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2 years ago

A student practicing for a track meet ran 250 m in 30 seconds. What was her average speed?

Physics

1 answer:

DochEvi [55]2 years ago

6 0

Answer:

8.33 meters/sec.

Explanation:

distance = 250 meters

time = 30 sec.

velocity = distance / time

= 250 meters

30 sec.

= 8.33 meters/sec.

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Answer:

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3 years ago

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A car traveling at 27.4 m/s hits a bridge abutment. A passenger in the car, who has a mass of 65.0 kg, moves forward a distance

Minchanka [31]

Answer:

$F=43570.9N$

Explanation:

We can calculate the acceleration experimented by the passenger using the formula $v_f^2=v_i^2+2ad$ , taking the initial direction of movement as the positive direction and considering it comes to a rest:

$a=\frac{v_f^2-v_i^2}{2d}=\frac{-v_i^2}{2d}$

Then we use Newton's 2nd Law to calculate the force the passenger of mass m experimented to have this acceleration:

$F=ma=\frac{-mv_i^2}{2d}$

Which for our values is:

$F=\frac{-(65kg)(27.4m/s)^2}{2(0.56m)}=43570.9N$

6 0

3 years ago

Suppose you have a 136-kg wooden crate resting on a wood floor. The coefficient of static friction is 0.4 and coefficient of kin

stealth61 [152]

Answer:

The the maximum force acting on the crate is 533.12 newtons.

Explanation:

It is given that,

Mass of the wooden crate, m = 136 kg

The coefficient of static friction, $\mu_s=0.4$

The coefficient of kinetic friction, $\mu_k=0.2$

We need to find the maximum force exerted horizontally on the crate without moving it. As the crate is not moving than the coefficient of static friction will act and the force is given by :

$F=\mu_s mg$

$F=0.4\times 136\ kg\times 9.8\ m/s^2$

F = 533.12 N

So, the maximum force acting on the crate is 533.12 newtons. Hence, this is the required solution.

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3 years ago