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otez555 [7]
3 years ago
14

Do you think scientists should continue to explore nuclear energy as a source of electricity? Explain your opinion and provide r

easons. Include an opposing opinion about another energy source.
Physics
1 answer:
Jet001 [13]3 years ago
3 0

Answer:

See explanation

Explanation:

Nuclear energy refers to energy trapped in the nucleus of atoms. Nuclear energy holds the greatest promise for large scale electricity generation. For instance, the amount of energy released by a given mass of uranium is 2.5 million times the energy released by combustion of an equal mass of carbon.One gram of uranium produces the same energy as 1 ton of coal, 17000 cubic feet of natural gas, 5000 pounds of wood and 149 gallons of oil.

Nuclear energy does not pollute the atmosphere as fossil fuels do. However, the challenges of effective control of the nuclear reactor and disposal of radioactive waste still remain a serious concern.

However, exploitation of nuclear energy as a source of electricity remain expensive and it is not quite easy to set up a nuclear power plant. Not many countries can afford to pay for the services of experts who operate nuclear energy plants talk more of providing the required equipment.

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Select the correct answer<br>What is the importance of writing a draft?​
givi [52]

Answer:to revise or edit

anything that can be made in the non draft one

Explanation:

4 0
3 years ago
What is most responsible for the uneven heating of the air in the atmosphere?
Phoenix [80]
<span>The answer would be convection currents. Convection happens when atoms with a lot of heat energy in a liquid or gas transfer and get the room of particles with fewer heat energy. Heat energy is transported from hot places to cooler places by convection.</span>
4 0
3 years ago
Which equation relates charge, time, and current?
Ierofanga [76]

Answer:

I = Δq / t

Explanation:

The quantity of electricity i.e charge is related to current and time according to the equation equation:

Q = It

Δq = It

Where:

Q => is the quantity of electricity i.e charge

I => is the current.

t => is the time.

Thus, we can rearrange the above expression to make 'I' the subject. This is illustrated below:

Δq = It

Divide both side by t

I = Δq / t

6 0
2 years ago
A circuit contains two 1.5 volt battery and a bulb with a resistance of 3 ohms. Calculate the current
horrorfan [7]

Answer:

<em>The current is 1 A</em>

Explanation:

<u>Current in a Series Connection </u>

When two or more elements are connected in series, all of them have the same current, and the sum of their individual voltages is the total voltage applied to the circuit.

According to Ohm's law:

V=R.I

Where V is the voltage, R is the resistance and I is the current of a circuit.

We have a voltage of V=1.5 V + 1.5 V = 3 V and a resistance of R=3 ohms.

We can calculate the current by solving for I:

\displaystyle I=\frac{V}{R}=\frac{3}{3}=1\ A

The current is 1 A

3 0
2 years ago
The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the
Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>

where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²

<em>P </em>sun =  2.863 10²³ × 1360

<em>P </em>sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars

where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

<em>I </em><em>sun-mars =  </em>3.85×10²⁶W / 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²

<em>I </em><em>sun-mars = </em>590 W/m²

6 0
3 years ago
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