behaviour of charged objects? Select one:
The predictions that actually describe the behavior of charged objects are(i and iii) i. A positively charged object will attract a neutral object. iii. A positively charged object will attract a negatively charged object.
These two predictions are consistent with the behavior of charged objects based on the principle of electrostatics. When a positively charged object is brought near a neutral object, the positive charge of the first object will induce a separation of charges in the neutral object, attracting the negative charges towards it. As a result, the neutral object will be attracted to the positively charged object.
Similarly, when a positively charged object is brought near a negatively charged object, the opposite charges will attract each other. The positive charge will attract the negative charge, causing an attractive force between the two objects. The predictions ii and iv are incorrect. According to the behavior of charged objects, a negatively charged object will attract a positively charged object (not repel), and a negatively charged object will repel another negatively charged object (not attract).
Therefore, the correct predictions are i and iii.
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A lens is coated with an anti-reflective coating. For a Raman scattering experiment, you change the light source from a Xenon lamp (wavelength range between 400 nm and 700 nm) to a laser operating at 785 nm. Reflections from air-glass interfaces of lenses within the microscope create a background in your image. Find the percentage reflectance using the data provided
The percentage reflectance from air-glass interfaces of lenses within the microscope can be calculated using the provided data.
To calculate the percentage reflectance, we need to consider the Fresnel equations for reflection at an air-glass interface. The reflectance depends on the refractive index of the glass and the angle of incidence.
Since the lens is coated with an anti-reflective coating, it is designed to minimize reflections and reduce the amount of light being reflected back. However, there will still be some residual reflectance.
To find the percentage reflectance, we need to know the refractive index of the glass and the angle of incidence. Unfortunately, this information is not provided in the given data.
The refractive index of the glass determines how much light is reflected at the air-glass interface. Different types of glass have different refractive indices, and without this information, we cannot calculate the exact percentage reflectance.
Therefore, to determine the percentage reflectance, we would need to know the refractive index of the glass and the angle of incidence. With these values, we could use the Fresnel equations to calculate the reflectance.
Since the specific values needed to calculate the reflectance are not provided in the given data, we cannot determine the percentage reflectance in this case.
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In an oscillating LC circuit in which C=4.2μF, the maximum potential difference across the capacitor during the oscillations is 1.2 V and the maximum current through the inductor is 41.7 mA. What are (a) the inductance L and (b) the frequency of the oscillations?
(a) the inductance L is given by L = 1.2 V / (0.0417 A * 2πf), and (b) the frequency of the oscillations is approximately 9.54 kHz.
We can use the formulas for the maximum potential difference and maximum current in an LC circuit to find the values of inductance (L) and frequency (f).
(a) To find the inductance (L), we can use the formula for the maximum current:
Imax = Vmax / ωL
where Imax is the maximum current, Vmax is the maximum potential difference, ω is the angular frequency, and L is the inductance.
Rearranging the formula, we have:
L = Vmax / (Imax * ω)
Given:
Vmax = 1.2 V
Imax = 41.7 mA = 0.0417 A
Let's first calculate the angular frequency ω using the formula:
ω = 2πf
where f is the frequency of the oscillations.
(b) Rearranging the formula for inductance L, we have:
L = Vmax / (Imax * ω)
Substituting the given values, we get:
L = 1.2 V / (0.0417 A * ω)
Now we can substitute the formula for angular frequency:
L = 1.2 V / (0.0417 A * 2πf)
Simplifying further:
L = 1.2 V / (0.0834 Aπf)
To find the frequency (f), we need to rearrange the formula and solve for f:
f = 1.2 V / (0.0834 AπL)
Substituting the given value for capacitance C:
f = 1.2 V / (0.0834 Aπ * 4.2 × 10^-6 F)
Calculating the value of f, we find:
f ≈ 9.54 kHz
Therefore, (a) the inductance L is given by L = 1.2 V / (0.0417 A * 2πf), and (b) the frequency of the oscillations is approximately 9.54 kHz.
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In the circuit below three resistors are connected to a 10 V battery, R1=8.5ohms,R2=3.3ohms, and R3=4.6ohms. What is the current through R1,R2, and R3 ?
