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Y_Kistochka [10]
2 years ago
8

A 2.0-kg ball is at rest when a horizontal force of 5.0 N is applied. In the absence of friction, what is the speed of the ball

after it ha gone 10 m?
Physics
1 answer:
Rzqust [24]2 years ago
7 0

Answer:

v = 7.1 m/s

Explanation:

Work applied will change kinetic energy

Fd = ½mv²

v = √(2Fd/m) = √(2(5.0)(10) / 2.0) = √50 = 7.07... m/s

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A natural atom can be negatively charged by
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By Gaining Electrons

Explanation:

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In a classroom, 4 out of every 6 students are female. If there are 27 students, how many of them are female?
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Which is not one of the forms that energy can take?
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A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr
mihalych1998 [28]

(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

f' = (\frac{v}{v\pm v_s})f

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

v_s is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)

f is the original frequency of the sound

Here we have

f = 350 Hz

When the train is approaching, we have

v_s = -40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s - 40.4 m/s})(350 Hz)=396.7 Hz

When the train has passed the platform, we have

v_s = +40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s + 40.4 m/s})(350 Hz)=313.1 Hz

Therefore the overall shift in frequency is

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And the negative sign means the frequency has decreased.

(b) 0.865 m

The wavelength and the frequency of a wave are related by the equation

v=\lambda f

where

v is the speed of the wave

\lambda is the wavelength

f is the frequency

When the train is approaching the platform, we have

v = 343 m/s (speed of sound)

f = f' = 396.7 Hz (apparent frequency)

Therefore the wavelength detected by a person on the platform is

\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m

5 0
3 years ago
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