All categories

11 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]11 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

#SPJ4

You might be interested in

. An object has a position given by ~r(t) = [3.0 m − (4.00 m/s)t]ˆı + [6.0 m − (8.00 m/s2 )t 2 ]ˆ , where all quantities are in

kupik [55]

Answer:

Explanation:

The position is $r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}$ .

The velocity is the first time-derivative of r(t).

$v(t) = \dfrac{d}{dt}r(t) = -4.00\,\hat{i} -16t\,\hat{j}$

The acceleration is the first time-derivative of the velocity.

$a(t) = \dfrac{d}{dt} v(t) = -16\hat{j}$

Since a(t) does not have the variable t, it is constant. Hence, at any time,

$a = -16\hat{j}$

Its magnitude is 16 m/s².

4 0

2 years ago

Consider a roller coaster begins 15m above the ground. If the cart has a mass of 75kg, what is the velocity of the cart halfway

SashulF [63]

Answer:

v = 12.12 m/s

Explanation:

Given that,

The mass of the cart, m = 75 kg

The roller coaster begins 15 m above the ground.

We need to find the velocity of the cart halfway to the ground. Let the velocity be v. Using the conservation of energy at this position, h = 15/2 = 7.5 m

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 7.5} \\\\v=12.12\ m/s$

So, the velocity of the cart is 12.12 m/s.

7 0

2 years ago

A car driving at a constant speed of 64 mi/h travels 68 miles. How many hours did this take?

babymother [125]

64 miles/hour
Therefore 1/64 hours/mile
68 miles * 1/64 hours/mile (notice how miles cancels out)
Therefore the answer is 68/64 hours = 1.0625 hours = 1 hour 3min and 45sec.

3 0

3 years ago

What percentage of the original kinetic energy is convertible to internal energy?

schepotkina [342]

There would be very less percentage loss of the kinetic energy during the conversion to internal energy, assuming that there is less air in the surroundings. Also, the friction will contribute to the conversion where if it is, the percentage loses is negligible.

7 0

3 years ago

Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the a

Naily [24]

Answer:

The angular acceleration of the pencil α = 17 rad·s⁻²

Explanation:

Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:

τ = I α (1)

W r = I α (2)

The weight is that the pencil has is,

sin 10 = r / (L/2)

r = L/2(sin(10))

The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:

I = 1/3 M L²

Thus,

mg(L / 2)sin(10) = (1/3 m L²)(α)

α(f) = 3/2(g) / Lsin(10)

α = 3/2(9.8) / 0.150sin(10)

α = 17 rad·s⁻²

Therefore, the angular acceleration of the pencil is 17 rad·s⁻²

3 0

2 years ago