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Vedmedyk [2.9K]
3 years ago
11

Select the Moon and use the Info view to determine which of the following statements is correct. The Last Quarter Moon.... rises

near noon and sets near midnight. rises at about 6am and sets at about 6pm. rises near midnight and sets near midday. rises and sets at the same time as the Sun. Submit Your Answer
Physics
1 answer:
Ann [662]3 years ago
5 0

Answer: The correct statement is that the Last Quarter Moon (rises near midnight and sets near midday).

Explanation:

Phases of the moon also called the LUNAR PHASE can be defined as the different shades of illumination on the moon as seen from the earth. The moon is the natural satellite of the earth that illuminates upon reflection of light from the sun. This means it doesn't have power to shine on its own. When carefully observed, there are times it gets dark and beings to glow brighter over a period of time. This occurs because as the moon completes its four weeks lunar cycle round the earth, how much of its face we see illuminated by sunlight depends on the angle the Sun makes with the Moon.

There are 8 main types of the moon phases these includes:

--> New moon: This is when the moon is not visible to the earth because it's between the earth and the sun. It rises at sunrise and sets at sunset.

--> The waxing crescent: At this phase the moon gets brighter and illuminated from the sun that a crescent shape is seen.

--> First quarter: this occurs one week after the new moon. The moon rises at noon and sets at midnight.

--> The waxing gibbous: This occurs after the first quarter phase where more than half of the lit part of the moon is seen.

--> Full moon: This occurs when the moon and the sun are opposite each other. That is way it is said to rise at sunset and sets at sun rise.

--> The waning gibbous: this occurs when more than half of the lit part of the moon gradually becomes darker

--> Third quarter ( Last Quarter): The moon rises at midnight and sets at noon. This occurs a week after the full moon.

--> The waning crescent: This occurs after the last quarter phase where a very thin fading crescent shaped moon is seen, just before the Moon is invisible again at the start of the cycle, the new moon.

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Which factor is the most imortant in performing the activities of daily living
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Answer:

Muscular endurance is the most important factor in performing the activities of daily living.

7 0
3 years ago
Harry Potter is chasing his nemesis Draco Malfoy during a quidditch match. Initially, Harry is 35m behind Draco, who has just sp
Sophie [7]

Answer:

the acceleration of harry is equal to 66.126 m/s²

Explanation:

given,

harry is 35 m behind Draco

speed of Draco = 40 m/s

original speed of harry = 50 m/s

acceleration = ?

time taken by the Draco

    t =\dfrac{r}{u} =\dfrac{75}{40}

     t = 1.875 s

distance covered by Harry

  d = 35 + 175 = 210 m

to calculate the acceleration of harry

s = u t+ \dfrac{1}{2}at^2

210 = 50\times 1.875+ \dfrac{1}{2}\times a\times 1.875^2

a × 3.516 × 0.5 = 116.25

a = 66.126 m/s²

hence, the acceleration of harry is equal to 66.126 m/s²

3 0
3 years ago
The mass of a spacecraft is about 435 kg . An engine designed to increase the speed of the spacecraft while in outer space provi
svlad2 [7]

Answer:

Δ v =  125 m/s

Explanation:

given,

mass of space craft = 435 Kg

thrust = 0.09 N

time = 1 week

       = 7 x 24  x 60 x 60 s

change in speed of craft = ?

Assuming no external force is exerted on the space craft

now,

T= m_s a

a=\dfrac{T}{m_s}

a =\dfrac{0.09}{435}

a = 2.068 x 10⁻⁴ m/s²

using equation of motion

Δ v = a t

Δ v = 2.068 x 10⁻⁴ x 7 x 24 x 60 x 60

Δ v =  125 m/s

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3 years ago
An airplane is flying overhead at a constant elevation of 3000ft. A man is viewing the plane from a position 4000ft from the bas
ziro4ka [17]

Answer: The distance between the man and the plane increasing at a rate of 400ft/s

Explanation: Please see the attachments below

4 0
3 years ago
A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image
Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}

=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}

Re-arrange to get i

\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

m=\frac{h'}{h} =-\frac{i}{o}

substitute the values to get the height of image h'

\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm

5 0
3 years ago
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