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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 capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.90 MΩ. A
iren2701 [21]

Explanation:

(a)  Formula to calculate the capacitance is as follows.

           V_{c} = V_{o} (e^(\frac{-t}{RC}))

Now, putting the given values into the above formula as follows.

          V_{c} = V_{o} (e^(\frac{-t}{RC}))

          3.5 V = 12 V (e^(\frac{-3.50 sec}{3.90 \times C}))

    e^(\frac{-3.50 sec}{3.90 \times C}) = 0.291

or,       \frac{-3.50 sec}{3.90 \times C} = ln (0.291)

      \frac{-3.50 sec}{3.90 \times C} = -1.23

              C = 0.729 F

Hence, the value of capacitance is 0.729 F.

(b)   Formula to calculate the constant of circuit is as follows.

               T = R \times C

                  = 3.1 \times 0.729

                  = 2.259 sec

Therefore, the time constant of the circuit is 2.259 sec.

5 0
4 years ago
HELP ASAP PLS. ILL GIVE BRAINLIEST
Blizzard [7]

Answer:

I'm not completely sure, but I believe the first and third of the three are mechanical.

Explanation:

Chemical potential isn't moving or about to go into motion. It can't be mechanical.

4 0
3 years ago
Two hockey players are about to collide on the ice, One player has a mass of
Furkat [3]

Answer:

19 kg x m/s East

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3 0
3 years ago
An artillery shell is launched on a flat, horizontal field at an angle of α = 31.7° with respect to the horizontal and with an i
kompoz [17]

Answer:193.90 m/s

Explanation:

Given

launch angle \theta =31.7^{\circ}

launch velocity v_0=202 m/s

Horizontal velocity of the shell after t=18.96 s

time of flight of Projectile T=\frac{2u\sin \theta }{g}

T={\frac2\times 202\sin 31.7}{9.8}

T=21.66 s

i.e. projectile is declining as t>\frac{T}{2}

but horizontal component of velocity will remain same as there is no opposing force in horizontal direction

Horizontal component of velocity is

u_x=v_0\cos \theta =202\cdot \cos 31.7=193.90 m/s  

8 0
3 years ago
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True [87]

Answer:

the answer is b

Explanation:

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3 years ago
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