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horsena [70]
3 years ago
10

PLS HELP. A flatbed car of a train moves 8 m/s to the east. A jogger runs along to top the flatbed car (which is not very safe)

with a velocity of 6 m/s to the west. How fast is the jogger moving?
Physics
1 answer:
jonny [76]3 years ago
5 0

Answer:

2 m/s

Explanation: Given that a flatbed car of a train moves 8 m/s to the east. A jogger runs along to top the flatbed car (which is not very safe) with a velocity of 6 m/s to the west.

Since the jogger is moving in an opposite direction to the direction of the train, and velocity is the distance covered in a specific direction, the jogger will be moving at a velocity relative to the velocity of the train.

Velocity = (8 - 6) m/s

Velocity = 2 m/s

Therefore, the jogger will be moving at the speed of 2 m/s

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Answer:

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3 years ago
Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the
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Answer:1.71 m/s

Explanation:

Given

mass of Susan m=12 kg

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Resolving Forces Along x axis

F_x=T\cos \theta -f_r

where f_r=friction\ Force  

F_y=mg-N-T\sin \theta

since there is no movement in Y direction therefore

N=mg-T\sin \theta

and f_r=\mu N

Thus F_x=T\cos \theta -\mu N

F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))                

F_x=25.114-18.558

F_x=6.556 N

Work done by applied Force is equal to change to kinetic Energy

F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2

6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2

v_f^2=\frac{6.556\times 2.7\times 2}{12}

v_f^2=2.95

v_f=1.717 m/s        

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