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Yuki888 [10]
3 years ago
7

Nuclear waste disposal is one of the largest issues with nuclear power. Cesium-137 is one of the high level waste products in an

uranium nuclear fission reaction. Safely disposing of Cs-137 is extremely important due both to this isotope's long half-life, 30.07 years, and it ability to substitute for potassium in biological reactions. A 5.2 mg sample of Cs-137 requires half-lives to decay to 0.65 mg. After 150 years, of Cs-137 remains. After 6 half-lives, of Cs-137 remains. After years, 0.0102 mg of Cs-137 remains. Finally, after approximately 601 years - half-lives, only of the Cs-137 remains.
Physics
1 answer:
vodomira [7]3 years ago
7 0

Answer:

A sample of 5.2 mg  decays to .65 mg or to 1/8 of its original amount.

1/8 = 1/2 * 1/2 * 1/2 or 3 half-lives.

3 * 30.07 = 90 yrs for 5.2 mg to decay to .65 mg

You can get these other numbers similarly:

5.2 / .0102 = 510  requires about 9  half-lives which is 30 * 9 = 270 yrs

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al aplicar una fuerza de 2 N sobre un muelle este se alarga 4cm.¿cuanto se alargara si la fuerza es el triple?¿que fuerza tendri
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1) 12 cm

2) 3 N

Explanation:

1)

The relationship between force and elongation in a spring is given by Hooke's law:

F=kx

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F=2 N

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2)

In this second problem, we know that the elongation of the spring now is

x=6 cm

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F=kx

And substituting k and x, we find:

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Answer:

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Th static friction force is a self adjusting force and comes into play when the body is at rest.

Here, the applied force is 110 N and the chest is not moving, that means a static friction force is acting between the chest and the floor. This static friction force is the force of contact between the chest and the floor. The static friction force is equal to the applied force when the body does not move.

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