Answer:The distinction between a permanent magnet and an electromagnet is essentially one in how the field is created, not the properties of the field afterwards. So electromagnets still have two poles, still attract ferromagnetic materials, and still have poles that repel other like poles and attract unlike poles.
Explanation:
Answer:
$$\begin{align*}
P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$
Explanation:
\eqalign{
P(Y-X=m\mid Y\gt X)
&=\sum_kP(Y-X=m,X=k\mid Y\gt X)\cr
&=\sum_kP(Y-X=m\mid X=k,Y\gt X)\,P(X=k\mid Y>X)\cr
&=\sum_kP(Y-k=m\mid Y\gt k)\,P(X=k\mid Y\gt X)\cr
}
P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$
Answer:
View Image
Explanation:
You didn't provide me a picture of the opamp.
I'm gonna assume that this is an ideal opamp, therefore the input impedance can be assumed to be ∞ . This basically implies that...
- no current will go in the inverting(-) and noninverting(+) side of the opamp
- V₊ = V₋ , so whatever voltage is at the noninverting side will also be the voltage at the inverting side
Since no current is going into the + and - side of the opamp, then
i₁ = i₂
Since V₊ is connected to ground (0V) then V₋ must also be 0V.
V₊ = V₋ = 0
Use whatever method you want to solve for v_out and v_in then divide them. There's so many different ways of solving this circuit.
You didn't give me what the input voltage was so I can't give you the entire answer. I'll just give you the equations needed to plug in your values to get your answers.
Answer:
R = 4.75 lb (↑)
Explanation:
Number of books = n = 19
Weight of each book = W = 1 lb
Length of the bookshelf = L = 40 inches
We can get the value of the distributed load as follows
q = n*W/L = 19*1 lb/ 40 inches = 0.475 lb/in
then the reactions at 4 ends (supports) of the bookshelf are
R = (q/2)/2 = 4.75 lb
We can see the bookshelf in the pic.
A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined
Assumptions :
1. Steady operating conditions exist.
2. Kinetic and potential energy changes are negligible.
Properties: The specific heat of geothermal water (
[) is taken to be 4.18 kJ/kg.ºC.
Analysis (a) We need properties of isobutane, we can obtain the properties from EES.
a. Turbine
P
=
b. Pump


c. 