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Serhud [2]
2 years ago
8

Why is nuclear energy an important discussion in today's world?

Physics
1 answer:
aksik [14]2 years ago
8 0

Answer:

Nuclear energy is not limited to the generation of electricity, but may equally well be used for such important tasks as desalination, production of hydrogen, space heating and process-heat applications in industry as well as for extraction of carbon from CO2 to combine with hydrogen to create synthetic liquid fuels.

Explanation:

You might be interested in
B. Complete the table to show the effect of each change on each electric quantity. (1 point)
notka56 [123]

Answer:

Effect on electric force

Multiply one charge by 2

The electric force is given by F=kq1q2/r2

From the equation, the force is directly proportional to the charge.

Hence if one charge is doubled, then the electric force is doubled.

Multiply distance by 2

The electric force is given by F=kq1q2/r2

From the equation, the force is inversely proportional to the square of the distance of separation.

If the distance is doubled, F is decreased by 22. This means that the force is multiplied by 1/4.

Effect on electric potential energy

Multiply one charge by 2.

The electric potential energy is given by U=kq1q2/r

From the equation, the electric potential energy is directly proportional to the charge q.

If one charge is doubled, the electric potential energy is doubled.

Multiply distance by 2

The electric force is given by U=kq1q2/r

From the equation, the electric potential energy is inversely proportional to the distance of separation r.

If the distance is doubled, U is divided by 2. This means that the electric potential energy is multiplied by 1/2.

Effect on potential difference

Potential difference is defined as the change in electric potential energy.

Increase in the charge causes an increase in the potential difference and an increase in the distance of separation decreases the potential difference.

4 0
2 years ago
Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exi
PIT_PIT [208]

Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

A.

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

B.

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

= 51.37455 kW

= 51.38 kW.

5 0
2 years ago
To practice Problem-Solving Strategy 25.1 Power and Energy in Circuits. A device for heating a cup of water in a car connects to
jeyben [28]

Answer:

The time to boil the water is 877 s

Explanation:

The first thing we must do is calculate the external resistance (R) of the circuit, from the description we notice that it is a series circuit, by which the resistors are added

    V = i (r + R)

We replace we calculate

     r + R = V / i

     R = v / i - r

     R = 10/12 -0.04

     R = 0.793 Ω

We calculate the power supplied

     P = V i = I² R

     P = 12² 0.793

     P = 114 W

This is the power dissipated in the external resistance

We use the relationship, that power is work per unit of time and that work is the variation of energy

     P = E / t

     t = E / P

     t = 100 10³/114

     t = 877 s

The time to boil the water is 877 s

4 0
2 years ago
A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen
saw5 [17]

Answer:

The electric field strength is 4.5\times 10^{4} N/C

Solution:

As per the question:

Area of the electrode, A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}

Charge, q = 50 nC = 50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}

Now, the electric field strength of the electrode is:

\vec{E} = \frac{\sigma}{2\epsilon_{o}}

where

\epsilon_{o} = 8.85\times 10^{- 12} F/m

\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}

\vec{E} = 4.5\times 10^{4} N/C

7 0
2 years ago
Consider two adjacent states, S1 and S2, that wish to control particulate emissions from power plants and cement plants; New Jer
natka813 [3]

Answer:

a. 7500

b. Yes

c. 2500

d. 7500

Explanation:

Please see attachment

7 0
2 years ago
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