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photoshop1234 [79]
2 years ago
8

A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking

from one end of the car to the other in the direction of motion, with speed 2.0 m/s with respect to the car. In the time it takes for him to reach the other end, how far has the flatcar moved?
Physics
1 answer:
Anna11 [10]2 years ago
5 0

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}

The total motion has a total velocity and is

Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}

The time the worker take walking is

t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m

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Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

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Response:

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<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

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M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

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<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

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S ?
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S=ut+0.5at^2
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