Answer:
The inducerd emf is 1.08 V
Solution:
As per the question:
Altitude of the satellite, H = 400 km
Length of the antenna, l = 1.76 m
Magnetic field, B = 
Now,
When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

Here, velocity v is perpendicular to the rod
Thus
e = lvB (1)
For the orbital velocity of the satellite at an altitude, H:

where
G = Gravitational constant
= mass of earth
= radius of earth

Using this value value in eqn (1):

HEYA MATE
YOUR ANSWER IS <em><u>D.PLACE</u></em><em><u> </u></em><em><u>IT</u></em><em><u> </u></em><em><u>IN</u></em><em><u> </u></em><em><u>AN</u></em><em><u> </u></em><em><u>ELECTRI</u></em><em><u>C</u></em><em><u> </u></em><em><u>FIELD</u></em>
<em><u>BE</u></em><em><u>CAUSE</u></em><em><u> </u></em><em><u>IT </u></em><em><u>makes</u></em><em><u> sense you can use alternating current to remove magnetism</u></em>
Density is defined as [mass] / [volume] .
The only choice listed with those physical dimensions is 'd' .
Answer: 33.7Ω
Explanation:
Since there are two resistors connected in series, the total resistance (Rtotal) of the circuit is the sum of each resistance.
i.e Rtotal = R1 + R2
R1 = 10Ω
R2 = 23.7Ω
Hence, Rtotal = 10Ω + 23.7Ω
Rtotal = 33.7Ω
Thus, the combined resistance for two resistors is 33.7Ω
Answer:
a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.
b) Average power
P(w)= 1062.07 [w]
P(hp)=1.42 [hp]
c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.
Explanation:
First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)
Work:

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )
![W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]](https://tex.z-dn.net/?f=W%3D%20F%2Ad%3D%20m%2Ag%2Ad%3D85%2A%209.8%2A12.75%3D10620.75%20%5BJ%5D)
Converting from Joules to Kcals:

Now lets take into account the efficiency of the human body (20%)
2.537 ---> 20%
x ---> 100%

So the student is consuming 12.685 KCals each time he runs up the stairs.
Now,
1 g --> 9 Kcals
1000 g --> 9000KCals
Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

He must run up the stairs 709.5 times, to burn 1 Kg of fat.
********************
For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)
![Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\](https://tex.z-dn.net/?f=Power%3D%5Cfrac%7BJoules%7D%7BSeconds%7D%20%3D%5Cfrac%7B53103.75%7D%7B50%7D%20%3D1062.075%20%5BW%5D%5C%5C)
![P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]](https://tex.z-dn.net/?f=P%28hp%29%3D%5Cfrac%7BP%28w%29%7D%7B745.7%7D%20%3D%5Cfrac%7B1062.075%7D%7B745.7%7D%20%3D1.42%5Bhp%5D)
*****