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1 year 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]1 year 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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A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the

olga2289 [7]

Answer:

a) F = 30 N, b) I = 12 N s , c) I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

p₀ = m v

The speed can be found by kinematics

v = v₀ - g t

v = 0 - 10 0.4

v = -4.0 m / s

Final after division

pf = m₁ v₁f + m₂ v₂f

p₀ = pf

M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

F = mg

F = 3 10 = 30 N

b) The expression for momentum is

I = Ft

Before the explosion the only force that acts is the weight

I = mg t

I = 3 10 0.4

I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

M v = m₁ v₁f + m₂ v₂f

v₂f = (M v - m₁ v₁f) / m₂

v₂f = (3 (-4) - 1 6) / 2

v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

I = F t = Dp

F t = pf –po)

F t= [m₁ v₁f + m₂ v₂f]

I = [1 6 + 2 (-9) -0]

I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

ΔI =If – I₀

ΔI = - 12 – (-12)

ΔI = -0 N s

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3 years ago

A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp

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Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

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