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Lelechka [254]
2 years ago
10

The layers of the atmosphere are divided into layers based on changes in __________.

Physics
1 answer:
Umnica [9.8K]2 years ago
6 0
A. altitude should be correct
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A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same di
Shalnov [3]

A distance of d is covered with 53 mile/hr initially. Time taken to cover this distance t1 = d/53 hour Next distance of d is covered with x mile hours. Time taken to cover this distance t2 = d/x hours. We have average speed = 26.5 mile / hour          

                                         = Total distance traveled/ total time taken                  

                                         = \frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} }  = \frac{106x}{x+53}

                              26.5 = \frac{106x}{x+53} \\ \\ 79.5 x = 1404.5\\ \\ x = 17.67 miles/hour

5 0
3 years ago
A short 6061-T6 aluminum cylindrical block, having an original diameter of 20mm and a length of 75mm, is placed in a compression
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Explanation:

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5 0
3 years ago
A dog of mass 4 kg runs up a hill of height 8 m. How much gravitational potential energy does the dog gain?
Genrish500 [490]
A. 314 because when you use the formula for the GPE ; GPE=MGH or means mass times gravity time height (4x8x9.8) and thats equivalent to 313.6 which rounds up to 314. Hope it helps 
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3 years ago
Read 2 more answers
The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr
Degger [83]

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

I(t)=0.88e^{-t/6\times3600s}

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

q=\int\limits {I} \, dt

Here the limits of integration is from 0 to infinite. So,

q=\int\limits {0.88e^{-t/6\times3600s}}\, dt

q=0.88\times(-6\times3600)(0-1)

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}

N = 1.2 x 10²³

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3 years ago
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