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Tatiana [17]
3 years ago
11

A radioactive sample has a count rate of 800 counts per minute. One hour later, the count rate has fallen to 100 counts per minu

te. What is the half-life of the sample?
Physics
1 answer:
katovenus [111]3 years ago
4 0
Radioactivity equation states:

N = No e^(-0.693t/t1/2) --- Where t = time, t1/2 = half-life, N = count after time t, No = Initial count

Using the values given;

100 = 800 e^(-0.693*1/t1/2)
100/800 = e^(-0.693t/t1/2)
0.125 = e^(-0.693/t1/2)
Taking natural logs on both sides;
log 0.125 = -0.693/t1/2 * log (e)
-0.9031 = -0.693/t1/2 * 0.4323
t1/2 = (-0.693*0.4323)/(-0.9031) = 0.3317 hours

Therefore, half-life of the sample is approximately 0.3317 hours

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Answer:

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During rescue missions, different types of energy can be devices for flashlight, this could be human powered energy such as squeezing or compressing. In flashlight electrical energy is converted to light and thermal energy.

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C

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- farmer has recently bought a 40 acre land and wishes to fence the perimeter. given that the Land is square, what lenght of fen
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3 0
3 years ago
Two equal point charges QQQ are separated by a distance ddd. One of the charges is released and moves away from the other due on
lys-0071 [83]

Answer:

The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

Explanation:

The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.

Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.

From the law of conservation of energy, U₁ + K₁ = U₂ + K₂

So, kQ²/d + 0 = kQ²/3d + K

K₂ = kQ²/d - kQ²/3d = 2kQ²/3d

So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

4 0
4 years ago
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