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Kamila [148]
3 years ago
11

What if m1 is initially moving at 3.4 m/s while m2 is initially at rest? (a) find the maximum spring compression in this case?

Physics
1 answer:
Lisa [10]3 years ago
7 0
<span>Ans : Initial E = KE = ½mv² = ½ * 1.2kg * (2.2m/s)² = 2.9 J max spring compression where both velocities are the same: conserve momentum: 1.2kg * 2.2m/s = (1.2 + 3.2)kg * v → v = 0.6 m/s which means the combined KE = ½ * (1.2 + 3.2)kg * (0.6m/s)² = 0.79 J The remaining energy went into the spring: U = (2.9 - 0.79) J = 2.1 J = ½kx² = ½ * 554N/m * x² x = 0.0076 m ↠(a)</span>
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Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite
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A motor has an armature resistance of 3.75 Ω . Part A If it draws 9.10 A when running at full speed and connected to a 120-V lin
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Back emf is 85.9 V.

<u>Explanation:</u>

Given-

Resistance, R = 3.75Ω

Current, I = 9.1 A

Supply Voltage, V = 120 V

Back emf = ?

Assumption - There is no effects of inductance.

A motor will have a back emf that opposes the supply voltage, as the motor speeds up the back emf increases and has the effect that the difference between the supply voltage and the back emf is what causes the current to flow through the armature resistance.

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  = 34.125 volts

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By plugging in the values,

120 V = back emf + 34.125 V

Back emf = 120 - 34.125

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