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9 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]9 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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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

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A bond between two copper atoms would be a<br> bond? *

mihalych1998 [28]

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A 70 ft rope hangs from a helicopter above this room. The rope has a mass per unit length of 2 lb/ft. In order to be rescued fro

Mrac [35]

Answer:

The work done to get you safely away from the test is 2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope = 2 lb/ft x 70 ft

= 140 lbs.

Total mass to be pulled to the helicopter, M = 120 lbs + 140 lbs

= 260 lbs

The work done is calculated from work-energy theorem as follows;

W = Mgh

where;

g is acceleration due gravity = 32.17 ft/s²

h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s² x 70 ft

W = 585494 lb.ft²/s²

1 lb.ft²/s² = 0.0421 J

W = 585494 lb.ft²/s² = 2.47 X 10⁴ J.

Therefore, the work done to get you safely away from the test is 2.47 X 10⁴ J.

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Delvig [45]

Answer: Globalization is the process of interaction and integration among people, companies, and governments worldwide.

Explanation:

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In the Bronsted-Lowry Theory of acids and bases, an acid is this.

madreJ [45]

Answer: An acid is a substance that donates a proton and produces a conjugate base.

Explanation:

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An example of Bronsted-Lowry acid and base is Ethanoic acid, CH3COOH and hydroxide ion, OH- respectively as shown in the reaction below

CH3COOH(aq) + OH-(aq) <---> CH3COO-(aq) + H2O(l)

Thus, ethanoic acid acts as an acid by donating a proton to the hydroxide ion which accepts it, thus producing ethanoate ion, CH3COO- as a conjugate base.

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