Answer
given,
resistance = 0.05 Ω
internal resistance of battery = 0.01 Ω
electromotive force = 12 V
a) ohm's law
V = IR
and volage
now,
inserting the values
I = 200 A
b) Voltage
V = I R
V = 200 x 0.05
V = 10 V
c) Power
P = I V
P = 200 x 10 = 2000 W
d) total resistance = 0.05 + 0.09 = 0.14 Ω
I = 80 A
V = 80 x 0.05 = 4 V
P = 4 x 80 = 320 W
Answer:
-5.8868501529 m/s² or -5.8868501529g
0.118909090909 s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
Dividing by g
The acceleration is -5.8868501529 m/s² or -5.8868501529g
The time taken is 0.118909090909 s
Answer:
The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
Explanation:
The magnetic field at the center of the solenoid is given by;
B = μ(N/L)I
Where;
μ is permeability of free space
N is the number of turn
L is the length of the solenoid
I is the current in the solenoid
The rate of change of the field is given by;
The induced emf in the shorter coil is calculated as;
where;
N is the number of turns in the shorter coil
A is the area of the shorter coil
Area of the shorter coil = πr²
The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m
Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²
E = 14 x 0.000491 x 0.02514
E = 1.728 x 10⁻⁴ V
Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V