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Ghella [55]
3 years ago
14

HOW TO ANSWER THIS QUESTION PLEASE ​

Physics
2 answers:
Over [174]3 years ago
5 0

Answer:

  • Radioactive decay is a long proxy
  • Use long tongs

↔ <em>for</em><em> </em><em>safety</em><em> </em><em>precautions</em><em> </em><em>the</em><em> </em><em>scientist</em><em> </em><em>should</em><em> </em><em>used</em><em> </em><em>this</em><em> </em><em>upon</em><em> </em><em>working</em><em> </em><em>with</em><em> </em><em>radioactive</em><em> </em><em>materials</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>laboratory</em><em>.</em>

Zigmanuir [339]3 years ago
4 0

Answer:

  • radioactive decay is a random process
  • use long tongs

Explanation:

Quantum theory has taught us that virtually everything to do with the nature of matter in the universe is a random process. The decay of atomic nuclei is something that makes that randomness "visible" with appropriate instrumentation. As such, the rate of decay will vary over the "short" term. Consequently, it is not surprising to see a variation of a few counts in an activity measurement over a period of a few minutes.

__

Safety with respect to radioactive materials involves staying as far away from them as possible. The use of long tongs for handling would be appropriate.

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The equation is balanced as it is.
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fomenos

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An 92-kg football player traveling 5.0m/s in stopped in 10s by a tackler. What is the original kinetic energy of the player? Exp
Artemon [7]

Explanation:

It is given that,

Mass of the football player, m = 92 kg

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(1) We need to find the original kinetic energy of the player. It is given by :

k=\dfrac{1}{2}mv^2

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In two significant figure, k=1.2\times 10^3\ J

(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0

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6 0
3 years ago
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7 0
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Masja [62]

For help with this answer, we look to Newton's second law of motion:

     Force = (mass) x (acceleration)

Since the question seems to focus on acceleration, let's get
'acceleration' all alone on one side of the equation, so we can
really see what's going on.

Here's the equation again:

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Divide each side by 'mass',
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5 0
3 years ago
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