Ideally, if all the magnetic of one winding cuts the other winding, and there isn't any loss in the transformer core or the resistance of the wire, then the voltage across each winding is proportional to the number of turns in its coil.
If you apply 100 V to a winding of 50 turns, then a winding that yields 20 volts
must be wound with
(20/100) of 50 turns = 10 turns
Answer:
the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers
Explanation:
The equilibrium wage rate and the equilibrium quantity of labor are found as the point where the equation of demand intercepts the equation of supply, so the equilibrium quantity of labor is:

15 - (1/200) L = 5 + (1/200) L
15 - 5 = (1/200) L + (1/200) L
10 = (2/200) L
(10*200)/2 = L
1000 = L
Then, the equilibrium wage rate is calculated using either the equation of demand for labor or the equation of supply of labor. If we use the equation of demand for labor, we get:
W = 15 - (1/200) L
W = 15 - (1/200) 1000
W = 10
Finally, the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers
To find the ratio of planetary speeds Va/Vb we need the orbital velocity formula:
V=√({G*M}/R), where G is the gravitational constant, M is the mass of the distant star and R is the distance of the planet from the star it is orbiting.
So Va/Vb=[√( {G*M}/Ra) ] / [√( {G*M}/Rb) ], in our case Ra = 7.8*Rb
Va/Vb=[ √( {G*M}/{7.8*Rb} ) ] / [√( {G*M}/Rb )], we put everything under one square root by the rule: (√a) / (√b) = √(a/b)
Va/Vb=√ [ { (G*M)/(7.8*Rb) } / { (G*M)/(Rb) } ], when we cancel out G, M and Rb we get:
Va/Vb=√(1/7.8)/(1/1)=√(1/7.8)=0.358 so the ratio of Va/Vb = 0.358.
Because of Surface tension
Answer:
So magnetic field will be equal to 0.1144 T
Explanation:
We have given frequency f = 60 Hz
Maximum emf e = 5500 volt
Number of turns N = 150
Area 
Emf generated in ac generator is given 
For maximum emf 
So maximum emf will be equal to 

So magnetic field will be equal to 0.1144 T