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timama [110]
3 years ago
7

A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work

done on the object? A. 0 joules B. 5.5 joules C. -5.5 joules D. 5.5 × 102 joules E. -5.5 × 102 joules Reset Next
Physics
1 answer:
kifflom [539]3 years ago
3 0

Answer: Option D: 5.5×10²Joules

Explanation:

Work done is the product of applied force and displacement of the object in the direction of force.

W = F.s = F s cosθ

It is given that the force applied is, F = 55 N

The displacement in the direction of force, s = 10 m

The angle between force and displacement, θ = 0°

Thus, work done on the object:

W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J

Hence, the correct option is D.

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You are in a car traveling at 20 m/s. An ambulance is behind you traveling 35 m/s in the same direction. What frequency do you h
arlik [135]

Answer:

The frequency heard is 576.78 Hz

Explanation:

The Doppler effect is defined as the apparent frequency change of a wave produced by the relative movement of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave motion when the sender and receiver, or observer, move relative to each other.

This is what happens in the first part of this problem, where the sender is train A and the receiver is train B. They are both moving in opposite directions. In this case, where both are in motion, the frequency perceived by the receiver will increase when receiver and transmitter increase their separation distance and will decrease whenever the separation distance between them is reduced. The following expression is considered the general case of the Doppler effect:

f'=f*\frac{v+-vR}{v+-vE}

Where:

f ', f: Frequency perceived by the receiver and frequency emitted by the issuer respectively. Its unit of measurement in the International System (S.I.) is the hertz (Hz), which is the inverse unit of the second (1 Hz = 1 s-1)

v: Velocity of propagation of the wave in the medium. It is constant and depends on the characteristics of the medium. In this case, the speed of sound in air is considered to be 343 m / s

vR, vE: Speed ​​of the receiver and the emitter respectively. Its unit of measure in the S.I. is the m / s

±, ∓:

We will use the + sign:

  • In the numerator if the receiver approaches the emitter
  • In the denominator if the emitter moves away from the receiver

We will use the sign -:

  • In the numerator if the receiver moves away from the emitter
  • In the denominator if the emitter approaches the receiver

In this case you are in a car traveling at 20 m/s and an ambulance is behind you traveling 35 m/s in the same direction.

In this case the receiver, you in the car, moves away from the emitter, while the emitter, the ambulance, approaches the receiver behind you in the same direction. So the frequency is calculated by the expression:

f'=f*\frac{v-vR}{v-vE}

Being:

  • f= 550 Hz
  • v=343 m/s
  • vR= 20 m/s
  • vE= 35 m/s

and replacing:

f'=550 Hz*\frac{343 m/s-20 m/s}{343 m/s-35 m/s}

you get:

f'= 576.78 Hz

The frequency heard is 576.78 Hz

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a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on
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Answer:

The object would weight 63 N on the Earth surface

Explanation:

We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}

Now, if the body is on the surface of the Earth, its weight (w) would be:

F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\

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