Answer:
Compared with the current in the first coil, the current in the second coil is unchanged.
Explanation:
All coils, inductors, chokes and transformers create a magnetic field around themselves consist of an Inductance in series with a Resistance forming an LR Series Circuit.
The steady state of current in the LR circuit is:
I= V/R (1 - e^-Rt/L)
Where I= current
R= Resistance
V= Voltage
Where R/L is the time constant.
For a conducting wire, it has a very small resistance. The time constant will be in microseconds. The current will be in a steady state after few second. The current is independent on the inductance and dependent on the resistance. The length of wire and the resistance here are the same. Therefore, the current remains unchanged.
Yes thats correct....becuase all of your weight is concentrated on a small area compared to the larger surface area of your feet!
is that what your question was?
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
Answer:
sorry I dont now the answer bro i am so sorry xd ;'(
Do you want me to translate it?
