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arlik [135]
3 years ago
15

PLEASE HELP ASAP!! FILL IN THE BLANKS. 99 POINTS!The force of _________ is what holds you down to earth. The ______________ of o

n a vector is indicated by the length of its arrow. A measure of the force of gravity on an object is called __________. Einstein explained gravity as a __________ in space. If you had the mass of 600 kg on the earth; you would have the mass of ______kg on the tiny Mercury.
Physics
2 answers:
WARRIOR [948]3 years ago
7 0

Answer:

Explanation:

The force of __gravity__ is what holds you down to earth. The __magnitude__ of on a vector is indicated by the length of its arrow. A measure of the force of gravity on an object is called __weight__. Einstein explained gravity as a __distortion__ in space. If you had the mass of 600 kg on the earth; you would have the mass of _600_kg on the tiny Mercury.

Dima020 [189]3 years ago
4 0

Answer:

The force of <u>Gravity</u> is what holds you down to earth.  

The <u>length</u> of on a vector is indicated by the length of its arrow.

A measure of the force of gravity on an object is called <u>weight</u>.

Einstein explained gravity as a <u>distortion </u>in space.

If you had the mass of 600 kg on the earth; you would have the mass of <u>226.8 kg</u> on the tiny Mercury.

I hope this helps you! :)

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Explanation:

As the given rod is attached to rigid supports as a result, the deformation occurring due to the change in temperature will cause stress in the rod.

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So,        \Delta T = \frac{\sigma_{y}}{E_{a}}

                          = \frac{36 \times 10^{3}}{(29 \times 10^{6} \times 6.5 \times 10^{-6})}

                          = 190.98^{o}F

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             \sigma' = \frac{P'}{A}

           \sigma' = -\frac{AE \alpha \Delta T}{A}

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                        = -29 \times 10^{6} \times 6.5 \times 10^{-6} \times 275

                        = -51.8375 \times 10^{3} psi

Hence, stress in the bar when temperature is raised to 320^{o}F is -51.8375 \times 10^{3} psi.

(b)  Now, we will calculate the residual stress as follows.

            \sigma_{r} = -\sigma_{y} - \sigma'

           \sigma_{r} = -36 + 51.837 ksi

                          = 15.837 ksi

Therefore, stress in the bar when the temperature has returned to 45^{o}F is 15.837 ksi.

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