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3 years ago

What causes moon phases, rotation or revolution

Physics

1 answer:

pav-90 [236]3 years ago

7 0

Answer:

The Moon's monthly

cycle of phases

results from the

changing angle of

its illumination by the Sun.

Because its period of

revolution is the same as its

period of rotation, the Moon

always keeps the same face

toward Earth.

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Find the volume of a rectangular prism that is 8m long, 4m wide, and 300cm high

konstantin123 [22]

L=8m
B=4m
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$\\ \bull\tt\longmapsto Volume=8(4)(3)$

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2 years ago

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What is cosmic background radiation?

Kobotan [32]

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An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elect

beks73 [17]

Answer:

Speed of the alpha particle is $v=1.8180\times 10^3m/sec$

Explanation:

We have given charge on alpha particle $q=3.2\times 10^{-19}C$

Mass of the alpha particle $m=6.68\times 10^{-27}kg$

Potential difference $V=-3.45\times 10^{-3}volt$

We have to find the speed of the alpha particle

From energy conservation we know that

$\frac{1}{2}mv^2=qV$

$\frac{1}{2}\times 6.68\times 10^{-27}\times v^2=3.2\times 10^{-19}\times 3.45\times 10^{-3}$

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3 years ago

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

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3 years ago

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mart [117]

Answer:children burn calories to being a student

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