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cestrela7 [59]
3 years ago
12

The phases of the moon are the changing appearances of the moon, as seen from Earth. Which phase happens immediately after a thi

rd quarter moon?
Physics
2 answers:
Drupady [299]3 years ago
6 0

The phases of the moon are the changing appearances of the moon, as seen from Earth. Which phase happens immediately after a third quarter moon are the following

Explanation:

  • After the full moon (maximum illumination), the light continually decreases. So the waning gibbous phase occurs next. Following the third quarter is the waning crescent, which wanes until the light is completely gone -- a new moon.

waning gibbous phase

  • The waning gibbous phase occurs between the full moon and third quarter phases. The last quarter moon (or a half moon) is when half of the lit portion of the Moon is visible after the waning gibbous phase.

Time takes by the moon to go through all the phases

about 29.5 days

  • It takes 27 days, 7 hours, and 43 minutes for our Moon to complete one full orbit around Earth. This is called the sidereal month, and is measured by our Moon's position relative to distant “fixed” stars. However, it takes our Moon about 29.5 days to complete one cycle of phases (from new Moon to new Moon).
  • At 3rd quarter, the moon rises at midnight and sets at noon. Then we see only a crescent. At new, the moon rises at sunrise and sets at sunset, and we don't see any of the illuminated side!
lisov135 [29]3 years ago
3 0

Answer:

Waning Gibbous

Explanation:

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Four pairs of objects have the masses as described below, along with the distances between
lord [1]

Answer:

<h2>Mass of 1 Kg and 2 Kg, 1 meters apart.</h2>

Explanation:

The gravitational force is defined as

F=G\frac{m_{1} m_{2} }{r^{2} }

By definition, the gravitational force depends directly on the product of the masses and indirectly on the distance between the masses, which means the further they are, the less gravitational force would be. And, the greater the masses, the greater the gravitational force.

Among the options, the pair that would have the greatest gravitational force is  Mass of 1 Kg and 2 Kg, with 1 meter between them.

Notice that the last choice includes the same masses but with a greater distance between them, that means it would be a weaker graviational force.

Therefore, the right answer is the second choice.

7 0
3 years ago
Use the dimensional analysis and check the correctness of given equation:-<br> PV= nRT
Harman [31]

PV=nRT

Here

P=Pressure

V=Volume

n=Molarity

R=universal gas constant

T=Temperature.

LHS

\\ \tt\bull\leadsto PV

\\ \tt\bull\leadsto [ML^2T^{-2}][M^0L^3T^0]

\\ \tt\bull\leadsto [ML^5T^{-2}]

RHS

\\ \tt\bull\leadsto nRT

\\ \tt\bull\leadsto [M^0L^{3}T^0][M^1 L^2 T^{-2}K^{-1}][M^0L^0T^0K^1]

\\ \tt\bull\leadsto [ML^5T^{-2}]

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6 0
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What parts of an atom determines which element it is?
uysha [10]

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4 0
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Read 2 more answers
Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat
Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e \frac{dv}{dt} =10 cm^3/s

Consider r be the radius of the balloon.

The volume of of a sphere is

v=\frac{4}{3} \pi r^3

Differentiate with respect to t

\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}

\Rightarrow 10 =4\pi r^2\frac{dr}{dt}

\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}

The surface of area of the balloon is(S) = 4\pi r^2

S=4\pi r^2

Differentiate with respect to t

\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}

Putting the value of \frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}

\Rightarrow \frac{dS}{dt} =\frac{20}{ r}

Given that r = 5 cm

[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}  =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0
3 years ago
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