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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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A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the s
leonid [27]

Complete question is;

A force stretches a wire by 0.60 mm. A second wire of the same material has the same cross section and twice the length.

a) How far will it be stretched by the same force?

b) A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force?

Answer:

0.15 mm

Explanation:

According to Hooke's Law,

E = Stress(σ)/Strain(ε)

Where E is youngs modulus

Formula for stress is;

Stress(σ) = Force(F)/Area(A)

Formula for strain is;

Strain(ε) = Change in length/original length = (Lf - Li)/Li

We are also told that a second wire of the same material has the same cross section and twice the length.

Thus;

Rearranging Hooke's Law to get the constants on one side, we have;

F/(AE) = ε

Thus from the conditions given;

ε1 = 0.6/Li

ε2 = (Change in length)/(2*Li)

And ε1 = ε2

Thus;

0.6/Li = Change in length/(2*Li)

Li will cancel out and we now have;

Change in length = 2 × 0.6 = 1.2 mm

Finally, we are told A third wire of the same material has the same length and twice the diameter as the first.

Area of a circle;A1 = πd²/4

Now, we are told d is doubled.

Thus, new area of the new circle is;

A2 = π(2d)²/4 = πd²

Rearranging Hooke's Law,we have;

F/A = εE

Since F and E are now constants, we have;

F/E = constant = Aε

Thus;

A1(ε1) = A2(ε2)

A1 = πd²/4

e1 = 0.60/Li

A2 = πd²

e2 = Change in length/Li

Thus;

((πd²/4) × 0.6)/Li = (πd² × Change in length)/ Li

Rearranging, Li and πd² will cancel out to give;

0.6/4 = Change in length

Change in length = 0.15 mm

4 0
2 years ago
How do you determine the specific heat of a substance?
svetlana [45]

Answer:

I think it's A

Explanation:

3 0
2 years ago
A mover pushes a 46.0kg crate 10.3m across a rough floor without acceleration. How much work did the mover do (horizontally) pus
Alchen [17]

Answer:

<h3>2,321.62Joules</h3>

Explanation:

The formula for calculating workdone is expressed as;

Workdone = Force * Distance

Get the force

F = nR

n is the coefficient of friction = 0.5

R is the reaction = mg

R = 46 ( 9.8)

R = 450.8N

F = 0.5 * 450.8

F = 225.4N

Distance = 10.3m

Get the workdone

Workdone = 225.4 * 10.3

Workdone  = 2,321.62Joules

<em>Hence the amount of work done is 2,321.62Joules</em>

3 0
3 years ago
Wood is an example of <br> A. Metalloid<br> B. Insulator<br> C. Nonmetal<br> D. Conductor
Colt1911 [192]
The Answer is b: insulator
4 0
2 years ago
Two cylinders with the same mass density rhoC = 713 kg / m3 are floating in a container of water (with mass density rhoW = 1025
Tanzania [10]

Answer:

Explanation:

Given

density of cylinder is \rho _c=713 kg/m^3

Length of first cylinder is L_1=20 cm

radius r_1=5 cm

For cylinder 2 L_2=10 cm

r_2=10 cm

h_1 and h_2 are the height above water

E

as object is floating so its weight must be balanced with buoyant force

\rho _c\frac{\pi }{4}d_1^2L_1g=\rho _w\frac{\pi }{4}d_1^2(L_1-h_1)g----1

For 2nd cylinder

\rho _c\frac{\pi }{4}d_2^2L_2g=\rho _w\frac{\pi }{4}d_2^2(L_2-h_2)g----2

Dividing 1 and 2 we get

\frac{L_1}{L_2}=\frac{L_1-h_1}{L_2-h_2}

\frac{20}{10}=\frac{20-h_1}{10-h_2}

2h_2=h_1

\\\Rightarrow\frac{h_2}{h_1}=\frac{1}{2}                            

5 0
3 years ago
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