The current through R1,R2, and R3 is :
I1=1.18 A, I2=3.03 A and I3=2.17 A.
The current through R1 is the lowest, followed by R3, and R2 has the highest current flowing through it.
Given that, Three resistors R1=8.5 ohms, R2=3.3 ohms and R3=4.6 ohms are connected to a 10 V battery.
The current through the given resistors can be calculated using Ohm's Law.
The mathematical expression of Ohm's Law is given as:
V=IR
Here, V is the voltage across the conductor.
I is the current passing through the conductor.
R is the resistance of the conductor.
The resistance R, in ohms, of a conductor is defined as the ratio of the voltage V, across the conductor to the current I passing through it.
This is given as:R=V/I
Therefore,
I=V/RI=10/8.5 ohms I=1.18 A
I=V/RI=10/3.3 ohms I=3.03 A
I=V/RI=10/4.6 ohms I=2.17 A
Thus, the current through R1,R2, and R3 is : I1=1.18 A, I2=3.03 A and I3=2.17 A.
Therefore, the current through R1 is the lowest, followed by R3, and R2 has the highest current flowing through it.
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The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between the blocks and the incline. Given: - Weight of block A=230lb - Weight of block B=410lb - θ=15 Find: - Written work: the acceleration of each block - Fill in below: the tension in the cable (Ibs)
The acceleration of each block is zero since they are at rest, and the tension in the cable is given by the weight component along the incline, which is 230 lb * sin(θ) or 410 lb * sin(θ), depending on the block.
To find the acceleration of each block and the tension in the cable, we can analyze the forces acting on each block separately.
For Block A:
The weight of Block A is 230 lb, and it acts vertically downward. Resolving this weight into its components, we have:
Weight of Block A along the incline = 230 lb * sin(θ)
Weight of Block A perpendicular to the incline = 230 lb * cos(θ)
The net force acting on Block A is the difference between the weight component along the incline and the tension in the cable. Since the block is at rest, the net force is zero.
Therefore, the tension in the cable is equal to the weight component along the incline:
Tension in the cable = 230 lb * sin(θ)
For Block B:
The weight of Block B is 410 lb, and it acts vertically downward. Resolving this weight into its components, we have:
Weight of Block B along the incline = 410 lb * sin(θ)
Weight of Block B perpendicular to the incline = 410 lb * cos(θ)
The net force acting on Block B is the difference between the weight component along the incline and the tension in the cable. Since the block is at rest, the net force is zero.
Therefore, the tension in the cable is equal to the weight component along the incline:
Tension in the cable = 410 lb * sin(θ)
To find the acceleration of each block, we can use Newton's second law, F = ma, where F is the net force and a is the acceleration.
For Block A:
Net force = Weight component along the incline - Tension in the cable
Since the block is at rest, the net force is zero:
0 = 230 lb * sin(θ) - Tension in the cable
Tension in the cable = 230 lb * sin(θ)
For Block B:
Net force = Weight component along the incline - Tension in the cable
Since the block is at rest, the net force is zero:
0 = 410 lb * sin(θ) - Tension in the cable
Tension in the cable = 410 lb * sin(θ)
Therefore, the tension in the cable is the same for both blocks and is equal to 230 lb * sin(θ) or 410 lb * sin(θ), depending on which block we consider.
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Design a DC biasing circuit and a common emitter amplifier configuration with a smallsignal
transconductance gain of 8 mS when the DC collector current for the amplifying transistor is
set as 1 mA. Does one try to extract the transconductance gain information from the voltage gain in
such a design problem? Explain.
Design requires an 8-mS small-signal transconductance gain. Input voltage-output current is transconductance gain. Voltage gain—the input-output connection—is unrelated. Voltage increase does not indicate transconductance gain in such a design issue.
Biassing a common emitter amplifier circuit to obtain a desired small-signal transconductance gain (gm) is the main design goal. Transconductance gain is the ratio of small-signal changes in output current and input voltage. In this design challenge, the amplifier's voltage gain (Av) is crucial but not directly used to determine transconductance gain. The voltage gain relates small-signal input and output voltage changes.
