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matrenka [14]
3 years ago
12

What is the potential energy of two charges of +4.6 μC and +1.0 μC that are separated by a distance of 10.0 cm?

Physics
1 answer:
Artist 52 [7]3 years ago
3 0

Answer:

U = 0.413 J

Explanation:

the potential energy between two charges q1 and q2 is given by the following formula:

U=k\frac{q_1q_2}{r}    (1)

k: Coulomb's constant = 8.98*10^9 NM^2/C^2

q1: first charge = 4.6 μC = 4.6*10^-6 C

q2: second charge = 1.0 μC*10^-6 C

r: distance between charges = 10.0 cm = 0.10 m

You replace the values of all variables in the equation (1):

U=(8.98*10^9Nm^2/C^2)\frac{(4.6*10^{-6}C)(1.0*10^{-6}C)}{0.10m}=0.413\ J

Hence, the energy between charges is 0.413 J

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A circuit contains two devices that are connected in parallel. If the resistance of one of these devices is 12 ohms and the resi
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<u>Answer</u>

3 Ohms

<u>Explanation</u>

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In this case the resistors are in parallel.

Total resistance = (1/12 + 1/4)⁻¹

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3 years ago
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Explanation:

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  To make height the subject of the formula, follow the following procedures;

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 M.E - \frac{1}{2}mv² = \frac{1}{2}mv² - \frac{1}{2}mv²+mgh

                  This gives:

                        M.E - \frac{1}{2}mv² = mgh

Multiply both sides of the expression by \frac{1}{mg}

  ( M.E -  \frac{1}{2}mv² ) x  \frac{1}{mg} =  \frac{1}{mg} x mgh

       h = ( M.E -  \frac{1}{2}mv² ) x  \frac{1}{mg}

Learn more:

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3 years ago
A football punter accelerates a .55 kg football
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Answer:

17.6 N

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\Delta t=0.25 s is the time elapsed

The change in momentum can be written as

\Delta p = m(v-u)

where

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v = 8.0 m/s is the final velocity

Combining the two equations and substituting the values, we  find the force exerted on the ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N

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