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3 years ago

An object, initially at rest, is subject to an acceleration of 34 m/s^2. How long will it take for that object to reach 3400m ?

Physics

1 answer:

Norma-Jean [14]3 years ago

6 0

Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m

Vf = 480.83 m/s

a=v/t
t=v/a
t=480.83/34
t=14.142 s

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vovikov84 [41]

Answer:8.75 s,

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Explanation:

Given

Initial velocity $=70 mph\approx 31.29 m/s$

velocity after 5 s is $30 mph\approx 13.41 m/s$

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$a=\frac{13.41-31.29}{5}=-3.576 m/s^2$

therefore time required to stop

v=u+at

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$0=31.29-3.576\times t$

$t=\frac{31.29}{3.576}=8.75 s$

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3 years ago

A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the fri

omeli [17]

Answer:

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Explanation:

When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:

a = v^2/r

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Therefore, a body with mass m, will feel a force f:

f = m v^2/r

Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:

fs = μN = μmg

The car will not slide if f = fs, i.e.

fs = μmg = m v^2/r

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