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lilavasa [31]
3 years ago
9

University Physics 9.68: Accelerating Compact Disc. A computer disc drive is turned on starting from rest and has constant angul

ar acceleration. If it took 0.0865 s for the drive to make its second complete revolution,
(a) how long did it take to make the first complete revolution?
(b) what is its angular acceleration, in rad/s2 ?
Physics
1 answer:
faust18 [17]3 years ago
5 0

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Definition of force equation
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Which of the following is not a property of displacement?
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Your correct answer is option C. Displacement is equal to the total distance traveled by an object in motion.

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You observe three carts moving to the left. Cart A moves to the left at nearly constant speed. Cart B moves to the left, gradual
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Answer:cart B

Explanation:

For cart A speed is constant therefore there is no acceleration because acceleration is rate of change of velocity

thus there is no net force

For cart B there is change in velocity in the left direction , so there is net acceleration towards left

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so there is net force in the left direction

For cart C there is decrease in velocity i.e. negative acceleration or deceleration . Therefore there is a net force towards right which opposes the motion                

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When you look at a transverse wave, the distance from the baseline to the crest of the wave is called the wave's ___________.
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2 years ago
Energy from the Sun arrives at the top of the Earth's atmosphere with an intensity of 1.36 kW/m 2 . How long does it take for 3.
alina1380 [7]

Answer:

The time taken by 3.55\times 10^9\ J to arrive on an area of 4.25\ m^2 is 6.14\times 10^5\ seconds.

Explanation:

Given the intensity of the energy is 1.36\ kW/m^2

And the arriving energy is 3.55\times 10^9\ J

Also, the area in which energy is being arriving is 4.25\ m^2

Now, we will use relation between energy (E), intensity of energy (p), area (A) and time (T).

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The equation is

E=pAT\\\\T=\frac{E}{pA}\\\\T=\frac{3.55\times 10^9}{1.36\times 10^3\times 4.25}\\\\T=0.614\times 10^6\ s\\T=6.14\times 10^5\ seconds

So, the time taken by 3.55\times 10^9\ J to arrive on an area of 4.25\ m^2 is 6.14\times 10^5\ seconds.

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3 years ago
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