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4 years ago

What is 4.6 billion years old in the big band theory

Physics

1 answer:

erastova [34]4 years ago

8 0

Answer:

solar system

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How can damage to living things occur from exposure to X-rays or gamma

OLEGan [10]

Answer:C

Explanation:

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3 years ago

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A racecar experiences a centripetal acceleration of 36.0 m/s2 as it travels at a constant speed of 27.0 m/s along a circular arc

alexdok [17]

Answer:

20.25 m

Explanation:

Centripetal acceleration is given by; the square of the velocity, divided by the radius of the circular path.

That is;

ac = v²/r

Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the radius, m

Therefore;

r = v²/ac

= 27²/36

= 20.25 m

Hence the radius is 20.25 meters

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3 years ago

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Which one is smaller a hectogram, dekagram, kilogram, centigram, or milligrams?

antoniya [11.8K]

A kilogram is smaller

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3 years ago

What is the chemical formula for copper(I)sulfide? A) CuS B) CuS2 C) Cu2S D) Cu2S3

weeeeeb [17]

Answer:

Cu2S

Explanation:

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3 years ago

A uniform steel rod of cross-sectional area A is attached to rigid supports and is unstressed at a temperature of 458F. The stee

Vedmedyk [2.9K]

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.

Let us assume that P is the compressive force in the rod due to change in temperature.

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

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

= $190.98^{o}F$

Now, we will calculate the actual change in temperature as follows.

$\Delta T = 320 - 45 = 275^{o}F$

This means that the actual change in temperature is more than required for yielding.

(a) Formula to calculate yielding stress is as follows.

$\sigma' = \frac{P'}{A}$

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

= $-E \alpha \Delta T$

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

= $-51.8375 \times 10^{3} psi$

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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3 years ago

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