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svetlana [45]
3 years ago
11

A charge of 31.0 μC is to be split into two parts that are then separated by 24.0 mm, what is the maximum possible magnitude of

the electrostatic force between those two parts?
Physics
1 answer:
miskamm [114]3 years ago
8 0

Answer:

1.72 x 10³ N.

Explanation:

When a charge is split equally and placed at a certain distance , maximum electrostatic force is possible.

So the charges will be each equal to

31/2 = 15.5 x 10⁻⁶ C

F = K Q q / r²

= \frac{9\times10^9\times(10.5)^2\times10^{-12}}{(24\times10^{-3})^2}

= 1.72 x 10³ N.

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fomenos

w=f×d

w=10N×25m

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2 years ago
Bill steps off a 3.0-m-high diving board and drops to the water below. At the same time, Ted jumps upward with a speed of 4.2 m/
Pani-rosa [81]

Answer:

equation of  motion for Bill is

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equation of  motion for Ted is

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g = 9.8 m/s^2

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y(t) = 2 + (-4.2)(t) +\frac{1}{2}\times (9.8t)^2

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Answer:

Last option in the list of possible answers, with U235 and n (neutron) in the left (originators) of the reaction diagram.

Explanation:

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