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4 years ago

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Two masses, each weighing 1.0 × 103 kilograms and moving with the same speed of 12.5 meters/second, are approaching each other.

They have a head-on collision and bounce off away from each other. Assuming this is a perfectly elastic collision, what will be the approximate kinetic energy of the system after the collision?
A. 1.6 × 105 joules

B. 2.5 × 105 joules

C. 1.2 × 103 joules

D. 2.5 × 103 joules

1 answer:

pav-90 [236]4 years ago

8 0

Answer:

A. 1.6 × 105 joules

Explanation:

As per the question, the data given in the question is as follows

Number of masses = 2

Each weightage = $1.0 \times 2.3$

Speed = 12.5 meters/ second

Based on the above information, the approximate kinetic energy after the collision is

A perfectly elastic collision is described as one where the collision does not cause any loss of kinetic energy.

So we sum the kinetic energy of each kind of system which is given below:

Kinetic energy is

$= 0.5(1.0 \times 10^3) (12.5)^2 + 0.5 (1.0 \times 10^3) (12.5 )^2$

= 156250 J

$= 1.6 \times 10^5 J$

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<h3>According to the given data:</h3>

the region to the left of the standard normal curve,

z=0.49

To the right of,

z = 2.05

So,

The area will be:

= P[z < 0.49] + P[ z >2.05]

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= .6879 + 1 - .9798

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Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

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I understand that the question you are looking for is:

Find the area under the standard normal curve to the left of z = 0.49 and to the right of z = 2.05. Round your answer to four decimal places, if necessary.

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