The question is incomplete. The complete question is :
In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?
Solution :
The underdamped RLC circuit
We know in one time period, v = 2v, at t = T,
so,
Now, Q value
∴
= 11.45
Answer:
a=g(sinθ-μkcosθ)
Explanation:
In an inclined plane the forces that interact with the object can be seen in the figure. The normal force, the weight w and the decomposition of the force vector of weight can be observed.
wx=m*g*sinθ
wy=m*g*cosθ
As the objects moves down an incline, acceleration in y axis is 0.
Then, by second Newton's Law:
Fy = m*ay
FN - m*g cos θ = 0,
FN=m*g cos θ
In x axis the forces that interacs are the x component of weight and friction force:
Fx = m*ax
mg sen u-FN*μk=m*a
Being friction force, Fr=FN*μk, we replace with its value in below formula:
m*g *sinθ-(m*g*cosθ*μk)=m*a
Then, isolating a:
a=(m*g sinθ-(m*g*cosθ*μk))/m
Solving, we have next equation:
a=g sinθ-(g*cosθ*μk)
Applying distributive property we have:
a=g*(sinθ-μk*cosθ)
