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bixtya [17]
3 years ago
5

A student is concerned about background radiation that she is exposed to. She claims that nuclear power plants have caused backg

round radiation, and that if they had not been built, she would not be exposed to radioactivity. What is incorrect about her reasoning
Physics
2 answers:
andreev551 [17]3 years ago
7 0

The answer is D) No, because background radiation comes from naturally occurring substances such as isotopes.

Explanation:

I took it answer c is wrong

denis-greek [22]3 years ago
5 0
Based on the given statement or reasoning above made by a student, what makes her reasoning incorrect is the part where she mentioned that "if they had not been built, she would not be exposed to radioactivity." This is definitely false. What makes this incorrect is that background radiation is a kind of radiation that is emitted by the sun, stars, and other bodies outside the universe and therefore, no one can escape from this. Hope this helps.
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Drupady [299]
F=ma, so 100=m×10. Solve for m by dividing by 10. The mass is 10 kg.
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Explanation:

Solid state: In this state, the molecules are closely packed and cannot move freely from one place to another that means no space between them and the intermolecular force of attraction between the molecules are strong.

In solid substance, the particles are very close to each other due to this the intermolecular forces of attraction are strongest.

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Alinara [238K]

Decelerate

<u>Explanation:</u>

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This could be understood by visualising a hypothetical situation where a charged object is moving against an electric force. Since the object is charged, it would exert a force in its direction of motion which would be opposed by the electric force, thus causing it to decelerate  

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an object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m. determine the image height (in cm)
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Given :

An object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m.

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the image height (in cm).

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\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}

Here, u = - 100 cm

f =  150 cm

\dfrac{1}{v} - \dfrac{1}{-100} = \dfrac{1}{150}\\\\v = - 300 \ cm

Now, magnification is given by :

m = \dfrac{v}{u} = \dfrac{h_i}{h_o}\\\\h_i = \dfrac{300}{100}\times 1\\\\h_i = 3 \ m

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