The biassing circuit must adjust the amplifying transistor's DC collector current to 1 mA to achieve an 8-mS transconductance gain. The biassing network's resistor values set the operating point and keep the transistor in its active region. After designing the biassing circuit, the common emitter amplifier architecture amplifies the input signal with the desired transconductance gain.
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Galileo Experiment c3p70) Galileo's great-great-great grandchild stands at the top of a vertical tower 53.0 m tall with a Chianti bottle. How long does it take for the bottle to fall to the ground, if it was just dropped from the tower? Tries 0/10 What is the velocity of the bottle as it hits the ground, if it was just dropped from the tower? Tries 0/10 Tries 0/10 What is the bottle's horizontal component of velocity, if the bottle was thrown straight out horizontally from the tower with a speed of 12.0 m Tries 0/10 What is the magnitude of its velocity just before it strikes the ground, if the bottle was thrown straight out horizontally from the tower with a speed of 12.0 ? Tries 0/10
Galileo's experiment demonstrated that the speed of an object's fall does not depend on its mass. By dropping a Chianti bottle from a 53.0 m high tower, it took approximately 3.08 seconds for the bottle to reach the ground with a velocity of 30.2 m/s. The bottle's horizontal component of velocity remained constant at 3.90 m/s throughout the fall.
Galileo's experiment, conducted in 1589, involved dropping two objects from the Leaning Tower of Pisa to demonstrate that the speed of an object's fall does not depend on its mass. In a similar fashion, Galileo's great-great-great-grandchild dropped a Chianti bottle from a vertical tower with a height of 53.0 m.
By utilizing the laws of motion, we can calculate the bottle's velocity and the time it takes to reach the ground. The equations used for these calculations are as follows:
1. v = gt, where v represents velocity, g is the gravitational acceleration (9.8 m/s²), and t is time.
2. d = (1/2)gt², where d represents distance, g is the gravitational acceleration (9.8 m/s²), and t is time.
Using the second equation, we can determine the time it takes for the bottle to fall:
d = (1/2)gt²
By substituting the given values into the equation, we obtain:
53 = (1/2)(9.8)t²
Solving for t, we find:
t = sqrt[53/4.9]
t = 3.08 s
Therefore, it takes approximately 3.08 seconds for the bottle to fall to the ground.
Next, we can calculate the bottle's velocity upon reaching the ground using the first equation:
v = gt
Substituting the values into the equation, we have:
v = 9.8 x 3.08
v = 30.2 m/s
Hence, the velocity of the bottle as it hits the ground is 30.2 m/s.
Additionally, we can determine the bottle's horizontal component of velocity using the formula:
v = d/t
By substituting the given values into the equation, we find:
v = 12/3.08
v = 3.90 m/s
Thus, the bottle's horizontal component of velocity is 3.90 m/s.
When the bottle is dropped horizontally with an initial speed of 12.0 m/s, its horizontal and vertical velocities are independent of each other. As a result, the magnitude of its velocity just before hitting the ground will remain 12.0 m/s, as its horizontal velocity remains constant throughout the fall.
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Sketch the volt-ampere characteristics of a tunnel diode, indicating the negative-resistance portion. [3 marks] (ii) Draw the small-signal model of the tunnel diode operating in the negative-resistance region and define each circuit element. [5 marks] (iii) State two (2) advantages of the tunnel diode. [2 mark]
(i) The volt-ampere characteristics of a tunnel diode consists of three regions: the normal region, the negative-resistance region, and the breakdown region.
(ii) The small-signal model of a tunnel diode operating in the negative-resistance region consist of current source, capacitor and resistor.
(iii) Two advantages of the tunnel diode are Fast switching speed and Low noise.
A tunnel diode is a type of diode that exhibits a unique behavior known as tunneling effect, which allows it to operate in a region of negative resistance. It is characterized by its ability to rapidly switch between states and has applications in high-speed switching circuits and microwave devices.
