Answer:
Ein: 2.75*10^-3 N/C
Explanation:
The induced electric field can be calculated by using the following path integral:
Where:
dl: diferencial of circumference of the ring
circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m
ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)
The electric field is always parallel to the dl vector. Then you have:
Next, you take into account that the area of the ring is constant and that dB/dt = - 0.220T/s. Thus, you obtain:
hence, the induced electric field is 2.75*10^-3 N/C
The cost of developing thermonuclear power with plasmabe defended because D. It can provide an inexpensive power source.
<h3>How did the
cost of developing t
hermonuclear power defended?</h3>
The cost of developing thermonuclear power defended becvause we can see in the paragraph how it was told that the generation of ths power can be donee through the understanding of the occurrence of plasmain nature,
It should be noted that this thermonuclear power with plasmabe posses the characteristics which make it to exist in the ionosphere, and it can be felt in the flames as well; as in the chemical and nuclearexplosions.
In conclusion the power can be seen as an inexpensive source power because the p[roduction of this power cn be found in most of the thing that can be found around us as discused above.
Therefore, option D is correct.
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Answer:
Slow
Explanation:
As we know that the time period of simple pendulum is given by
Where, l be the effective length of the pendulum and g be the acceleration due to gravity
As the length of the pendulum increase, then the tie period of the pendulum will also increase.
It means the clock run slow.
Answer: E = 0.85
Therefore the efficiency is: E = 0.85 or 85%
Explanation:
The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input heat QH
E = W/QH.
W=QH – QC,
Where Qc is the output heat.
That is,
E=1 - Qc/QH
E =1 - Tc/TH
where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir.
Note: The unit of temperature must be in Kelvin.
Tc = 300K
TH = 2000K
Substituting the values of E, we have;
E = 1 - 300K/2000K
E = 1 - 0.15
E = 0.85
Therefore the efficiency is: E = 0.85 or 85%