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skelet666 [1.2K]
3 years ago
7

Different forces were applied to five balls, and each force was applied for the same amount of time. The data is in the table. A

4 column table with 3 rows. The first column is labeled Object with entries Ball A, Ball B, Ball C, Ball D. The second column is labeled Force in Newtons with entries 50 N, 60 N, 70 N, 80 N. The third column is labeled Time in seconds with entries 1. 25, 1. 25, 1. 25, 1. 25. The fourth column is labeled Impulse in Newtons times seconds with entries 62. 5, 75, 87. 5, 100. If the force on ball A doubles while the time remains the same, what happens to the impulse? It divides in half. It does not change. It doubles. It quadruples.
Physics
1 answer:
Annette [7]3 years ago
7 0

The magnitude of impulse on Ball A will become twice (doubles). Hence, option (c) is correct.

Impulse:

The effect of applied force acting for a very small interval of time is known as Impulse.

Given data:

The initial magnitude of the force on Ball A is, F = 50 N.

The time interval for the applied force on Ball A is, t = 1.25 s.

And the impulse at initial on ball A is, I = 62.5 N-s.

Now, if force on ball A doubles, which means new magnitude of force is,

F' = 2F

F' = 2 × 50

F' = 100 N

For same time interval, the new impulse on Ball A is given as,

I' = F' × t

Solving as,

I' = 100 × 1.25

I' = 125 N-s

Taking the ratio of both the impulses as,

I' / I = 125/62.5

I' = 2 I

Thus, we can conclude that the magnitude of impulse on Ball A will become twice (doubles). Hence, option (c) is correct.

Learn more about the impulse here:

brainly.com/question/904448

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Given that,

His friend design has a turnable disk of radius 1.5m

R = 1.5m

The mass is twice the fully loaded elevator.

Let the mass of the full loaded elevator be M

Then, mass of the turn able

Mt = ½M

Radius of the disk that serves as a vertical pulley is ¼ radius of turntable and 1/16 of the mass.

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I_p = 3M / 256

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For the elevator

Fnet = ma

Mg - T1 = Ma

a = (Mg-T1) / M

For vertical pulley,

The torque is given as

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τ_p = 3M/256 × α_p = (T2-T1)•r

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The torque is given as

τ_t = I_t × α_t = T2•r

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So, the torque are equal

τ_t = τ_p

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M cancel out

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48•α_t = α_p

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