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3 years ago

A car is pulled with a force of 10,000 N. The car's mass is 1267 kg. But, the car covers 394.6 m in 15 seconds

Physics

1 answer:

Rasek [7]3 years ago

6 0

F=m*a=>a=F/m=10000/1267=7.89 m/s²

d=v₀t+a't²/2<=>394.6=112.5*a'=>a'=3.5

a-a'=7.89-3.5=4.39 m/s²

This difference causes friction forces

We apply the second principle of dynamics: vector: F + N + G + Ff = ma (vector vectors, I can not here)

Scalar: Ox: F-Ff = ma

Oy N-mg = 0

Ff = -ma+ F =-1267*7.89+10000=-8869+10000=1131 N

This frictional force (Ff) is opposite to the traction (F)

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A rugby player runs with the ball directly toward his opponent's goal, along the positive direction of an x axis. He can legally

Over [174]

Answer:

minimum angle is 128.69°

Explanation:

given data

player velocity with respect ground v1 = 3.5 m/s

ball velocity with respect himself v2 = 5.6 m/s

to find out

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solution

we know ball velocity with respect field will be

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we consider angle that player hit ball is θ

then by as per figure triangle

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cosθ = $\frac{3.5}{5.6}$

θ = 51.31

so minimum angle is 180 - 51.31 = 128.69°

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3 years ago

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3 years ago

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3 years ago

A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.

JulsSmile [24]

Answer:

$\mu =0.75$

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

$F_c=m.a_c$

The centripetal acceleration a_c is computed as

$\displaystyle a_c=\frac{v^2}{r}$

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

$F_c=m.\frac{v^2}{r}$

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

$F_r=\mu N$