Answer:
32.76 Volt
Explanation:
frequency, f = 400 Hz
Area of crossection, A = 13 cm²
Maximum flux density, B = 0.9 tesla
Number of turns in secondary coil, N = 70
Let the maximum induced voltage is e.
According to the Faraday's law of electromagnetic induction, the induced emf is equal to the rate of change of magnetic flux.
e = dФ/dt

Time is defined as the reciprocal of frequency.
So, e = N B A f
e = 70 x 0.9 x 13 x 10^-4 x 400
e = 32.76 volt
Answer:
35%
Explanation:
The car's engine gives off 65% thermal energy
So only 35 % is converted into mechanical energy .
input heat = Q₁ = 100
output heat = Q₂ = 65
Work output = Q₁ - Q₂ = W
W = 100 - 65 = 35
Efficiency = W / Q₁ X 100
= (35/ 100) X 100
= 35%.
Answer:
Your question was incomplete so here is the complete question and answer.
Q. When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g., 10K race)
a) plain water
b) 5-7 percent glucose solution
c) Glucose polymer solution of 6-8 percent
d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.
Ans. d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.
Explanation:
Temperature Regulation is an important phenomenon for the person exposed to extreme hot conditions or weather. Exercising in hot conditions increase the body temperature. Greater and intense exercise, greater the production of heat. Then the heat dissipation takes place in the form of excessive sweating which results in dehydration. That was just the brief overview of temperature regulation. Above mentioned techniques are equally good hydration techniques so there is no difference at all. You can have a plain water or glucose solutions of above mentioned percentages.
Answer:
Induced current, I = 18.88 A
Explanation:
It is given that,
Number of turns, N = 78
Radius of the circular coil, r = 34 cm = 0.34 m
Magnetic field changes from 2.4 T to 0.4 T in 2 s.
Resistance of the coil, R = 1.5 ohms
We need to find the magnitude of the induced current in the coil. The induced emf is given by :

Where
is the rate of change of magnetic flux,
And 



Using Ohm's law, 
Induced current, 

I = 18.88 A
So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.