Answer:
emf will also be 10 times less as compared to when it has fallen 
Explanation:
We know, from faraday's law-
and 
So, as the height increases the velocity with which it will cross the ring will also increase. 
Given
Now, from 
From equation a and b we see that velocity when dropped from is 10 times greater when height is 40 
so, emf will also be 10 times less as compared to when it has fallen
Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J
Explanation: In order to solve this problem we have to used the second Newton law given by:
∑F= m*a
F-f=m*a where f is the friction force (uk*Normal), from this we have
F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N
then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N
the net Force = (34.5-24.5)N= 10 N
Finally the work done by the net force is equal to kinetic energy change so
W=∫Force net*dr= 10 N* 0.1 m= 1J
I don’t think I will have any time to go
Answer:
20 Ω
Explanation:
Voltage, current, and resistance are related by Ohm's law:
V = IR
40 V = (4 A) R
R = 10 Ω
The total resistance of the circuit is 10 Ω.
Resistors in parallel have a total resistance of:
1/R = 1/R₁ + 1/R₂
1 / (10 Ω) = 1 / (20 Ω) + 1/R₂
R₂ = 20 Ω
Answer:
a.) O.3A and 0.15A
b.) P1 = 36W, P2 = 18W
c.) 54W
Explanation:
Constant resistance R1 = 400 ohms, R2 = 800 ohms
a.)
The current through each bulb
While in parallel.
I = 120/800/3
= 0.45A
For 400= 800/400+800
= 800/1200 x 0.45
= 0.3A
I-800 = 400 /400+800
= 400/1200*0.45
= 0.15A
B.)
The power dissipated in each bulb
P409 = I²400r
= 0.3A²x400 ohms
= 36 W
P800 = I²800R
= 0.15A²*800 ohms
= 18W
C.)
The total power dissipated
= 36W + 18 W
= 54W
