All categories

3 years ago

Why does the moon go through phases?

Physics

2 answers:

givi [52]3 years ago

8 0

Answer:

The Moon has phases because it orbits Earth, which causes the portion we see illuminated to change.

Explanation:

The Moon takes 27.3 days to orbit Earth, but the lunar phase cycle (from new Moon to new Moon) is 29.5 days. ... As the Sun sets, the Moon rises with the side that faces Earth fully exposed to sunlight (5).

Anit [1.1K]3 years ago

4 0

Answer: it has phase because it orbits and rotates causing the sunlight to hit different parts of the moon making it light up a certain way

Explanation:

You might be interested in

As the speed of a fluid increases, ____.

Fofino [41]

The correct answer for the question that is being presented above is this one: "a. the pressure decreases." As the speed of a fluid increases, the pressure decreases." The relationship of the speed and the pressure is inversely proportional. As the pressure increases, the speed decreases.

3 0

3 years ago

Read 2 more answers

What is black matter?

Musya8 [376]

I think you meant what is "Dark matter" if so the answer is Dark matter is a hypothetical form of matter that is thought to account for approximately 85% of the matter in the universe.

5 0

3 years ago

At which location would an objects weight be the greatest. 1.pluto 2.earth 3.sun 4.moon

Maslowich

Weight of the object is given as

$Weight = mg$

so weight of the object will be maximum where gravity(g) will be maximum

so here out of all given four the gravity "g" will be maximum at the surface of SUN as we know that sun will be of maximum mass out of all four planets

as we know that

$g = \frac{GM}{R^2}$

so correct answer will be

3. SUN

8 0

4 years ago

Read 2 more answers

An iron wire has length 8.0m and a diameter 0.50mm. The sir has a resistance R.

Rudik [331]

The re<span>sistance of the second wire is 16 R.
where R is the resistance of the first wire.

R = </span>ρ $\frac{l}{A}$
where l = length of the wire
A = area of the wire
A = $\pi r^{2}$ where, r = $\frac{diameter of wire}{2}$

Thus, on finding the ratio of resistance of the two wires, we get,

$\frac{R1}{R2} = \frac{l1A2}{l2A1}$

here, R1 = R
l1 = 8m
l2 = 2m
A1=π $0.25^{2}$
A1=π $0.50^{2}$

we get. R2 = 16R

7 0

4 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago