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BlackZzzverrR [31]
3 years ago
14

A charge of 70 A ·h (ampere-hours) moves through a poten- difference of 25 V. What are (a) the charge in coulombs and(b) the mag

nitude of the change in the potential energy of the charge?
Physics
1 answer:
kenny6666 [7]3 years ago
3 0

Answer

Given,

Charge, q = 70 A.h

Potential difference, V = 25 V

a) Charge, q = ?

we know that

  q= I t

    = 70 A.h

    = 70 x 60 x 60 A.s

    = 252,000 C

b) Potential energy of the charge

      U = q V

      U =  252,000 x 25

      U = 6,300,000 J

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Calculate the work done by the cyclist when his power output is 200 W for 1800 seconds.
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3.6 × 10^5 J

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a 0.500 kg mass is oscillating on a spring with k=330 N/m. the total energy of its oscillation is 3.24 J .what is the amplitude
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6 0
3 years ago
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ear the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of
sesenic [268]

Answer:

a) V_{2/1}=0.8m/s

b) The second runner will win

c) d = 10.54m

Explanation:

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V_{2/1} = V_{2} - V_{1} = 0.8m/s

For part (b) we will calculate the amount of time that takes both runners to cross the finish line:

t_{1} = \frac{X_{1}}{V_{1}}=\frac{250}{3.45}=72.46s

t_{2} = \frac{X_{2}}{V_{2}}=\frac{250+45}{4.25}=69.41s

Since it takes less time to the second runner to cross the finish line, we can say the she won the race.

For part (c), we know how much time it takes the second runner to win, so we just need the position of the first runner in that moment:

X1 = V1*t2 = 239.46m  Since the finish line was 250m away:

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6 0
3 years ago
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To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where
BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

\int E.dA = \frac{q}{\epcilon_0}

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

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q = \rho * \pi r^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r}{2 \epcilon_0}

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

q = \int \rho dV

q = \rho * \pi r_0^2 *L

now using the equation of Gauss law we will have

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E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r_0^2}{2 \epcilon_0 r}


7 0
4 years ago
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