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Citrus2011 [14]
3 years ago
11

Why does the moon appear to wax grow larger and then wane or get smaller

Physics
1 answer:
Umnica [9.8K]3 years ago
5 0
When the moon is waxing it means that the sunlit fraction we can see from earth is getting larger. when it is wanning, the sunlit fraction is getting smaller and even as the phases of the moon change the total amount of sunlight the moon gets remains the same. half the moon is always in sunlight just as half the earth is always in sunlight. but because the period of rotation for the moon is the same as its period of revolution, on earth we always see the same side of the moon. If you lived on the far side of the moon, you would see the sun for half of each lunar day, but you would never see the earth.
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A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at
storchak [24]

Answer

given,

wavelength = λ = 18.7 cm

                    = 0.187 m

amplitude , A = 2.34 cm

v = 0.38 m/s

A)  angular frequency = ?

     f = \dfrac{v}{\lambda}

     f = \dfrac{0.38}{0.187}

     f =2.03\ Hz

angular frequency ,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) the wave number ,

      K = \dfrac{2\pi}{\lambda}

     K= \dfrac{2\pi}{0.187}

    K =33.59\ m^{-1}

C)

as the wave is propagating in -x direction, the sign is positive between x and t

y ( x ,t) = A sin(k  x - ω t)

y ( x ,t) = 2.34  x  sin(33.59 x - 12.75 t)

4 0
3 years ago
According to the theory of plate tectonics, the lithosphere is separated into sections that are called tectonic plates. These pl
loris [4]

Answer:

Transform :j

Explanation:

4 0
4 years ago
Estimate the average power output of the sun, given that about 1350 w/m2 reaches the upper atmosphere of the earth.
fenix001 [56]

the <u>average powe</u><u>r</u> output of the sun is <u>3.8 × </u>10^{26}<u> W</u> given that about 1350 w/m2 reaches the upper atmosphere of the earth.

As we know intensity falling over a surface is equal to Energy falling per unit area per unit volume.

I = \frac{Energy}{area*time}

Since Power is equal to energy/time

I = power/area

or

Power = Intensity × area

Now, Distance from Sun to Earth , r = 149.6 ×10^{9}  m

So, Surface area of the sphere of radius is:

A = 4 π r^{2}

= 4 × 3.14 × (149.6 × 10^{9})²

=  2.81 × 10^{23}m²

Thus Average Power output of the sun will be:

P = 1350 × 2.81 × 10^{23}

= 3.7935 × 10^{26}

P = 3.8 ×    10^{26}  W

So the average power output of the sun is 3.8 × 10^{26} W.

If you need to learn more about about the average power of a resistor, click here.

brainly.com/question/12972958

#SPJ4

7 0
2 years ago
Read 2 more answers
A friend claims that throwing a baseball up towards the school roof illustrates gravitational potential energy transforming into
Cloud [144]

Answer:

The Statement is wrong because the reverse is the case as it is the  kinetic energy that is being transformed to gravitational potential energy.

Explanation:

As your friend throws  the baseball into the air the ball gains an initial velocity (u) and this makes the Kinetic energy to be equal to

                    KE = \frac{1}{2} mu^2

Here  m is the mass of the baseball

       Now as this ball moves further upward the that velocity it gained reduce due to the gravitational force and this in turn reduces the kinetic energy of the ball and this kinetic energy lost is being converted to gravitational potential energy which is mathematically represented as (m×g×h)

as energy can not be destroyed but converted to a different form according to the first law of thermodynamics

Looking a the formula for gravitational potential energy we see that the higher the ball goes the grater the gravitational potential energy.

4 0
4 years ago
Why the earth is not in thermal equilibrium with the sun?
mr_godi [17]
Changes in the earths ability to reflect sunlight (called the 'albeido') due to clouds and other weather like ice formations set up feedback loops that prevent a perfect equlibrium with the sun.
4 0
4 years ago
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