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bekas [8.4K]
3 years ago
8

An explosion inside a nuclear plant resulted in an exposure of 250 Rem/hr. (2 miles away). How far will you have to move away to

decrease your exposure to 2 Rem/hr.?
Physics
1 answer:
makkiz [27]3 years ago
7 0

Answer:

f something happens to go wrong at a nuclear reactor, anyone living in a 10-mile radius of the plant may have to evacuate. This map also shows a 50-mile evacuation zone, the safe distance that the U.S. government recommended to Americans who were near

because

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Which method do acoustic use to limit reverberations in music halls
dalvyx [7]

Answer:

reverberation time appropriate to the use and size of the room, adequate balance between direct and reverberant sound, intimacy and good sound diffusion in the room to obtain a uniform sound.

Explanation:The process of ... second method measured the speed of sound propagation by the phase shift.

8 0
3 years ago
Quick please and will give Brainliest!!!
masha68 [24]

25 nC

That is the answer

3 0
3 years ago
Read 2 more answers
What are two different examples of positive exeleration
sdas [7]

Answer:Accelerating your car to get up to speed on a freeway

An airplane accelerating to take off

Accelerating out of the starting blocks at a track meet

Explanation:

5 0
3 years ago
The Coulomb force between two charges q1 and q2 at separation r in the air is 10N. If half of the separation is filled with medi
adoni [48]

Answer:

The value of new coulomb force is 1.43 N.

Explanation:

Given;

Coulomb's force in vacuum (air), F_v = 10 N

dielectric constant, K = 7

The Coulomb's force between two charges separated by a distance r in a vacuum is given as;

F_v = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2}

The Coulomb's force between two charges separated by a distance r in a medium with dielectric constant is given as

F_m = \frac{1}{4\pi K\epsilon_0} \frac{q_1q_2}{r^2}

Take the ratio of the two forces;

\frac{F_v}{F_m} = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2} \ \times \ \frac{4\pi K\epsilon_0 r^2}{q_1q_2} = K\\\\\frac{F_v}{F_m} = K\\\\\frac{10}{F_m} = 7\\\\F_m = \frac{10}{7} \\\\F_m = 1.43 \ N

Therefore, the value of new coulomb force is 1.43 N.

5 0
3 years ago
At a certain location, Earth's magnetic field of 34 µT is horizontal and directed due north. Suppose the net field is zero exact
zheka24 [161]

Answer:

a) I = 13.77 A

b) 0 ° or to the East

Explanation:

Part a

The magnetic field by properties would be 0 at the radius on this case r =8.1 cm.Analyzing the situation the wirde would produce a magnetic field equals in magnitude to the magnetic field on Earth by with the inverse direction.

The formula for the magnetic field due to a wire with current is:

B = \frac{\mu_0 I}{2 \pi r}

In order to have a value of 0 for the magnetic field at the radius then we need to have this balance

B (r=8.1) = B (Earth)

Replacing:

B = \frac{\mu_0 I}{2 \pi r)}= B_{Earth}

Solving from I, from the last equation we got:

I = \frac{2 \pi r B_{earth}}{\mu_0}

I=\frac{2 \pi 0.081 m (34 x 10^{-6} T)}{4 \pi x 10^{-7} Tm/A} = 13.77 A

Part b

We can use the right hand rule for this case.

The magnetic field of the wire would point to the South, because the magnetic field of the earth given points to the North. Based on this the current need's to flow from West to East in order to create a magnetic field pointing to the south, because the current would be perpendicular to the magnetic field created.

8 0
3 years ago
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