Homework Statement
Assume 1.500 mol of a monatomic ideal gas is compressed from 3.00 L to 1.00 L.
a. If the initial and final temperature is 10.0 °C, what are the initial and final pressures (in atm)?
b. How much work input (in kJ) is required if a reversible isothermal path at 10.0 °C is...
Homework Statement
Consider the beam shown in (Figure 1) . Suppose that a = 15 in. , b = 8 in. , c = 1 in., and d = 4 in.
Determine the moment of inertia for the beam's cross-sectional area about the x axis...
Homework Statement
A radio broadcast antenna is located at the top of a steep tall mountain. The antenna is broadcasting 104.3 FM (in Megahertz) with a power of 5.00 kilowatts.
What is the peak intensity of the signal at a receiving antenna located 25.0 km away?
Homework Equations
Honestly...
That's the expression for flux.
I would need to use the area of the small coil because that is what experiencing the induced emf.
So Na*Nb*pi*Rb^2*μ*5e^(-(t-6)/2))/2Ra is the flux, and the time derivative will be the induced emf in the small coil
Ok...
So would the correct equation be:
Na*Nb*pi*Ra^2*μ*5e^(-(t-6)/2))/2Ra
And then take the derivative with respect to time and plug in whatever value for t?
I think I understand the concept.
Oh! The magnetic field created by the large coil. So I need to multiple in dA, which would be 2*pi*r to get flux. Then I can take the time derivative to get emf
Oh ok, I must have missed that.
So it would be d/dt(μ*5e^(-(t-6)/2))/2Ra?
Does that make sense? That would be the emf in the large coil, but how do I translate that into induced emf for the smaller one? Or is it all the same?