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In-s [12.5K]
3 years ago
6

A hydrogen fuel cell supplies power for a small motor. the fuel cell delivers a current of 0.5 a and a voltage of 0.43 v. what i

s the power supplied to the motor by the fuel cell?
Physics
1 answer:
gogolik [260]3 years ago
8 0
We want to know what is the power supplied by the power cell if the current I=0.5 A and the voltage V=0.43 V. The equation for power P is P= I*V, so:

P=I*V=0.5*0.43=0.215 W

So the correct answer is that the power cell is supplying the motor with P=0.215 W of power. 
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A string is attached to a ball that has a mass of 0.11 kg. A student pulls up on the string so that the ball accelerates upward
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Answer:

T=+1.133N

Explanation:

Tension and weight are forces that have opposite directions

Weight is negative (downward)

W=m*g= 0.11kg*(-9.8m/s^2)

W= -1.078N

Tension is possitive (upward)

The total force will be the sum of both (the difference taking in consideration the direction)

Ft= T+W

Also the total force is the product of the mass due to acceleration:

Ft=m*a

Ft= +0.11kg*0.5m/s^2

Ft=+0.055N (upward)

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T= Ft-W

T=+0.055N-(-1.078N)

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As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th
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Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek

Conservation of Momentum:

P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1

Energy Balance:

\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2}  ....... Eq 2

Substitute Eq 2 into Eq 1

0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V  \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}

Using Eq 1

0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m

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