A 220g mass is on a frictionless horizontal surface at the end of a spring that has a force constant of 7.0N/m The mass is displ
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Answer:
The solution and the explanation are in the Explanation section.
Explanation:
According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:
TA = W * (a * cosθ)
The torque due to effort at C point is:
TC = E * (b * cosθ)
The net torque is equal to 0, we have:
Tnet = 0
W * (a * cosθ) - E * (b * cosθ) = 0
From the figure, you can observe that a/b < 1, thus E < W
Answer:
236.3 x C
Explanation:
Given:
B(0)=1.60T and B(t)=-1.60T
No. of turns 'N' =100
cross-sectional area 'A'= 1.2 x m²
Resistance 'R'= 1.3Ω
According to Faraday's law, the induced emf is given by,
ℰ=-NdΦ/dt
The current given by resistance and induced emf as
I = ℰ/R
I= -NdΦ/dtR
By converting the current to differential form(the time derivative of charge), we get
= -NdΦ/dtR
dq= -N dΦ/R
The change in the flux dФ =Ф(t)-Ф(0)
therefore, dq = (Ф(0)-Ф(t))
Also, flux is equal to the magnetic field multiplied with the area of the coil
dq = NA(B(0)-B(t))/R
dq= (100)(1.2 x )(1.6+1.6)/1.3
dq= 236.3 x C