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3 years ago

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The half-life of a certain isotope is 15 minutes. How much of a 400 g sample will remain after 90 minutes?12.5 g6.25 g26.7 g66.7

g25 g

1 answer:

Virty [35]3 years ago

4 0

There is a total of 6 half lives that need to take place.

ONE HALF LIFE = 200

TWO HALF LIFES = 100

THREE HALF LIFES = 50

FOUR HALF LIFES = 25

FIVE HALF LIFES = 12.5

SIX HALF LIFES = 6.25

The answer is 6.25g

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A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi

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Answer:

a

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b

The value is $u = 0.2569 \ m/s$

Explanation:

From the question we are told that

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The spring constant is $k = 17.5 \ N/m$

The instantaneous speed is $v = 39.0 \ cm/s= 0.39 \ m/s$

The position consider is x = 0.750A meters from equilibrium point

Generally from the law of energy conservation we have that

The kinetic energy induced by the hammer = The energy stored in the spring

So

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k * A^2$

Here a is the amplitude of the subsequent oscillations

=> $A = \sqrt{\frac{m * v^ 2 }{ k} }$

=> $A = \sqrt{\frac{0.750 * 0.39 ^ 2 }{17.5} }$

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Generally from the law of energy conservation we have that

The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k x^2 + \frac{1}{2} * m * u^2$

=> $\frac{1}{2} * 0.750 * 0.39^2 = \frac{1}{2} * 17.5* 0.750(0.081 )^2 + \frac{1}{2} * 0.750 * u^2$

=> $u = 0.2569 \ m/s$

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