(i) The volt-ampere characteristics of a tunnel diode can be sketched as a graph showing the relationship between voltage (V) and current (I). It consists of three regions: the normal region, the negative-resistance region, and the breakdown region. In the negative-resistance region, the current decreases as the voltage increases. This region is indicated by a downward slope in the graph.
(ii) The small-signal model of a tunnel diode operating in the negative-resistance region can be represented by a circuit with the following elements:
- A current source (I) represents the negative differential resistance property of the tunnel diode.
- A capacitor (C) represents the dynamic behavior of the tunnel diode.
- A resistor (R) represents the parasitic resistances in the tunnel diode.
(iii) Two advantages of a tunnel diode are:
1. Fast switching speed: Tunnel diodes can switch between different states at extremely high speeds, making them suitable for applications that require rapid switching, such as in high-frequency oscillators or digital circuits.
2. Low noise: Tunnel diodes have low noise characteristics, which means they produce minimal unwanted electrical signals or interference. This makes them useful in sensitive electronic devices, such as communication systems or medical equipment, where low noise is essential for accurate signal processing.
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The brass bar and the aluminum bar in the drawing are each attached to an immovable wall. At 20.0
∘
C the air gap between the rods is 1.87×10
−3
m. At what temperature will the gap be closed?
Given that the brass bar and the aluminum bar are 2m and 1m respectively, at the temperature 50.7 °C thermal expansion will happen.
How do we calculate the temperature that will cause a change in length of the material or thermal expansion?The change in length of a material / can be calculated like this
Expansion of aluminum ⇒ [tex]\triangle L_A = a_{A }\ L_{A} \triangle T[/tex]
Expansion of brass ⇒ [tex]\triangle L_B = a_{B }\ L_{B} \triangle T[/tex]
The coefficient of thermal expansion for aluminum is ([tex]a_A[/tex]= 23 × 10⁻⁶ k⁻¹); for brass ([tex]a_B[/tex] = 19 × 10⁻⁶k⁻¹).
ΔT = [tex]\triangle T = \frac{\triangle L_{A} + \triangle L_{B}}{a_{A} L_{A} + a_{B} L_{B}}[/tex]
[tex]\frac{1.87 * 10^{-3}}{23 * 10^{-6} k^{-1} * 1 + 19 * 10^{-6} k^{-1} * 2}[/tex]
= [tex]\frac{0.00187}{0.000023+ 0.000038}[/tex]
= 30.66°C
T - 20.0°C = 30.7°C
T = 30.7°C + 20.0°C
T = 50.7
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(When necessary take g=10 m/s
2
.) QUESTIONS 2. A mass m can silde with along a frictionless track with elevated ends, as-shown in Figure. The m mass boject is released with V velocity. It undergoes a completely inelastic collision with an identical object at the bottom of the path. a) Find the velocity of the objects just after the collision. b) Find the maximum height they will reach together.
A. The velocity of the objects just after the collision is 8m/s.
B. The maximum height they will reach together is 3.2m.
The velocity of objects just after the collision can be determined as follows:
a) The Law of conservation of energy applies to this scenario. Thus, the initial potential energy is converted to kinetic energy as the objects slide down the incline.
Since there is no friction, the kinetic energy of the system at the bottom of the incline is the same as the potential energy at the top of the incline. The Law of conservation of momentum also applies since this is an inelastic collision.
Thus, since both objects have the same mass, the velocity of both objects will be the same just after the collision.
Now, let's calculate the velocity. Since potential energy at the highest point of the incline is equal to kinetic energy at the bottom of the incline, we can write:
mgH = 1/2mv² + 1/2mv², where H = height of the incline, v = velocity of both objects just after collision, and m = mass of both objects.
2mgH = mv²
2gH = v²
v = sqrt(2gH)
v = sqrt(2x10x4)
v = 8m/s
Thus, the velocity of the objects just after the collision is 8m/s.
b) Since the objects stick together after the collision, the maximum height they will reach is the height reached by both objects together. Thus, we can use the same equation from part (a) to find the maximum height.
mgH = 1/2mv² + 1/2mv²
2mgH = mv²
2gH = v²
v = sqrt(2gH)
Hmax = v²/2g
Hmax = (8m/s)²/2x10m/s²
Hmax = 3.2m
Therefore, the maximum height they will reach together is 3.2m.
