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horsena [70]
2 years ago
7

People are constantly being barraged with low-level radiation. Which radiation source below contributes the most low-level radia

tion exposure to the average person?a) cosmic raysb) medical X-raysc) radond) rocks, soils, and food
Physics
2 answers:
Gennadij [26K]2 years ago
7 0

Answer:

C) Radon

Explanation:

Humans are terrified of radiation attack but we are unaware that we face radiations from a lot of natural sources which includes food and water. But their level is low and thus do not have any major impact on our body.

As per US Nuclear Regulatory Commission,on an average, an American faces a radiation of 620 millirem each year. Out of this 50% is because of Natural background radiation. Majority of this radiation is due to Radon in the air, with small contribution (around 30 millirem) from Cosmic rays and even smaller contribution from Earth itself.

Sati [7]2 years ago
3 0

I think the answer is C) Radond

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In the northern hemisphere planetary winds deflect to the
umka2103 [35]

Answer:

Planetary are deflected to right due to Coriolis effect.

Explanation:

   The term Coriolis effect is defined as an effect in which the rotating object experience a force called as coriolis force which acts perpendicular to the axis of rotation and direction of motion. The effect talks about how the moving objects like ocean currents, wind are deflected due to the rotation of earth.

      Winds and ocean currents are strongly affect by this effect.

   

8 0
2 years ago
When you have a pot of water on the stove, heat is transferred to the water. Describe the behavior of the water molecules and ho
NikAS [45]
<span>Temperature causes water molecules to move more quickly, because each individual molecule has more energy as it gets hotter (according to Kinetic molecular theory). If you get water hot enough, the molecules move so much that the hydrogen bonds that hold them together start to break and the water becomes a gas ... water vapor. This your answer unless there are choices.</span>
3 0
3 years ago
Read 2 more answers
2. One mole of a monatomic ideal gas undergoes a reversible expansion at constant pressure, during which the entropy of the gas
Anna35 [415]

Answer:

The initial and final temperatures of the gas is 300 K and 600 K.

Explanation:

Given that,

Entropy of the gas = 14.41 J/K

Absorb gas = 6236 J

We know that,

ds=\dfrac{dQ}{dt}

At constant pressure,

dQ=C_{p}dt

\Delta s=\int_{T_{1}}^{T_{2}}{\dfrac{C_{p}dT}{T}}

\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}

Put the value into the formula

14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})

\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}

0.693=ln\dfrac{T_{2}}{T_{1}}

ln2=ln\dfrac{T_{2}}{T_{1}}

T_{2}=2T_{1}...(I)

We need to calculate the initial and final temperatures of the gas

Using formula of energy

\Delta Q=C_{p}\Delta T

Put the value into the formula

6236=2.5\times8.3144(T_{2}-T_{1})

6236=20.786(T_{2}-T_{1})

T_{2}-T_{1}=\dfrac{6236}{20.786}

T_{2}-T_{1}=300

Put the value of T₂

2T_{1}-T_{1}=300

T_{1}=300\ K

Put the value of T₁ in equation (I)

T_{2}=2\times300

T_{2}=600\ K

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

8 0
2 years ago
An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 17601760 × 103 seconds (ab
Yuliya22 [10]

Answer:

The value is a_r  = 3.81 *10^{-3} m/s^2

Explanation:

Generally the moon's radial acceleration is mathematically represented as

a_r  =  r *  w^2

Here w is the angular velocity which is mathematically represented as

w =\frac{2 \pi }{ T}

substituting 1760 * 10^3 \ seconds for T(i.e the period of the moon ) we have

w =\frac{2  * 3.142 }{  1760 * 10^3}

=> w = 3.57 *10^{-6} \  rad/s

From the question r(which is the radius of the orbit ) is evaluated as

r =  R + H

substitute 3.60 * 10^6 m for R and 295.0 * 10^6  \  m H

       r =  295.0 * 10^6   +3.60 * 10^6

=>   r =  2.986 *10^{8} \  m

So

    a_r  =   2.986 *10^{8} *  ( 3.57 *10^{-6} )^2

      a_r  = 3.81 *10^{-3} m/s^2

4 0
3 years ago
A battery with an emf of 12.0 V shows a terminal voltage of 11.4Vwhen operating in a circuit with two lightbulbs, each rated at
FinnZ [79.3K]

Answer:

0.95 A

Explanation:

P = Power of bulb = 4 W

V = Voltage = 12 V

Power is given by

P=\dfrac{V^2}{R}\\\Rightarrow R=\dfrac{V^2}{P}\\\Rightarrow R=\dfrac{12^2}{4}\\\Rightarrow R=36\ \Omega

The connections are in Parallel

Equivalent resistance of these two bulb

\dfrac{36}{2}=18\ \Omega

Current from the source

I=\dfrac{11.4}{18}=0.63\ A

The voltage drop is

12-11.4=0.6\ V

Voltage drop is given by

\Delta V=R_iI\\\Rightarrow R_i=\dfrac{\Delta V}{I}\\\Rightarrow R_i=\dfrac{0.6}{0.63}\\\Rightarrow R_i=0.952380952381\ A

The internal resistance is 0.95 A

3 0
3 years ago
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