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denis-greek [22]
3 years ago
5

The mass of the basketball shown below is 4 times the mass of the baseball.

Physics
1 answer:
igomit [66]3 years ago
8 0

Answer:

<h3>40.0 Joules. None of the answers is correct</h3>

Explanation:

Kinetic energy of an object is expressed as;

KE = 1/2mv² where;

m is the mass of the object

V is the velocity of the object.

Let Ma be the mass of the basketball and Mb be the mass of baseball.

If the mass of the basketball shown below is 4 times the mass of the baseball, then Ma = 4Mb

KE of the basket ball = 1/2MaVa²

Va² = 2(KE)a/Ma ...... 1

KE of the baseball = 1/2MbVb²

Vb² = 2(KE)b/Mb .......... 2

Since their velocities are the same, hence Va²= Vb²

2(KE)a/Ma = 2(KE)b/Mb

Substituting Ma = 4Mb into the resulting expression.

2(KE)a/4Mb = 2(KE)b/Mb

2(KE)a/4 = 2(KE)b

Given Kinetic energy of baseball (KE)b = 10.0J, the expression becomes;

2(KE)a/4 = 2(10)

2(KE)a = 4*20

2(KE)a = 80

(KE)a = 80/2

(KE)a = 40.0J

Hence the kinetic energy of the basketball is 40.0Joules

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Answer:

T = 2.83701481512 seconds

Explanation:

Hi!

The formula that you will want to use to solve this question is:

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g --> acceleration due to gravity (9.8m/s^2)

since we know that the mass of the bob at the end of the pendulum does not affect the period of the pendulum, we can go ahead and ignore that bit of information (unless, of course, the weight causes the pendulum to stretch)

so now we can plug in our given info into the formula above and solve!

T = 2*pi * sqrt(2/9.8)

T = 2.83701481512 seconds

*Note*

- I used 3.14 to pi, if you need to use a different value for pi (a longer version, etc) your answer will be slightly different

I hope this helped!

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A spherical balloon is inflating with helium at a rate of 72 ft2/min . How fast is the​ balloon's radius increasing at the insta
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The volume of the balloon is given by:

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Differentiate both sides with respect to time t:

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Plug in and solve for dr/dt:

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A = 4πr²

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Differentiate both sides with respect to time t:

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Given values:

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Plug in and solve for dA/dt:

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The surface area is changing at a rate of 48.25ft²/min

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