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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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Hi there!

We can use Newton's Second Law:

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a = acceleration (m/s²)

We are given the mass and acceleration, so:

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Once an object enters orbit, what keeps the object moving sideways?
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3 years ago
Which benefit outweighs the risks in the technological design of headphones?
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Calculate the speed of an 8.0x10^4 kg airliner with a kinetic energy of 1.1x10^9 j ...?
Over [174]
Kinetic energy is the energy possessed by an object on motion. it is expressed as follows:

KE = 0.5mv^2

where m is the mass and v is the velocity of the object. We calculate as follows:


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6 0
3 years ago
A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega

so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

\omega = 0.23 rad/s

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0
3 years ago
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