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gayaneshka [121]
3 years ago
7

A crate is sliding down a ramp that is inclined at an angle 30.6 ° above the horizontal. The coefficient of kinetic friction bet

ween the crate and the ramp is 0.360. Find the acceleration of the moving crate.

Physics
1 answer:
Vlada [557]3 years ago
6 0

Answer:1.95 m/s^2

Explanation:

Given

inclination \theta =30.6^{\circ}

coefficient of kinetic friction \mu =0.36

As crate is moving Down therefore friction will oppose the motion

using FBD

mg\sin \theta -f_r=ma

f_r=\mu N

f_r=\mu mg\cos \theta

mg\sin \theta -\mu mg\cos \theta =ma

a=g\sin \theta -\mu g\cos \theta

a=g(\sin (30.6)-0.36\cdot \cos (30.6))

a=9.8\times 0.199

a=1.95 m/s^2          

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Option 3: -48 cm

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The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 m/s (about 45 mi/h). Initially, the car
Alex73 [517]

Answer:

15.8640053791 s

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0 denotes initial

x denotes displacement

c denotes car

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x_0_{cr}=-49.5\ m

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v_0_{c}=v_0_{t}

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x_t=v_tt

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x_c-x_t=x_0+v_0_{c}+\dfrac{1}{2}at-v_{t}t=26\\\Rightarrow 26=x_0+\dfrac{1}{2}at^2\\\Rightarrow 26+49.5=\dfrac{1}{2}0.6t^2\\\Rightarrow t=\sqrt{\dfrac{2(26+49.5)}{0.6}}\\\Rightarrow t=15.8640053791\ s

The time taken is 15.8640053791 s

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The distance the car travels is 392.780107582 m

v=v_0+at\\\Rightarrow v=20+0.6\times 15.8640053791\\\Rightarrow v=29.5184032275\ m/s

The velocity of the car is 29.5184032275 m/s

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