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3 years ago

A ball was rolling downhill at 2 m/s. After 5s, it was rolling at 90 m/s. What is its acceleration?

Physics

1 answer:

olga55 [171]3 years ago

4 0

Answer:

17.6 m/s²

Explanation:

Given:

$v_{f}$ = 90 m/s (final velocity)

$v_{i}$ = 2 m/s (initial velocity)

Δt = 5s (change in time)

The formula for acceleration is:

$a_{avg}$ = Δv / Δt

We can find Δv by doing

Δv = $v_{f}$ - $v_{i}$

Replace the values

Δv = 90m/s - 2m/s

Δv= 88m/s

Using the equation from earlier, we can find the acceleration by dividing the average velocity by time.

$a_{avg}$ = Δv / Δt

$a_{avg}$ = $\frac{88m/s}{5/s}$

acceleration = 17.6 $m/s^{2}$

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Answer:

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Explanation:

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3 years ago

A radar station sends out a 250000 Hz sound wave at a speed of 340 m/s. The sound wave bounces off a weather ballon and returns

Eddi Din [679]

Answers:

a)The balloon is 68 m away of the radar station

b) The direction of the balloon is towards the radar station

Explanation:

We can solve this problem with the Doppler shift equation:

$f'=\frac{V+V_{o}}{V-V_{s}} f$ (1)

Where:

$f=250,000 Hz$ is the actual frequency of the sound wave

$f'=240,000 Hz$ is the "observed" frequency

$V=340 m/s$ is the velocity of sound

$V_{o}=0 m/s$ is the velocity of the observer, which is stationary

$V_{s}$ is the velocity of the source, which is the balloon

Isolating $V_{s}$ :

$V_{s}=\frac{V(f'-f)}{f'}$ (2)

$V_{s}=\frac{340 m/s(240,000 Hz-250,000 Hz)}{240,000 Hz}$ (3)

$V_{s}=-14.16 m/s$ (4) This is the velocity of the balloon, note the negative sign indicates the direction of motion of the balloon: It is moving towards the radar station.

Now that we have the velocity of the balloon (hence its speed, the positive value) and the time ( $t=4.8 s$ ) given as data, we can find the distance: