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maksim [4K]
3 years ago
13

Which of the following describes the bending of light due to a change in its speed?

Physics
2 answers:
Akimi4 [234]3 years ago
5 0
It is refraction because if you look in a glass of water with a straw, it looks as if there are 2 straws.

Artyom0805 [142]3 years ago
3 0
I think the answer is reflection but I might be wrong
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What is 16.558 m/s rounded to three significant figures?
In-s [12.5K]

Answer:

Option C. 16.6 m/s

Explanation:

To round this 16.558 m/s to 3sf, we need to count the number beginning from 1. When we get to the 3rd number( ie 5), we'll examine the fourth number(i.e 5)to see if it less than five or greater. If it less than five, then we'll discard it. But if it five or greater, we'll approximate it and add it to the 3rd number.

So.

16.558 m/s = 16.6m/s to 3sf

3 0
4 years ago
!!PLEASE ANSWER ASAP!!
ikadub [295]
Work done = force * distance
2m * 9.8
work done = 19.6 J
5 0
3 years ago
What changes a ball velocity
Grace [21]

Answer/Explanation: Speed and direction can change with time. When you throw a ball into the air, it leaves your hand at a certain speed. As the ball rises, it slows down. Then, as the ball falls back toward the ground, it speeds up again. When the ball hits the ground, its direction of motion changes and it bounces back up into the air.

5 0
4 years ago
A satellite of mass 5600 kg orbits the Earth and has a period of 6200 s.
Troyanec [42]

Answer:

(a)  Radius of orbit will be =7.32\times10^6m

(b) Earth gravitational force will be =4.18\times 10^4N

(C) Height will be 0.92\times 10^6m

Explanation:

We have given

Mass of the earth, M=6\times 10^{24}kg

Mass of the satellite, m = 5600 kg

Radius of earth, R=6.4\times 10^6m

Time period T = 6200 sec

We know that \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{6200}=0.00101rad/sec

Now

(a) We know that \omega ^2=\frac{GM}{R^3}

R^3=\frac{GM}{\omega ^2}  

R^3=\frac{6.67\times 10^{-11}\times 6\times 10^{24}}{0.00101 ^2}

R^3=3.92\times 10^{20}

Radius of the orbit R=7.32\times 10^6m

(b)

Force F=\frac{GMm}{R^2}=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 5600}{(7.32\times 10^6)^2}=4.18\times 10^4N

(c)

Altitude h=radius\ of\ orbit-radius\ of\ earth=7.32\times 10^6-6.4\times 10^6=0.92\times 10^6m

8 0
3 years ago
What is the potential energy of a 2500 g object suspended 5 kg above the earth's surface?
Alinara [238K]

Answer:

potential \: energy = mgh \\  = ( \frac{2500}{1000} ) \times 10 \times 5 \\  = 125 \: newtons

if height is 5 m

8 0
3 years ago
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