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makkiz [27]
3 years ago
12

Earth’s atmosphere is in hydrostatic equilibrium. What this means is that the pressure at any point in the atmosphere must be hi

gh enough to support the weight of air above it. How would you expect the pressure on Mt. Everest to differ from the pressure in your classroom? Explain why.
Physics
1 answer:
Misha Larkins [42]3 years ago
6 0

Answer: The pressure that one experiences on the Mount Everest will be different from the one, in a classroom. It is because pressure and height are inversely proportional to each other. This means that as we move up, the height keeps on increasing but the pressure will keep on decreasing. This is the case that will be observed when one stands on the Mount Everest as the pressure is comparatively much lower there.

It is because as we move up, the amount of air molecules keeps on decreasing but all of the air molecules are concentrated on the lower part of the atmosphere or on the earth's surface.

Thus a person in a low altitude inside a classroom will experience high pressure and a person standing on the Mount Everest will experience low pressure.

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An car is brought to rest in a distance of 652 m using a constant acceleration of -2.7 m/s2. What was the
Goshia [24]

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I’m so sorry I tried solving it but I don’t understand it can you explain the question a little bit more ty

Explanation:

7 0
3 years ago
A small 18 kilogram canoe is floating downriver at a speed of 1 m/s. What is the canoe's kinetic energy?
In-s [12.5K]

Kinetic energy = (1/2) (mass) (speed²).

A Physicist in the canoe, or on a raft floating downriver next to the canoe, will say that the canoe's kinetic energy is zero.

A Physicist on the riverbank, watching the canoe drift by at 1 m/s, will say that its kinetic energy is 9 Joules.

They're both correct.

8 0
3 years ago
What is a element that tends to be shiny, easy shaped, and a good conductor of heat?
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The answer to your question is Metal
8 0
3 years ago
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
3 years ago
A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal
Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, a=g=-9.8\ m/s^2(Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

S=ut+\frac{1}{2}at^2

Plug in the given values and solve for 'u'. This gives,

0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0
3 years ago
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