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kati45 [8]
2 years ago
12

The following diagram shows various positions of the moon in its orbit around Earth. The image of the moon shows its phase as se

en from Earth.
In which position will three-fourths of the illuminated side of the moon be visible from Earth?

A) A
B) B
C) C
D) D

Physics
2 answers:
inysia [295]2 years ago
3 0
<span>Three-fourths of the illuminated side of the moon be visible
from Earth when the moon is at position-B.  (choice-B)


Notice the particularly egregious screw-up on the drawing at the
"Third Quarter" position.  There, the drawing shows the illuminated 
side of the moon AWAY from the sun, and the dark side of the moon
TOWARD the sun.  That's about as silly as you can get.  </span>
kykrilka [37]2 years ago
3 0

Answer:

B) B      

Explanation:

As the moon revolves about the Earth, we see different shapes of the moon everyday.  Phases of moon occur as different portion of illuminated side of the moon is visible from the earth.

According to diagram, at position A, there is full moon phase.

At position C, the illuminated side does not face the earth, so new phase occurs.

At position D, the moon is moving from new moon phase to half moon, so it will be crescent phase.

At position B, the moon is moving from full moon to half moon phase. So at B, it will be in gibbous phase - three-fourths of the illuminated side of the moon would be visible.

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A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.500 m2 .
vesna_86 [32]

Answer:

The energy of the wave is 1.435 x 10⁻⁴ J

Explanation:

Given;

area of the window, A = 0.5 m²

the rms value of the field, E = 0.06 V/m

The peak value of electric field is given by;

E_o = \sqrt{2} *E_{rms}\\\\E_o = \sqrt{2}*0.06\\\\E_o = 0.0849 \ V/m

The average intensity of the wave is given by;

I_{avg} = \frac{c \epsilon_o E_o^2 }{2}\\\\I_{avg} =  \frac{(3*10^8)( 8.85*10^{-12}) (0.0849)^2 }{2}\\\\I_{avg} = 9.569*10^{-6} \ W/m^2

The average power of the wave is given by;

P = I x A

P = (9.569 x 10⁻⁶ W/m²) (0.5 m²)

P = 4.784 x 10⁻⁶ W

The energy of the wave is given by;

E = P x t

E = (4.784 x 10⁻⁶ W)(30 s)

E = 1.435 x 10⁻⁴ J

Therefore, the energy of the wave is 1.435 x 10⁻⁴ J

3 0
2 years ago
Describe how an oscilloscope should be used to measure the frequency of the sound wave from the sonometer
ratelena [41]

Answer:

            T = reading (cm) time base (s / cm)

            f = 1 / T

Explanation:

An oscilloscope is a piece of equipment that allows you to visualize and measure a wave that reaches you, in the case of having a sonometer this transforms the sound wave into an electrical signal to be introduced through one of the voltage channels of the equipment, on the screen we will see the oscillating alternating signal, if it is fixed we can make the reading, if it is moving the time base and the trigger must be adjusted to stop it.

In the oscilloscope we can read the period of the signal, this is the time it takes for the signal to repeat itself with this value, we can calculate the frequency with the formula, for the reading of the period the distance is measured on the labeled screen and multiplied by the time base

         

            T = reading (cm) time base (s / cm)

            f = 1 / T

6 0
2 years ago
Resistance is seen as part of ______________--- how a person understands his or her being and relationship to the world at large
aleksley [76]

Answer:

The self and world construct

Explanation:

in physics

Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.

Self psychology views resistances as protecting a vulnerable self. Resistances are seen as efforts to maintain levels of organization that patients have achieved within the context of their traumatic life situation.

3 0
3 years ago
An incident light ray strikes water at an angle of 20 degrees. The index of refraction of air is 1.0003, and the index of refrac
sammy [17]
In order to determine the angle of the refracted ray, we may apply Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is constant for a given wave when it passes through two different media. Mathematically, this is:
n₁sin(∅₁) = n₂sin(∅₂)
Where n is the refractive index. Substituting the values given into the equation:
1.0003 * sin(20°) = 1.33 * sin(∅)
∅ = 14.91

The angle of the refracted ray is 15°.
4 0
3 years ago
Read 2 more answers
5. The Weeks family took a trip to the Dallas 200 for
zheka24 [161]

Answer: idk

Explanation: hahahaha

4 0
2 years ago
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