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andrew-mc [135]
3 years ago
11

Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

light to get from the Sun to the Earth (in s) is written in scientific notation. Note: the speed of light is approximately 3.0×108 m/s.
Physics
1 answer:
schepotkina [342]3 years ago
5 0
If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!
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If a 42kg rolling object slows from 11.5m/s to 3.33m/s how much work did friction do
geniusboy [140]

Answer: 1608.39 J

Explanation: Given that the

mass M = 42kg

U = 11.5m/s

V = 3.33m/s

how much work did friction do

Work done = Force × distance

Work done = Ma × distance

But acceleration a = V/t

Work done = M × V/t × d

Work done = M × V × d/t

Where d/t = velocity

Therefore,

Work done = M × U × V

Work done = 42 × 11.5 × 3.33

Work done = 1608.39 J

8 0
3 years ago
A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di
Lyrx [107]

Answer:

D

Explanation:

From the information given:

The angular speed for the block \omega = 50 \ rad/s

Disk radius (r) = 0.2 m

The block Initial velocity is:

v = r \omega \\ \\  v = (0.2  \times 50) \\ \\  v= 10 \ m/s

Change in the block's angular speed is:

\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s

However, on the disk, moment of inertIa is:

I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2

The time t = 10s

∴

Frictional torques by the wall on the disk is:

T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10})  \\ \\ =0.6 \ N.m

Finally, the frictional force is calculated as:

F = \dfrac{T}r{}

F= \dfrac{0.6}{0.2} \\ \\ F = 3N

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