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Citrus2011 [14]
4 years ago
7

A 9.0-V battery moves 20 mC of charge through a circuit running from its positive terminal to its negative terminal. How much en

ergy was delivered to the circuit?
Physics
1 answer:
dem82 [27]4 years ago
7 0

Answer:

E = 0.18 J

Explanation:

given,

Potential of the battery,V = 9 V

Charge on the circuit, Q = 20 m C

                                        = 20 x 10⁻³ C

energy delivered in the circuit

E = Q V

E = 20 x 10⁻³ x 9

E =  180 x 10⁻³

E = 0.18 J

Energy delivered in the circuit is equal to E = 0.18 J

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You have two windlasses (winches) named X and Y. Windlass X drum diameter is 50cm and windlass Y drum diameter is 75cm. Both are
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You have two windlasses  named X and Y. Windlass X drum diameter is 50cm and windlass Y drum diameter is 75cm. Both are hoisting a 50kg weight, Hence, X will lift faster and easier than Y

<h3>What is windlasses (winches) will lift faster?</h3>

Generally, the equation for the Torque is  mathematically given as

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T=122.625 Nm

For, Torque (Y)

T=50x9.81x0.375

T=183.93 Nm

Generally, the equation for the Equation of Power is  mathematically given as

P= Torque * Angular velocity

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According to the previous statement, windlass X has a speed of lift that is at its maximum, but windlass Y has a speed that is very low.

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