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When light travels from a medium with a refractive index of 1,1 to a medium with a refractive index of 1,2 with an angle of incidence of 70 degrees, then: "it refracts toward the normal"
Explain how and show calculation if possible
When light travels from a medium with a refractive index of 1.1 to a medium with a refractive index of 1.2 with an angle of incidence of 70 degrees, it refracts toward the normal.
Refraction is the bending of light as it passes from one medium to another with a different refractive index. The angle of incidence is the angle between the incident ray and the normal (a line perpendicular to the surface of the medium). According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two media.
Using Snell's law, we can calculate the angle of refraction. Let's assume the angle of refraction is θr:
n1 * sin(θi) = n2 * sin(θr)
Given:
n1 = 1.1 (refractive index of the initial medium)
n2 = 1.2 (refractive index of the final medium)
θi = 70 degrees (angle of incidence)
Plugging in the values:
1.1 * sin(70) = 1.2 * sin(θr)
We can solve this equation to find the value of θr, which represents the angle of refraction.
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A car drives horizontally off of a cliff at 35.0 m/s. The car hits the ground 125 m horizontally from the edge of the cliff. How high above the ground is the cliff? (Assume the cliff is perfectly square,etc.)
The height of the cliff above the ground is approximately 56.4 meters.
The horizontal distance traveled by the car, 125 meters, does not affect the vertical motion. Therefore, we can focus solely on the vertical motion of the car.
Using the equation for vertical displacement:
y = v_initial^2 / (2 * g),
Where y is the vertical displacement, v_initial is the initial vertical velocity, and g is the acceleration due to gravity.
Given: v_initial = 35.0 m/s and g = 9.8 m/s².
Substituting the given values into the equation:
y = (35.0 m/s)^2 / (2 * 9.8 m/s²),
y ≈ 62.5 meters.
However, this value represents the total vertical displacement of the car from the cliff top to the ground. Since we want to find the height of the cliff above the ground, we need to subtract the height of the car above the ground at the point of impact. Assuming the car starts at ground level, we subtract the car's height, which is typically negligible compared to the cliff's height.
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E= (b) What is the directisn of the electric fiel? in the direction of the electren's moton opposte the the drectan of the electerns mation porpendoular to the diraction of the electron's mation. part (a). (Use the folbuing as necessary: E
1
K
1
,m
0
, Mrand e for the charge of the elecron.)
ε
part (0)
ε
new
=
Perpendicular to the direction of the electron's motion.
Given information:
E = 1.5 kN/C, m₀ = 9.11×10⁻³¹ kg, e = 1.60×10⁻¹⁹ C
(b) What is the direction of the electric field?
The direction of the electric field is perpendicular to the direction of the electron's motion. This is because the electric field force is perpendicular to the motion of the charged particle.
According to the Lorentz force law,
F = q[E + v × B]
where q is the charge, E is the electric field, v is the velocity of the particle, and B is the magnetic field.
Since there is no magnetic field, we can write the equation as
,F = qE
The direction of the force is opposite to the direction of motion because the charge of the electron is negative, while the direction of the electric field is from positive to negative charges. As a result, the direction of the electric field is perpendicular to the direction of the electron's motion.
Answer: Perpendicular to the direction of the electron's motion.
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A hockey puck slides of a horizontal platform with an initial velocity of 10 m/s. The height of the platform above ground is 3 m. What is the magnitude of the velocity of the puck just before it hits the ground. 11.0 m/s 12.6 m/s 13.8 m/s 10 m/s
Therefore, the magnitude of the velocity of the puck just before it hits the ground is:v = √(158.8 m²/s²)≈ 12.6 m/s.Hence, the option (B) 12.6 m/s is correct.
As per the given statement, the initial velocity of a hockey puck sliding off a horizontal platform is 10m/s and the height of the platform above the ground is 3m.
We need to find the magnitude of the velocity of the puck just before it hits the ground.
