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5 months ago

The weight of air in a column 1-m2 in cross section that extends from sea level to the top of the atmosphere is?

Physics

1 answer:

Sauron [17]5 months ago

3 0

From sea level to the top of the atmosphere, a column of air with a 1-m2 cross section weighs 101,000 N.

To find the answer we have to know about the pressure.

<h3>What is Pressure?</h3>

The pressure at a point on a surface is the thrust acting per unit area around that point.

$Pressure, P=\frac{Thrust}{Area}=\frac{F}{A}$

A particular mass of air is contained in a column that rises from the ocean to the top of the atmosphere. Then,

$P=P_a=1.013*10^5 pascal$

Pa is the atmospheric pressure at the sea level.

By combining both the equations, we get the weight of air,

$\frac{F}{A}=P_a$

$F=W=P_a*A=1.013*10^5*1=101300N$

Thus, we can conclude that, the weight of air in a column 1-m2 in cross section that extends from sea level to the top of the atmosphere is 101300N.

Learn more about the pressure here:

brainly.com/question/24818722

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Bill and Nageen have each built an electric motor and they are testing them by having them lift boxes. Bill's motor lifts a box

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Answer:

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=> 0.194F > 0.175F => Nageen's motor applied more power to the box than Bill's motor.

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A plucked guitar string produces a sound wave with a wavelength of 0.15 m and a velocity of 10.5 m/s, what is the frequency of t

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2 years ago

An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode

sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell $v_{0}$ in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

$v_{x}=v_{0} \times \cos \theta$

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of $v_{x}$ and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

$v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}$

$v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}$

For motion with constant acceleration, we know

$s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}$

Along the horizontal, x-axis, we might write this as

$x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}$

Measuring distances relative to the firing point means

$x_{0}=0$

we know that,

$a_{x}=0$

or,

$v_{x}=v_{x 0}=\text { constant }$

By applying the values, we get,

$x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}$

The acceleration of gravity is vertically downward and is $g=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

$y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}$

we know, $y_{0}=0$ and $a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , so,

$y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)$

y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0

2 years ago