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swat32
3 years ago
8

The idea is to get as much EMF produced from the sprinter running through it. If you were the Olympic coach on a year when there

happens to be a global energy crisis, and medals were assigned based on how much EMF (or current) were produced by the sprinters, what 3 pieces of advice (one is quite obvious; the other two involve the fixed orientation of the baton and the maintained position of the baton within the circular solenoid cross-section) would you give your sprinters in order to win?
Physics
1 answer:
never [62]3 years ago
3 0

Answer:

the greater the speed, the greater the electromotive force

* The metal pole must be parallel to the field

* you must keep the ball of the field

Explanation:

To determine the advice to the runners, let's use the Farad equation to and

           fem = -N \frac{ d \phi}{dt}  = -N \frac{ B A Cos \theta }{dt}

how the runners are moving

                 fi = B l x

            fem = -N B l v

therefore the advice we can give are:

* the greater the speed, the greater the electromotive force

* The metal pole must be parallel to the field

* you must keep the ball of the field

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Answer: de?

Explanation:

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A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then
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Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

(f). -138.6J.

Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.

Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J

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