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3 years ago

Thomas is climbing Mt. Everest. What happens as he climbs farther up the mountain?

Physics

2 answers:

Eva8 [605]3 years ago

6 0

Answer:

B: The air pressure decreases

Explanation:

Pressure at any given height in the atmosphere is the total weight of the air above a unit area at any height.

Increase in elevation leads to decrease or lesser air molecules above a given surface than a similar surface at lower levels.

lakkis [162]3 years ago

4 0

Answer:

B. the air pressure decreases

Explanation:

As elevation increases, there is less overlying atmospheric pressure mass, so that atmospheric pressure decreases with increasing elevaton.

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How much energy is required to change a 44 g

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Answer:

Ang answer units of J heat of fusion is 3.33 x 105

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3 years ago

How are hurricanes and thunderstorms similar? A. Both cover an area as large as 650 km. B. Both involve the formation of large c

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3 years ago

Read 2 more answers

A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the f

hram777 [196]

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

$h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2$ .... (2)

By equation the equation (1) and (2), we get

$41.7=1.12 u +4.9 \times 1.12^{2}$

u = 31.75 m/s

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3 years ago

A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w

trapecia [35]

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A) Conservation of Energy,

$KE = PE$