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Studentka2010 [4]
3 years ago
5

Which option below correctly compares the average annual dose of background radiation to the dose liked to an increased cancer r

isk?
A. The average annual dose of background radiation is 250 times larger than the dose linked to increased cancer risk.

B. The average annual dose of background radiation is 250 times smaller than the dose linked to increased cancer risk.

C. The average annual dose of background radiation is 100,000 times smaller than the dose linked to increased cancer risk.

D. The average annual dose of background radiation is 100,000 times larger than the dose linked to increased cancer risk.

Physics
1 answer:
julia-pushkina [17]3 years ago
6 0
The answer for this question would be choice "<span>B. The average annual dose of background radiation is 250 times smaller than the dose linked to increased cancer risk."

You only have to compare 4.0 x 10^-4 and 1.0 x 10^-1. And if you can observe carefully, when you try to multiply the average annual dose of background radiation by 250, you would get 0.1 which is equivalent to the amount of annual dose linked to increased cancer risk. Therefore, the answer is B.</span>
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lesantik [10]
<h3><u>Answer</u>;</h3>

1600 years

<h3><u>Explanation</u>;</h3>
  • Half life is the time taken for a radioactive isotope to decay by half of its original amount.
  • We can use the formula; N = O × (1/2)^n ; where N is the new mass, O is the original amount and n is the number of half lives.
  • A sample of radium-226 takes 3200 years to decay to 1/4 of its original amount.

Therefore;

<em>1/4 = 1 × (1/2)^n</em>

<em>1/4 = (1/2)^n </em>

<em>n = 2 </em>

Thus; <em>3200 years is equivalent to 2 half lives.</em>

<em>Hence, the half life of radium-226 is 1600 years</em>

3 0
3 years ago
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77julia77 [94]

Answer:

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4 0
3 years ago
Which of these observations would most likely be seen in stage N
Free_Kalibri [48]
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3 0
3 years ago
a. How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter t
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Answer:

Explanation:

The electric field outside the sphere is given as,

E = k Q /r²

here Q = n x 1.6 x 10⁻¹⁹ C

where n is the number of electons

if the dimeter of sphere d= 25 cm= 0.25 m

then the radius r = 0.125 m

we get

n= E r²/ k x 1.6 x 10⁻¹⁹ C

n = 1350N/C x (0.125m)² / (8.99 x 10⁹ N m²/C² x 1.6 x 10⁻¹⁹ C)

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A 10.0 kg weather rocket generates a thrust of 230 NN . The rocket, pointing upward, is clamped to the top of a vertical spring.
blondinia [14]

Answer: 0.2m

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F_r=M_r*g\\F_r=10*9.81\\F_r=98.1N

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F_r=k*x\\x=F_r/k\\x=98.1/480\\x=0.2m

The part b and c of this question is done in the attachment

7 0
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