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3 years ago

It's a cloudy and rainy day. The air pressure is most likely _____.

1 answer:

kramer3 years ago

5 0

The air pressure ( or atmospheric pressure ) is the force of air over a unit of area. Changes in the air pressure causes the weather changes. High pressure usually brings good weather with dry and cool air. But in a low pressure zone warm air is rising up. This vertical movements are caused by winds high in the troposphere. Water molecules stay as a gas in warmer air. After the vertical movement they condense and bring steady continuous rain. Therefore the low pressure brings cloudly and rainy weather. Answer: The air pressure is most likely low<span>. </span>

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1)Pitch is measure of the ____________ of sound.

amid [387]

Answer:

1) wave length

2) parallel reflection

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2 years ago

Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis

WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{Mm}{r^{2}}$ (1)

Where:

$F$ is the gravitational force between Earth and Moon

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$m=7.349(10)^{22} kg$ is the mass of the Moon

$r=3.9(10)^{8} m$ is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to $F$ :

$F=m.a_{C}$ (2)

Where $a_{C}$ is the centripetal acceleration given by:

$a_{C}=\frac{V^{2}}{r}$ (3)

Being $V$ the orbital velocity of the moon

Making (1)=(2):

$m.a_{C}=G\frac{Mm}{r^{2}}$ (4)

Simplifying:

$a_{C}=G\frac{M}{r^{2}}$ (5)

Making (5)=(3):

$\frac{V^{2}}{r}=G\frac{M}{r^{2}}$ (6)

Finding $V$ :

$V=\sqrt{\frac{GM}{r}}$ (7)

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}}$ (8)

Finally:

$V=1010.92 m/s$

5 0

3 years ago

Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po

gregori [183]

Answer:

E = 0 r <R₁

Explanation:

If we use Gauss's law

Ф = ∫ E. dA = $q_{int}$ / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

E = 0 r <R₁

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3 years ago

Galaxies are separated by?

Inessa [10]

It is number c I think

8 0

3 years ago

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