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allsm [11]
3 years ago
13

You are bungee jumping from a bridge. Initially, while you are falling the slack bungee cord isn’t exerting any forces or torque

s on you. Earth’s gravity is effectively pulling downward on you at your center of gravity. Consider a net force and a net torque about your center of mass. Which of these two influences are acting on you as you fall?
a. The net force is zero, but the net torque is nonzero.
b. The net force is nonzero, but the net torque is zero.
c. The net force is zero and the net torque is zero as well.
d. Both the net force and the net torque are nonzero.
Physics
1 answer:
harina [27]3 years ago
4 0

Answer:

he fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

Explanation:

Let's pose the solution of this problem, to be able to analyze the firm affirmations.

When the person is falling, the weight acts on them all the time, initially the rope has no force, but at the moment it begins to lash it exerts a force towards the top that is proportional to the lengthening of the rope.

The equation for this part is

                 Fe - W = m a  

                 k x - mg = m a

As the axis of rotation is located at the top where they jump, there is a torque.

What is it

                Fe y - W y = I α

angular and linear acceleration are related

       a = α r

       Fe y - W y = I a / r

In the fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

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A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.2m/s and the hatchback ca
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Answer:

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

Let us assume that both vehicles are at origin at the start means initial position is zero i.e. s_{o} = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.

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Using s =s_{o}+v_{i}t+1/2at^{2}

s = 33.2t .......... eq (1)

Hatchback:

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Using s =s_{o}+v_{i}t+1/2at^{2}

putting in the data we will get

s=(1/2)(5)t^{2}

now putting 's' value from eq (1)

2.5t^{2}-33.2t=0

which will give,

t = 13.28 s

so both vehicles will meet up gain after 13.28 sec.

putting t = 13.28 in eq (1) will give

s = 440.896 m

So, both vehicles will meet up again at 440.896 m.

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