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Tamiku [17]
3 years ago
15

A nuclear accident (intentional or unintentional) can cause significant harm to those living nearby or at a distance. Harmful le

vels of invisible gamma radiation penetrate the body, not only causing devastating injuries but possibly contaminating others. What type of transmission precaution prevents such person-to-person contamination?
(A) Airborne
(B) Droplet
(C) Contact
(D) Standard
Physics
1 answer:
Vladimir [108]3 years ago
4 0

Answer:

(C) Contact

Explanation:

Radiation Poisoning is a horrific disease that is transmitted in many ways. After a nuclear accident radiation is released into the atmosphere becoming airborne and affecting everyone that breathes it in. It is also transmitted in droplets from acid rainfall. Once the gamma radiation penetrates the body of a person it can contaminate other people by having skin to skin contact with the person that is contaminated. Therefore it can be prevented by having precaution of direct contact with anyone contaminated.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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You’re still in a redecorating mood and want to move a 35kg bookcase to a different place in the living room. You exert 58 N on
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Answer: 0.16

Explanation:

Newton's second law states that the resultant of the forces exerted on the bookcase is equal to the product between the mass (m) and the acceleration (a):

F-F_f = ma

where

F = 58 N is the force applied

Ff is the frictional force

Substituting, we find the frictional force

F_f = F-ma=58 N-(35 kg)(0.12 m/s^2)=53.8 N

The frictional force has the form:

F_f = \mu mg

where \mu is the coefficient of kinetic friction. Re-arranging the formula, we can find the coefficient:

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a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres
shusha [124]

Answer:

Part a) When collision is perfectly inelastic

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b) When collision is perfectly elastic

v_m = \frac{m + M}{2m}\sqrt{5Rg}

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

mv_m = (m + M)v

so we have

v = \frac{mv_m}{m + M}

now we know that in order to complete the circle we will have

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\frac{mv_m}{m + M} = \sqrt{5Rg}

now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

Now we know that collision is perfectly elastic

so we will have

v = \frac{2mv_m}{m + M}

now we have

\sqrt{5Rg} = \frac{2mv_m}{m + M}

v_m = \frac{m + M}{2m}\sqrt{5Rg}

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