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jasenka [17]
3 years ago
6

This is the change in kinetic energy of a system in which a 16 kg object moving at 25 m/s slows to a velocity of 20 m/s

Physics
1 answer:
Dennis_Churaev [7]3 years ago
5 0

The kinetic energy of an object is given by

KE = 0.5mv²

where m is the mass and v is the velocity.

To calculate the change in kinetic energy...

Initial KE:

KEi = 0.5mVi²

where Vi is the initial velocity.

Final KE:

KEf = 0.5mVf²

where Vf is the final velocity.

ΔKE = KEf - KEi

ΔKE = 0.5mVi² - 0.5mVf²

ΔKE = 0.5m(Vf²-Vi²)

Given values:

m = 16kg

Vi = 25m/s

Vf = 20m/s

Plug in the given values and solve for ΔKE:

ΔKE = 0.5×16×(20²-25²)

ΔKE = -1800J

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What is the mean of 22, 72, 79, 72, and 70
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Answer:

63

Explanation:

You first have to add all the numbers together.

22+72+79+72+70 = 315

You divide the total by the amount of numbers (5)

315/5 = 63

The mean is 63

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The two pucks of equal mass did not move linearly (they came to a stop) after the collision due to the conservation of linear mo
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Compared to the pucks given, the pair of pucks will rotate at the same rate.

Answer: Option A

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By twice the mass yet keeping the speeds unaltered, also twice the angular momentum's to the two-puck framework.  Be that as it may, we likewise double the moment of inertia. Since L=I \times \omega, the turning rate of the two-puck framework must stay unaltered.

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3 years ago
What is your least favorite candy? Explain
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A vocalist with a bass voice can sing as low as 92 Hz.
Inessa05 [86]

Answer:

  • 3.26 x 10 to the power of 6

Explanation:

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5 0
3 years ago
Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long
Ksivusya [100]

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

<u>A) Determine how long the clock run on the battery. use the relation below</u>

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

8 0
3 years ago
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