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wel
4 years ago
8

A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts

per minute for a standard of zero age. You may want to reference (Pages 913 - 916) Section 21.4 while completing this problem. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact. Express your answer using two significant figures. t
Physics
1 answer:
prohojiy [21]4 years ago
7 0

Answer:

Explanation:

The relation between activity and number of radioactive atom in the sample is as follows

dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms

For the beginning period

dN₀ / dt = λ N₀

58.2 = λ N₀

similarly

41 = λ N

dividing

58.2 / 41 = N₀ / N

N = N₀ x .70446

formula of radioactive decay

N=N_0e^{-\lambda t }

.70446 =e^{-\lambda t }

- λ t = ln .70446 =   - .35

t = .35 / λ

λ = .693 / half life

= .693 / 5715

= .00012126

t = .35 / .00012126

= 2886.36

= 2900 years ( rounding it in two significant figures )

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6 0
4 years ago
A lamp is connected to the power supply.
Vika [28.1K]

Answer:

There are 45 turns in the secondary coil.

Explanation:

Given that,

Input potential of the lamp, V_{in}=5\ V

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We need to find the number of turns needed on the secondary coil. We know that the ratio for a transformer is as follows :

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5 0
3 years ago
A merry-go-round is spinning at a rate of 4.04.0 revolutions per minute. Cora is sitting 0.50.5 m from the center of the merry-g
dsp73

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

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7 0
3 years ago
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