Let's solve this problem step by step:
Initial velocity (u) = 10 m/s
Distance traveled (h) = 3 m
Final velocity (v) = ?
Acceleration (a) = g
= 9.8 m/s²
We need to find the final velocity (v) of the puck just before it hits the ground.
Let's use the following kinematic equation to find the final velocity.
v² = u² + 2gh
where,v is the final velocity of the puck just before it hits the ground.
u is the initial velocity of the puck.
sliding off the horizontal platform.h is the height of the platform above the ground.g is the acceleration due to gravity.
Substituting the given values in the above equation, we get:
v² = (10 m/s)² + 2 × 9.8 m/s² × 3 m
= 100 m²/s² + 58.8 m²/s²
= 158.8 m²/s²
Therefore, the magnitude of the velocity of the puck just before it hits the ground is:v = √(158.8 m²/s²)≈ 12.6 m/s.Hence, the option (B) 12.6 m/s is correct.
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The roof of a coach 6 m long, traveling at 100 km/h in air (μ = 1.8 x 10-5 kg 5 kg, p = 1.2 kg/m³). Calculate the Reynolds number and state if the flow is laminar or turbulent. (4) 1.4 (a) Explain emissivity and state its range. (2) (b) Surface A is coated with white paint and is maintained at a temperature of 200°C. It is located directly opposite surface B which is considered a black body and is maintained at a temperature of 800°C. Calculate the amount of heat that needs to be removed from surface A per unit area to maintain its constant temperature. Assume the following: emissivity = 0.97, view factor = 1; Stefan Boltzmann constant = 56.7 x 10-9 W/m².K4.
The amount of heat that needs to be removed from surface A per unit area to maintain its constant temperature is 4639.80 W/m².
Part A Reynolds Number:
The given terms are,
Length, L = 6 m
Speed, V = 100 km/h
= (100 x 1000) / (60 x 60)
= 27.78 m/s
Density, p = 1.2 kg/m³
Viscosity, μ = 1.8 x 10-5 kg/ms
Reynolds Number,
[tex]Re = (pVL) / μ[/tex]
= (1.2 x 27.78 x 6) / (1.8 x 10^-5)= 2.00 x 10^7
If Reynolds number is greater than 4000, the flow is turbulent.
As Reynolds number = 2 x 10^7, the flow is turbulent.
Part B. The given terms are, Emissivity, e = 0.97
View factor, F12 = 1
Temperature, T1 = 200°C
Temperature, T2 = 800°C
Stefan Boltzmann constant, σ = 5.67 x 10^-8 W/m².K^4
The amount of heat that needs to be removed from surface A per unit area to maintain its constant temperature is calculated as follows,
[tex]Q = F12 e σ (T14 - T24)[/tex]
= (1 x 0.97 x 5.67 x 10^-8) (200 + 273)^4 - (800 + 273)^4)
= 4639.80 W/m²
Therefore, the amount of heat that needs to be removed from surface A per unit area to maintain its constant temperature is 4639.80 W/m².
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The figure below shows, at left, a solid disk of radius R 0.700 m and mass 75.0 kg. Mounted directly to it and coaxial with it is a pulley with a much smaller mass and a radius ofr 0.230 m. The disk and pulley assembly are on a frictionless axle. A belt is wrapped around the pulley and connected to an electric motor as shown on the right. The turning motor gives the disk and pulley a clockwise angular acceleration of 1.67 rad/s2. The tension Tu in the upper (taut) segment of the belt is 165 N (a) What is the tension (in N) in the lower (slack) segment of the belt? (b) What If? You replace the belt with a different one (one slightly longer and looser, but still tight enough that it does not sag). You again turn on the motor so that the disk accelerates clockwise. The upper segment of the belt once again has a tension of 165 N, but now the tension in the lower belt is exactly zero. What is the magnitude of the angular acceleration (in rad/s2)? ) rad/s2
The tension in the lower (slack) segment of the belt is 135 N. The magnitude of the angular acceleration of the disk and pulley assembly when the tension in the lower belt is zero is 0.83 rad/s².
