According to Newtons first law, an object in motion will stay in motion unles.....

A 1 uF parallel-plate capacitor is charged to 12 V and then disconnected from the voltage supply. The plate separation distance

nlexa [21]

Answer:

22

Explanation:

6 0

3 years ago

Two satellites are in circular orbits around a planet that has radius 9.00×106m. One satellite has mass 68.0 kg, orbital radius

34kurt

Answer: 6782 m/s

Explanation:

Given

Radius of the planet, r = 9*10^6 m

Mass of satellite 1, m1 = 68 kg

Radius of satellite 1, r1 = 6*10^7 m

Orbital speed of satellite 1, vs1 = 4800 m/s

Mass of satellite 2, m2 = 84 kg

Radius of satellite 2, r2 = 3*10^7 m

Orbital speed of satellite 2, vs2 = ?

We know that magnitude of gravitational force, F = (G.m.m•) / r²

Where,

m = mass of satellite

m• = mass of planet

r = radius of orbit

If we consider Newton's second law that states that, F = ma, thus

F(g) = ma(rad)

Where, a(rad) = v²/r

F(g) = mv²/r

Substituting in the initial equation

mv²/r = (G.m.m•) / r²

v² = (G.m•) / r

v = √[G.m•/r]

To find vs2, we first need to find mass of the planet, m• we know that G is a gravitational constant, so we plug in the values

vs1 = √[G.m•/r1]

4800 = √[(6.67*10^-11 * m•) / 6*10^7]

4800² = (6.67*10^-11 * m•) / 6*10^7

2.3*10^7 * 6*10^7 = 6.67*10^-11 * m•

1.38*10^15 = 6.67*10^-11 * m•

m• = 1.38*10^15 / 6.67*10^-11

m• =2.07*10^25 kg

Having found that, we use the value to find our vs2

vs2 = √[(G.m•) / r2]

vs2 = √[(6.67*10^-11 * 2.07*10^25) / 3*10^7]

vs2 = √(1.38*10^15 / 3*10^7)

vs2 = √4.6*10^7

vs2 = 6782.33 m/s

Therefore, the orbital speed of the second satellite is 6782 m/s

4 0

4 years ago

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Plzzz someone help me!!!!.

Vlad [161]

Answer:

He could re read his answear.

Explanation:

4 0

3 years ago

To what pitch of the sound related?

jekas [21]

Hi friend
---------------
Your answer
---------------------

★ Pitch => It is the shrillness or coarseness of the sound.

Higher frequency = Shrill sound

Lower frequency = coarse sound

So, Final answer -

A. Frequency of sound waves

HOPE IT HELPS

6 0

4 years ago

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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

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According to Newtons first law, an object in motion will stay in motion unles.....​

According to Newtons first law, an object in motion will stay in motion unles.....