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Sauron [17]
3 years ago
13

Write one general properties of non contact force​

Physics
1 answer:
Gekata [30.6K]3 years ago
3 0

Answer:    Gravitational force between the two masses does not depend on the medium separating two masses.

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Ursula is a student of Kantian philosophy who argues that we should never use people as a means to an end unless
ANTONII [103]

Answer: we embrace and are simultaneously respecting their rational autonomy

Explanation:

Ursula is a student of Kantian philosophy who argues that we should never use people as a means to an end unless we embrace and are simultaneously respecting their rational autonomy

8 0
4 years ago
Which description BEST explains the direction of the moving force of air?
marin [14]
Out of the choices given, the best choice to explain the direction of the moving force of air is from area o high pressure to areas of low pressure. 

3 0
3 years ago
Which of the following BEST demonstrates a change in weight? *
Radda [10]

Answer:

A person changes from 100 kg to 75 kg after dieting

Explanation:

3 0
3 years ago
Read 2 more answers
A mercury barometer reads 745.0 mm on the roof of a building and 760.0 mm on the ground. Assuming a constant value of 1.29 kg/m3
Ipatiy [6.2K]

Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

\Delta P \propto \rho \cdot \Delta h

\Delta P = k \cdot \rho \cdot \Delta h

Where:

\Delta P - Manometric pressure difference, measured in kilopascals.

\rho - Fluid density, measured in kilograms per cubic meter.

\Delta h - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}

\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}

Where:

\Delta h_{air} - Height difference of the air column, measured in meters.

\Delta h_{Hg} - Height difference of the mercury column, measured in meters.

\rho_{air} - Density of air, measured in kilograms per cubic meter.

\rho_{Hg} - Density of mercury, measured in kilograms per cubic meter.

If \Delta h_{Hg} = 0.015\,m, \rho_{air} = 1.29\,\frac{kg}{m^{3}} and \rho_{Hg} = 13600\,\frac{kg}{m^{3}}, the height difference of the air column is:

\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)

\Delta h_{air} = 158.140\,m

The height of the building is 158.140 meters.

5 0
3 years ago
Read 2 more answers
A plane traveling horizontally at 120 ​m/s over flat ground at an elevation of 3610 m must drop an emergency packet on a target
antoniya [11.8K]

Answer:

Explanation:

Horizontal displacement

x = 120 t

Vertical position

y = 3610 - 4.9 t²

y = 0 for the ground

0 = 3610 - 4.9 t²

t = 27.14 s

This is the time it will take to reach the ground .

During this period , horizontal displacement

x = 120 x 27.14 m

= 3256.8 m

So packet should be released 3256.8 m before the target.

8 0
3 years ago
Read 2 more answers
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