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3 years ago

The greater the mass of an object? I’ll give brainliest!*

Physics

1 answer:

liraira [26]3 years ago

6 0

between B and C im thinking its more C though

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El movimiento de los cuerpos y la formación de imágenes son ejemplos de esta clase de fenómeno

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Answer:

depende de que fenómenos nos referimos de acuerdo al los cuerpos de formación puede aver movimiento contante

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3 years ago

A ball is thrown against a wall and bounces back toward the thrower with the same speed as it had before hitting the wall. Does

lorasvet [3.4K]

Answer:

A) Although the speed is the same, the direction has changed. Therefore, the velocity has changed.

Explanation:

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3 years ago

You are 12 miles north of your base camp when you begin walking north at a speed of 2 miles per hour. What is your location, rel

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If you walk at a pace of 2 miles per hour for 5 hours, you should have walked 10 miles. You would be 2 miles away from your base camp.

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3 years ago

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Julie is cycling at a speed of 3.4 meters/second. If the combined mass of the bicycle and Julie is 30 kilograms, what is the kin

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Velocity=3.4m/sec
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3 years ago

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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

$\Delta f = 313.1 Hz - 396.7 Hz = -83.6 Hz$

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