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Tasya [4]
3 years ago
13

Half of the moon is always illuminated by the sun. Given that this is true, then what causes the moon to change phases throughou

t the month? Give a detailed explanation, please.
Physics
1 answer:
ollegr [7]3 years ago
8 0

It is true that only half of the moon is always illuminated by the sun.  The reason why the moon changes it phases is because Moon orbits around the Earth, and due to this some areas of the half of the moon will not be lit by the sun.

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ANSWER ASAP!!!!
trapecia [35]
As you increase the temperature, the matter begins to expand. Due to this, the distance between matter particles decreases and they are no more compact. Hence, density decreases. 
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3 years ago
Which statements correctly characterize a lunar eclipse? Select all that apply.
Zepler [3.9K]

A) it occurs when earth is between the sun and the moon

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3 years ago
A rock is rolling down a hill, and it’s halfway down. Would it have both Kintectic engery and Potenial engery? PLEASE HELP
dimaraw [331]

Answer:

Only kinetic.

Explanation:

Potential energy means it has the potential to move. Not something already in motion.

6 0
3 years ago
A camcorder has a power rating of 15 watts. If the output voltage from its battery is 5 volts, what current does it use?
MrMuchimi

Formula

W = E * I

Givens

E = 5 volts

W = 15 watts

I = ?

Solution

W = E * I

15 = 5 * I

15/5 = I

I = 3 amps.   Answer

5 0
3 years ago
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit
lapo4ka [179]

Answer:

The  angle is  \theta  =  15.48^o

Explanation:

From the question we are told that  

     The distance of the dartboard from the dart is  d  =  3.66  \ m

     The time taken is  t =  0.455 \ s

   

The  horizontal component of the speed of the dart is mathematically represented as

      u_x =  ucos \theta

where u is the the velocity at dart is lunched

  so

      distance =  velocity \ in \ the\  x-direction  *  time

substituting values

      3.66 =   ucos  \theta *  (0.455)

 =>   ucos \theta =  8.04  \ m/s

From projectile kinematics the time taken by the dart can be mathematically represented as

         t  =  \frac{2usin \theta }{g}

=>    usin \theta =  \frac{g  * t}{2 }

       usin \theta =  \frac{9.8  * 0.455}{2 }

      usin \theta = 2.23

=>   tan \theta =  \frac{usin\theta }{ucos \theta }  =  \frac{2.23}{8.04}

       \theta  =  tan^{-1} [0.277]

      \theta  =  15.48^o

     

4 0
3 years ago
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