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3 years ago

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Car a is traveling at a constant speed 90 m/s east. a second car b is initially at rest when car a passes it at t=0. if car b st

arts immediately to chase car a at t=0, and catches car a at t=25 s, what was car b's acceleration (assume a constant acceleration)?

1 answer:

MatroZZZ [7]3 years ago

4 0

The answer choice is 32

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A car slows down from speed of 72

Nastasia [14]

Explanation:

Given parameters:

Initial velocity = 72km/hr

Final velocity = 0km/hr

Time taken = 25s

Unknown:

Acceleration = ?

Solution:

To solve this problem, convert km/hr to m/s;

1000m = 1km

3600s = 1hr

72km/hr;

1km/hr = 0.278m/s

72km/hr = 0.278 x 72 = 20.02m/s

Acceleration is the change in velocity divided by the time taken;

Acceleration = $\frac{final velocity - initial velocity }{time}$

Acceleration = $\frac{0 - 20.02}{25}$ = -0.8m/s

The car is actually decelerating at a rate of 0.8m/s

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Answer:

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Explanation:

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What should a free-body diagram look like for a skydiver who has opened his parachute and is now slowing down as he falls?

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3 years ago

Read 2 more answers

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

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2 years ago