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igomit [66]
3 years ago
8

To catch a fast-moving ball, you extend your hand forward before contact with the ball and let it ride backward in the direction

of the ball's motion. Doing this reduces the force of contact on your hand principally because the Group of answer choices force of contact is reduced relative velocity is less time of contact is increased time of contact is decreased none of these
Physics
1 answer:
const2013 [10]3 years ago
7 0

Answer:

C: time of contact is increased

Explanation:

When a ball is falling from height, it maintains the same mass but velocity may change depending on the conditions in that environment.

Now, to catch the ball on your palm, you need to find a way to reduce the momentum of the ball upon hitting your hand.

Now, to reduce the momentum, we need to apply impulse.

This is because we know that;

Impulse(I) = F × t = change in momentum (m(v - u))

Thus;

I = Ft

F = I/t

We are told that extending the hand reduces the force of contact on your hand.

Thus;

For F which is the force to be reduced, the denominator (t) has to be increased.

Thus, the correct answer is that time of contact is increased

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A rock thrown with speed 12.0 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 15.5 m bef
Dimas [21]

Supposing there's no air resistance, horizontal velocity is constant, which makes it very easy to solve for the amount of time that the rock was in the air.


Initial horizontal velocity is: <span>
cos(30 degrees) * 12m/s = 10.3923m/s 

15.5m / 10.3923m/s = 1.49s 

So the rock was in the air for 1.49 seconds. </span>

<span>

Now that we know that, we can use the following kinematics equation: 

d = v i * t + 1/2 * a * t^2 

Where d is the difference in y position, t is the time that the rock was in the air, and a is the vertical acceleration: -9.80m/s^2. </span>

<span>
Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s 

So: 

d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2 
d = 8.94 + -10.89</span>

d = -1.95<span>

<span>This means that the initial y position is 1.95 m higher than where the rock lands. </span></span>

5 0
3 years ago
A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the
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Answer:

+16 J

Explanation:

We can solve the problem by using the 1st law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change of the internal energy of the system

Q is the heat (positive if supplied to the system, negative if dissipated by the system)

W is the work done (positive if done by the system, negative if done by the surroundings on the system)

In this case we have:

Q = -12 J is the heat dissipated by the system

W = -28 J is the work done ON the system

Substituting into the equation, we find the change in internal energy of the system:

\Delta U=-12 J-(-28 J)=+16 J

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