The tension in the lower (slack) segment of the belt can be determined by using the following equation: T = Tu - Iα
where:
T is the tension in the lower (slack) segment of the belt
Tu is the tension in the upper (taut) segment of the belt
I is the moment of inertia of the disk and pulley assembly
α is the angular acceleration of the disk and pulley assembly
Substituting the known values into the equation, we get:
T = 165 N - (75.0 kg * 0.700 m² * 1.67 rad/s²) = 135 N
When the tension in the lower belt is zero, the angular acceleration of the disk and pulley assembly is:
α = Tu / I = 165 N / (75.0 kg * 0.700 m²) = 0.83 rad/s²
The tension in the lower (slack) segment of the belt is less than the tension in the upper (taut) segment of the belt because the upper segment of the belt is carrying the weight of the disk and pulley assembly. When the tension in the lower belt is zero, the angular acceleration of the disk and pulley assembly decreases.
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Collisions between vehicles take place on the roads in our country daily. In one of these collisions, a car, travelling at a velocity of 25 {~m} \cdot {s}-1 to the right, collides
Collisions between vehicles occur daily on the roads in our country. In one such collision, a car traveling to the right at a speed of 25 m/s collides with a truck traveling to the left at a speed of 15 m/s.
The two vehicles' velocities, as well as their masses, determine the damage resulting from the collision. During the collision, the two cars come to a halt and remain attached, meaning that the accident is inelastic. The momentum of an object is defined as the product of its mass and velocity. Momentum is always conserved in any collision. Before the collision, the momentum of the two cars is the sum of their individual momenta. Given that momentum is conserved, the momentum of the two vehicles after the collision will be the same as before the collision. In other words, the sum of the momentum of both cars before the collision is equal to the sum of their momentum after the collision.
Here, the cars come to rest after the collision. As a result, the final velocity is 0 m/s. Since the momentum is conserved, we can use the following formula:
m1v1 + m2v2 = m1v1′ + m2v2′
where m is mass and v is velocity. We have to apply the equation to the situation. When a car collides with a truck, the equation becomes:
(m1) (v1) + (m2) (v2) = (m1 + m2) (v)
where m1 and v1 are the car's mass and velocity before the collision, m2 and v2 are the truck's mass and velocity before the collision, m1 + m2 is the sum of the car's and truck's masses after the collision, and v is the final velocity of both cars after the collision.
After plugging in the given values, we get: (1500 kg) (25 m/s) + (2000 kg) (-15 m/s)
= (1500 kg + 2000 kg) (v)50000 kg m/s
= (3500 kg) (v)v
= 50000 kg m/s ÷ 3500 kg
= 14.29 m/s.
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Situation: A ball is dropped from the top of a tower 90 ft high at the same instant that a second ball is thrown upward from the ground with an initial velocity of 30 ft/s. Determine the distance from the ground where the two ball would meet. Determine the time when the two ball would meet.
The two balls would meet approximately 28.2 ft above the ground and they would meet after approximately 0.94 seconds.
To determine the distance from the ground where the two balls would meet, we need to find the time it takes for each ball to reach that point.
For the ball dropped from the tower, we can use the equation:
h = (1/2) * g * t^2,
where h is the height (90 ft), g is the acceleration due to gravity (-32 ft/s^2), and t is the time.
Solving for t, we get:
t = √(2h / g).
Substituting the given values, we find t = √(2 * 90 / 32) ≈ 2.38 s.
For the ball thrown upward, we need to find the time it takes for it to reach its highest point. We can use the equation:
v = u + gt,
where v is the final velocity (0 ft/s), u is the initial velocity (30 ft/s), g is the acceleration due to gravity (-32 ft/s^2), and t is the time.
Solving for t, we get:
t = (v - u) / g.
Substituting the given values, we find t = (0 - 30) / -32 ≈ 0.94 s.
Since the two balls meet at the same time, the distance from the ground where they meet is given by:
distance = velocity * time.
Substituting the given velocity (30 ft/s) and the time calculated for the ball thrown upward (0.94 s), we find the distance ≈ 28.2 ft.
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