The emf induced in the second coil is given by:
V = -M(di/dt)
V = emf, M = mutual indutance, di/dt = change of current in the first coil over time
The current in the first coil is given by:
i = i₀
i₀ = 5.0A, a = 2.0×10³s⁻¹
i = 5.0e^(-2.0×10³t)
Calculate di/dt by differentiating i with respect to t.
di/dt = -1.0×10⁴e^(-2.0×10³t)
Calculate a general formula for V. Givens:
M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)
Plug in and solve for V:
V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))
V = 320e^(-2.0×10³t)
We want to find the induced emf right after the current starts to decay. Plug in t = 0s:
V = 320e^(-2.0×10³(0))
V = 320e^0
V = 320 volts
We want to find the induced emf at t = 1.0×10⁻³s:
V = 320e^(-2.0×10³(1.0×10⁻³))
V = 43 volts
Answer:
I belive it would be ture
Explanation:
It's been a while since I learned this but I think that is right.
Answer:
Explanation:
When a body is held against a vertical wall , to keep them in balanced position , normal force is applied on their surface . this force creates normal reaction which acts against the normal force and it is equal to the normal force as per newton's third law . Ultimately friction force is created which is proportional to normal force and it acts in vertically upward direction . It prevents the body from falling down .
Hence normal force = reaction force .
From second law also net force is zero , so if normal force is N and reaction force is R
R - N = mass x acceleration = mass x 0 = 0
R = N .
Ranking normal force from highest to smallest
150 N , 130 N , 120 N
B )
Frictional force is equal to the weight of the body because the body is held at rest .
Ranking of frictional force form largest to smallest
7 kg , 5 kg , 3 kg , 1 kg .
Here frictional force is irrespective of the normal force acting on the body because frictional force adjusts itself so that it becomes equal to weight in all cases here because it always balances the weight of the body .
Answer:
Distance travelled is 7 meters and the displacement is 3 meters
Given Information:
Wavelength = λ = 39.1 cm = 0.391 m
speed of sound = v = 344 m/s
linear density = μ = 0.660 g/m = 0.00066 kg/m
tension = T = 160 N
Required Information:
Length of the vibrating string = L = ?
Answer:
Length of the vibrating string = 0.28 m
Explanation:
The frequency of beautiful note is
f = v/λ
f = 344/0.391
f = 879.79 Hz
As we know, the speed of the wave is
v = √T/μ
v = √160/0.00066
v = 492.36 m/s
The wavelength of the string is
λ = v/f
λ = 492.36/879.79
λ = 0.5596 m
and finally the length of the vibrating string is
λ = 2L
L = λ/2
L = 0.5596/2
L = 0.28 m
Therefore, the vibrating section of the violin string is 0.28 m long.
