Answer:
Induced emf in the coil, E = 0.157 volts
Explanation:
It is given that,
Number of turns, N = 100
Diameter of the coil, d = 3 cm = 0.03 m
Radius of the coil, r = 0.015 m
A uniform magnetic field increases from 0.5 T to 2.5 T in 0.9 s.
Due to this change in magnetic field, an emf is induced in the coil which is given by :


E = -0.157 volts
Minus sign shows the direction of induced emf in the coil. Hence, the induced emf in the coil is 0.157 volts.
Answer:
The punnet chart indicate the cross between a black cat and a spotted cat.
Explanation:
As the punnet chart was not given with the question, the question is searched and found online which is as attached herewith.
As seen from the punned table the column head has two capital H which indicate that the cat is a black cat.
White the row heads indicate a Capital and a small h thus this is a spotted cat
Thus the punnet chart indicate the cross between a black cat and a spotted cat.
Answer:
F = 132N
Explanation:
Let the pulling force be F
Given the mass of the barrel is 12 Kg and The acceleration is 1.2 m/
We know that Net Force = 
Now net force = F - mg
F - mg = ma
F = mg + ma
F = m(a+g) =
Note that there will be a downward force acting on the body which is equal to mg
Answer:
Explanation:
Assuming that velocities remain negligible,
The work will be equal to the change in potential energy of the center of mass
mass is 30 m(5 kg/m) = 150 kg
W = 150(9.8)(30/2) = 22,050 J ≈ 22 kJ