PHYSICS COLLEGE

A disgruntled autoworker pushes a small foreign import offacliff with a height of h. the vehicle lands a distance away
fromthe cliff. Determine how fast the vehicle was pushed off
thecliff.

Answers

Answer 1
Answer:

Answer:

v = a/√(2h/g) m/s

Explanation:

Lets say the distance away from the cliff is a.

then, a = v t

where v is velocity with which it was thrown and t is time taken to fall.

Using equations of motion, we can also say that

h=1/2gt^2

where h is the height of the cliff

Thus, t^2 = 2h/g and t = √(2h/g)

Thus, v = a/√(2h/g).

the vehicle was pushed off  the cliff with the velocity , v = a/√(2h/g). m/s

Related Questions

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg), who is sitting at the very bottom end, and whom he holds onto when he arrives. Laughing, John & William leave the slide horizontally and land in the muddy ground near the foot of the slide. (A) If John starts out 1.8 m above William, and the slide is essentially frictionless, how fast are they going when they leave the slide? (B) Thanks to the mud he acquired, John will now experience an average frictional force of 105 N as he slides down. How much slower is he going when he reaches the bottom than when friction was absent?
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As a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 2.43 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 24 m/s. One horsepower equals 746 W.

Answers

Answer:

F=75.53N

Explanation:

To calculate the power we define the equation,

P=Fv

Where,

F= Force

V= Velocity,

Here we have that 2.43hp is equal to 1812.78W,

clearing F,

F=(P)/(v) = (1812)/(24)\nF=75.53N

The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.1. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe?
2. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now:______.
a. the same as before.
b. lower than before.
c. higher than before.
3. If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound that can be produced in the pipe?
4. What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?
4-1. Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present.
A. Only the odd multiples of the fundamental frequency.
B. Only the even multiples of the fundamental frequency.
C. All integer multiples of the fundamental frequency.
E. What length of open-closed pipe would you need to achieve the same fundamental frequency as the open pipe discussed in Part A?
A. Half the length of the open-open pipe.
B. Twice the length of the open-open pipe.
C. One-fourth the length of the open-open pipe.
D. Four times the length of the open-open pipe.
E. The same as the length of the open-open pipe.
F. What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?
F-1. Recall that possible frequencies of standing waves that can be generated in an open-closed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the third
harmonic.

Answers

Final answer:

The physics of wind instruments is based on standing waves. The lowest frequency of a sound wave produced in an open-open pipe can be calculated. When a hole is drilled through the pipe, the fundamental frequency is lower than before. Only odd multiples of the fundamental frequency will be present in a pipe with a hole halfway down its length. An open-closed pipe needs to be twice the length of an open-open pipe to achieve the same fundamental frequency. The first possible harmonic after the fundamental frequency in an open-closed pipe is the third harmonic.

Explanation:

The lowest frequency f of the sound wave produced when blowing into an open-open pipe can be calculated using the formula f = v/2L, where v is the speed of sound and L is the length of the pipe. Plugging in the values, we get f = 343/(2*0.8), which equals 214.375 Hz.

When a hole is drilled through the side of the pipe, the fundamental frequency of the sound wave generated in the pipe is lower than before. This is because the effective length of the pipe has been changed, resulting in a lower frequency.

The fundamental frequency of the sound that can be produced in the original pipe with a hole drilled halfway down its length can be calculated as f = v/L, where L is the new effective length of the pipe. Since the hole is halfway down, the effective length becomes half of the original length, resulting in a frequency equal to the original fundamental frequency.

When blowing air into the pipe with a hole halfway down its length, only the odd multiples of the fundamental frequency will be present. Therefore, the frequencies that can be created are only the odd multiples of the fundamental frequency.

The length of an open-closed pipe needed to achieve the same fundamental frequency as an open-open pipe is twice the length of the open-open pipe. This is because an open-closed pipe has only odd harmonics, which are spaced twice as far apart as the harmonics in an open-open pipe.

The first possible harmonic after the fundamental frequency in an open-closed pipe is the third harmonic.

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A virtual image produced by a lens is always

Answers

The correct answer is:

"located in front of lens"

just took PF test and this was right answer

If a coil stays at rest in a very large static magnetic field, no emf is induced in this coil. Group of answer choices True False

Answers

Answer:

True

Explanation:

  • Faraday's Law says that there is a emf induced in a conductor when the vector flux of the magnetic field across it changes in time.
  • This can be true due to one of two facts, either the magnitude of the magnetic field changes in time, or the area through which the flux occurs changes due to the movement of the object.
  • In this case, due to the magnetic field is constant, and the coil stays at rest, there is no possible change in flux, so emf induced is zero.

Madelin fires a bullet horizontally. The rifle is 1.4 meters above the ground. The bullet travels 168 meters horizont before it hits the ground. What speed did Madelin's bullet have when it exited the rifle?

Answers

The position vector of the bullet has components

x=v_0t

y=1.4\,\mathrm m-\frac g2t^2

The bullet hits the ground when y=0, which corresponds to time t:

1.4\,\mathrm m-\frac g2t^2=0\implies t=0.53\,\mathrm s

The bullet travels 168 m horizontally, which would require a muzzle velocity v_0 such that

168\,\mathrm m=v_0(0.53\,\mathrm s)

\implies v_0\approx320\,(\mathrm m)/(\mathrm s)

Final answer:

In the given physics problem, the bullet travels horizontally 168 meters before hitting the ground from a height of 1.4 meters. By calculating the time it takes for the bullet to fall to the ground due to gravity and then applying that time to the horizontal distance traveled, we find that the speed of the bullet when it exited the rifle was approximately 313.43 m/s.

Explanation:

The scenario defined is a classic Physics problem where an object is fired horizontally and falls to the ground due to gravity. We can calculate the horizontal speed of the bullet using the equations of motion associated with the vertical, free-fall motion of the bullet.

Gravity causes the bullet to fall to the ground. As we know that the height from the ground is 1.4 meters, we can calculate the time taken for the bullet to hit the ground using the equation: time = sqrt(2 * height / g), where g is the gravitational constant (approx. 9.8 m/s^2).

Substituting the given value, we get time = sqrt(2 * 1.4 / 9.8), which is around 0.536 seconds. The bullet travels 168 meters in this time horizontally, therefore its horizontal speed will be distance / time, which is 168 meters / 0.536 seconds = 313.43 m/s. So, Madelin's bullet had a speed of around 313.43 m/s when it exited the rifle.

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The magnetic flux that passes through one turn of a 11-turn coil of wire changes to 5.60 from 9.69 Wb in a time of 0.0657 s. The average induced current in the coil is 297 A. What is the resistance of the wire?

Answers

Answer:

2.31 Ω

Explanation:

According to the Faraday's law of electromagnetic induction,

Induced emf = - N (dΦ/dt)

Emf = -N (ΔΦ/t)

where N = number of turns = 11

Φ = magnetic flux

ΔΦ = change in magnetic flux = 9.69 - 5.60 = 4.09 Wb

t = time taken for the change = 0.0657 s

Emf = 11(4.09/0.0657)

Emf = - 684.78 V (the minus sign indicates that the direction of the induced emf is opposite to the direction of change of magnetic flux)

From Ohm's law,

Emf = IR

R = (Emf)/I

I = current = 297 A

R = (684.78)/297

R = 2.31 Ω

Hope this Helps!!

Explanation:

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