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3 years ago

As altitude increases, what happens to air pressure?

Physics

2 answers:

dlinn [17]3 years ago

4 0

Answer:

Explanation:

Mashcka [7]3 years ago

3 0

Answer:

It decreases

Explanation:

The air pressure tends to be higher on the places with the lowest altitudes, and lower at the places with higher altitudes. Basically, the air pressure is the wight of the air, and since the air is denser and heavier at the lower altitudes, the air pressure is higher, while on the higher altitudes the air is less dense, thus the air pressure is lower. So in practice we can expect that the air pressure in a low valley will be higher than the air pressure at the top of higher mountain.

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1. What causes velocity to change?

tresset_1 [31]

The amount of force an object has will change the velocity

4 0

3 years ago

Read 2 more answers

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp

MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

$V_{Ax}=V_{A}cos\theta$

$V_{Ax}=(21m/s)cos(-14^{o})$

$V_{Ax}=20.38 m/s$

$V_{Ay}=V_{A}sin\theta$

$V_{Ay}=(21m/s)sin(-14^{o})$

$V_{Ay}=-5.08 m/s$

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

$x_{A}=V_{Ax}t$

$x_{A}=(20.38m/s)(0.317s)$

$x_{A}=6.46m$

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

$V_{Bx}=20.88 m/s$

$V_{By}=-2.195 m/s$

$y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}$

$0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}$

$-4.9t^{2}-2.195t+2.1=0$

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}$

t= -0.9159s or t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

$x_{B}=V_{Bx}t$

$x_{B}=(20.88m/s)(0.468s)$

$x_{B}=9.77m$

So once we got the two distances we can now find the difference between them:

$x_{B}-x_{A}=9.77m-6.46m=3.31m$

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0

2 years ago

Give reason for the following:

Zigmanuir [339]

a) An inflated balloon was pressed against a wall after it has been rubbed with a piece of synthetic cloth. It was found that the balloon sticks to the wall. <u>This is because a positive and negative electric charge is produced, therefore the balloon sticks to the wall.</u>

b) When an object is thrown up, it comes back to ground <u>because of gravitational attraction force of earth</u>.

c) Mountaineers suffer nose bleeding at higher altitudes <u>because the oxygen level decreases with increase in altitude, which the body cannot adjust.</u>

d) Foundations of high rise buildings are kept wide <u>because more is the area of contact, less is the pressure efforts. So, foundations are wide so as to decrease the possibility of the building from falling down.</u>

e) Deep sea divers or high altitude fliers wear special suits <u>so as prevent their body from being crushed by the water pressure. Since water pressure is maximum at deep seas and oceans, therefore, more is the risk of being injured.</u>

f) Walls of a dam are thickened near the base <u>so that the dam can handle the kinetic energy pressure and prevent itself from breaking down, which if not, can lead to flooding</u>.

HOPE IT HELPS...

5 0

2 years ago

A gas in a cylinder expands from a volume of 0.110 m³ to 0.320 m³. heat flows into the gas just rapidly enough to keep the press

Elden [556K]

80000 Joule is the change in the internal energy of the gas.

<h3>In Thermodynamics, work done by the gas during expansion at constant pressure:</h3>

ΔW = -pdV

ΔW = -pd (V₂ -V₁)

ΔW = - 1.65×10⁵ pa (0.320m³ - 0.110m³)

= - 0.35×10⁵ pa.m³

= - 35000 (N/m³)(m³)

= -35000 Nm

ΔW = -35000 Joule

Therefore, work done by the system = -35000 Joule

<h3>Change in the internal energy of the gas,</h3>

ΔV = ΔQ + ΔW

Given:

ΔQ = 1.15×10⁵ Joule

ΔW = -35000 Joule

ΔU = 1.15×10⁵ Joule - 35000 Joule

= 80000 Joule.

Therefore, the change in the internal energy of the gas= 80000 Joule.

Learn more about thermodynamics here:

brainly.com/question/14265296

#SPJ4

3 0

1 year ago

Any help is appreciated please

stiv31 [10]

Answer:

C. It speeds up, and the angle increases

Explanation:

We can answer by using the Snell's law:

$n_i sin \theta_i = n_r sin \theta_r$

where

$n_i, n_r$ are the refractive index of the first and second medium

$\theta_i$ is the angle of incidence (measured between the incident ray and the normal to the surface)

$\theta_r$ is the angle of refraction (measured between the refracted ray and the normal to the surface)

In this problem, light moves into a medium that has lower index of refraction, so

$n_r < n_i$

We can rewrite Snell's law as

$sin \theta_r =\frac{n_i}{n_r}sin \theta_i$

and since

$\frac{n_i}{n_r}>1$

this means that

$sin \theta_r > sin \theta_i$

which implies

$\theta_r > \theta_i$

so, the angle increases.

Also, the speed of light in a medium is given by

$v=\frac{c}{n}$

where c is the speed of light and v the refractive index: we see that the speed is inversely proportional to n, therefore the lower the index of refraction, the higher the speed. So, in this problem, the light will speed up, since it moves into a medium with lower index of refraction.

4 0

3 years ago