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tigry1 [53]
3 years ago
15

A battery with an emf of 4 V and an internal resistance of 0.7 capital omega is connected to a variable resistance R. Find the c

urrent and power delivered by the battery when R is (a) 0, I = 5.714285714 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 5.714285714 OK P = 0 W * [1.25 points] 3 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0 OK (b) 5 capital omega, I = 0.701754386 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0.701754386 OK P = 2.462296091 W * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 2.462296091 OK (c) 10 capital omega, and I = 0.3738317757 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0.3738317757 OK P = 1.397501965 W * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 1.397501965 OK (d) infinite. I = 0 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0 OK P = W
Physics
1 answer:
Zinaida [17]3 years ago
4 0

Answer:

E = I(R + r)

Making I the subject of the formular by dividing both sides by R + r,

I = E/(R + r)

E = 4V, r = 0.7Ohm, R = 0

I = 4/(0 + 0.7) = 4/0.7

I = 5.174285714A

Explanation:

For a cell of emf E, internal resistance r, connected to an external resistance R, the current flowing through the circuit will be given as:

I = E/(R + r). I is measured in Amperes(A), emf in volts(V), R in Ohms and internal resistance r also in ohms

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Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

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\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

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