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3 years ago

During which lunar phase is the moon's far side entirely lit

Physics

2 answers:

Alla [95]3 years ago

7 0

The phase where the moons far side is entirely lit is called the full moon

yulyashka [42]3 years ago

4 0

During which lunar phase is the moon's far side entirely lit?

"Near side entirely lit" means we see the whole side facing us. The moon looks like a complete circle.

Hope This Helped! :3

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31. Draw a free body diagram for a 15.5N box that is being pushed to the right with a 18. N force while experiencing 4.30 N of r

posledela

Answer:

See answers below

Explanation:

F = mg,

15.5 N = m(9.8 m/s²)

m = 1.58 kg

Fnet = Applied force - resistance,

Fnet = 18 N - 4.30 N,

Fnet = 13.70 N

Fnet = ma

13.70 N = (1.58 kg)a

a = 8.67 m/s²

For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.

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2 years ago

As a gaseous element condenses, the atoms become ________ and they have ________ attraction for one another.

romanna [79]

Answer:

blank 1: close together

blank 2: more

3 0

1 year ago

HELP PLS

melisa1 [442]

Answer:

red I think

Explanation:

it's on red so I googled some of it and the closest was red

4 0

3 years ago

Barney walks at a velocity of 1.7 meters/second on an inclined plane, which has an angle of 18.5° with the ground. What is the h

Viktor [21]

The velocities and the speed build a triangle, where the 1.7 m/s are the hypotenuse and the x-velocity and y-velocity are the other sides.

So the x-velocity is: speed*cos(angle) 

now plug in 
x=1.7 m/s * cos(18.5)=1.597 m/s

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3 years ago

Two cars cover the same distance in a straight line. Car A covers the distance at a constant velocity. Car B starts from rest an

SpyIntel [72]

Answer:

a) 2.43 m/s

b) 4.83 m/s

c) 0.023 m/s²

Explanation:

a) Both cars cover a distance of 510 m in 210 s. Since car A has no acceleration

Speed = Distance / Time

$\text{Speed}=\frac{510}{210}=2.43\ m/s$

Velocity of car A is 2.43 m/s

t = Time taken = 210 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 510 m

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 510=0\times 210+\frac{1}{2}\times a\times 210^2\\\Rightarrow a=\frac{510\times 2}{210^2}\\\Rightarrow a=0.023\ m/s^2$

Acceleration of car B is 0.023 m/s²

$v=u+at\\\Rightarrow v=0+0.023\times 210\\\Rightarrow v=4.83\ m/s$

Final velocity of car B is 4.83 m/s

6 0

3 years ago