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2 years ago

what is the mass of vertical column of air that exerts pressure of 1 atm over a10m by 10m square surface

Physics

1 answer:

Oksana_A [137]2 years ago

4 0

Answer:

$m=1.01\times 10^6\ kg$

Explanation:

Given that,

Pressure, P = 1 atm = 101325 Pa

Area of the square surface, A = 10² = 100 m²

We need to find the mass of vertical column of air. We know that, pressure is equal to the force acting per unit area. So,

$P=\dfrac{mg}{A}\\\\m=\dfrac{PA}{g}\\\\m=\dfrac{101325\times 10^2}{10}\\\\m=1.01\times 10^6\ kg$

So, the required mass of the vertical column of air is $1.01\times 10^6\ kg$ .

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Difference between diurnal and annual motion in two points

Valentin [98]

Answer:

Explanation below:

Explanation:

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3 years ago

Mr. Ben drove from Town A to Town B. For the first 3 h, he traveled at an average speed of

Vanyuwa [196]

Answer:

It’s 7 hours

Explanation:

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3 years ago

An object whose specific gravity is 0.850 is placed in water. What fraction of the object is below the surface of the water?

Fynjy0 [20]

Answer:

The fraction of the object that is below the surface of the water is ¹⁷/₂₀

Explanation:

Given;

specific gravity of the object, γ = 0.850

Specific gravity is given as;

$specific \ gravity = \frac{density \ of the \ object}{density \ of \ water}\\\\0.85= \frac{density \ of the \ object}{1000 \ kg/m^3} \\\\density \ of the \ object = 850 \ kg/m^3$

Fraction of the object's weight below the surface of water is calculated as;

$= \frac{850}{1000} \ \times\ 100\%\\\\= 85 \% \\\\= \frac{17}{20}$

Therefore, the fraction of the object that is below the surface of the water is ¹⁷/₂₀

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3 years ago

A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin

Mashcka [7]

Answer:

$r_{cm}=[12.73,12.73]cm$

Explanation:

The general equation to calculate the center of mass is:

$r_{cm}=1/M*\int\limits {r} \, dm$

Any differential of mass can be calculated as:

$dm = \lambda*a*d\theta$ Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

$\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}$

So, the differential of mass is:

$dm = \frac{2M}{a\pi}*a*d\theta$

$dm = \frac{2M}{\pi}*d\theta$

Now we proceed to calculate X and Y coordinates of the center of mass separately:

$X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta$

$Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta$

Solving both integrals, we get:

$X_{cm}=2*a/\pi=12.73cm$

$Y_{cm}=2*a/\pi=12.73cm$

Therefore, the position of the center of mass is:

$r_{cm}=[12.73,12.73]cm$

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2 years ago

How much heat does it take to raise the temperature of 10.0 kg of water by 1.0 C?

fomenos

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2 years ago