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kati45 [8]
4 years ago
14

In the illustration above the moon is in the position called

Physics
1 answer:
ZanzabumX [31]4 years ago
4 0

Answer:

new moon

Explanation:

  • The new moon is the first lunar phase and is seen as a thin waxing crescent and its shape is formed the direct sunlight as viewed from the earth and stays form 6 am to 6 pm.
  • It varies slightly due to its elliptical orbit and these phases are due to the angles from were the observer can see the moon illumination by the sun as it orbits the planet and the moon reflects the light received from the sun.
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Newton’s third law is the law of
matrenka [14]

Answer:

B. interactive

Explanation:

newtons first law is inertia his second is acceleration and third is interactive.

6 0
3 years ago
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A plane electromagnetic wave varies sinusoidally at 90.0 MHz as it travels along the x direction. The peak value of the electric
mr_godi [17]

Answer:

Explanation:

frequency n = 90 x 10⁶ Hz .

time period T = 1 / n

= 1 / 90 x 10⁶

= 1.11 x 10⁻⁸ S.

wavelength = velocity of light / frequency

= 3 x 10⁸ / 90 x 10⁶

= 3.33 m

maximum value of the magnetic field. ( B₀ )

E₀ / B₀ = c where E₀ and  B₀ are maximum electric and magnetic field .

E₀ / c=  B₀

200/ 3 x 10⁸

= 66.67 x 10⁻⁸ T .

expressions in SI units for the space and time variations of the electric field

E=E_{0y}sin(2\pi nt - \frac{2\pi x}{\lambda} )

E=200sin(180\times 10^6\pi t - \frac{2\pi x}{\lambda} ) N/C

B=B_{0z}sin(2\pi nt - \frac{2\pi x}{\lambda} )

B=66.67\times 10^{-8}sin(180\times 10^6\pi t - \frac{2\pi x}{\lambda} ) T

4 0
3 years ago
Which is the correct equation for calculating the kinetic energy of an object?
Solnce55 [7]

Answer:

KE = 1/2(m)(v^2)

Explanation:

5 0
3 years ago
Which are advantages of reflecting telescopes? Check all that apply.
Mashcka [7]

On e2020 the answers are:

1. There is no rainbow-like halo around the image.

2. Reflecting telescopes can be made very large.

4. Only the reflecting side of the primary mirror needs to be perfectly shaped and smooth.

I hope that helps!!

3 0
3 years ago
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A volume of air increases 0.227 m^3 at a net pressure of 2.07 x 10^7 Pa. How much work is done on the air?
aliya0001 [1]

Answer:

The work done on the air is 4.699 x 10⁶ Joules

Explanation:

Given;

increase in air volume, ΔV = 0.227 m³

net pressure of the air, P = 2.07 x 10⁷ Pa

The work done on the air is given by;

W = PΔV

Where;

W is the work done on the air

P is the net pressure

ΔV  is the increase in air volume

Substitute the given values and solve for work done;

W = (2.07 x 10⁷ Pa) (0.227 m³)

W = 4.699 x 10⁶ Joules

Therefore, the work done on the air is 4.699 x 10⁶ Joules

7 0
3 years ago
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