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hram777 [196]
3 years ago
13

If you were to separate all of the electrons and protons in 1.00 g (0.001 kg) of matter, you’d have about 96,000 C of positive c

harge and the same amount of negative charge. If you placed these charges 8.00 m apart, how strong would the attractive forces between them be?
Physics
1 answer:
natima [27]3 years ago
3 0

Answer:

The attractive force between them is 1.296 \times 10^{18} N

Explanation:

Given:

Charge q = 96000 C

Distance between two charges r = 8 m

According to the coulomb's law,

    F = \frac{kq^{2} }{r^{2} }

Where k = 9 \times 10^{9} = force constant.

   F = \frac{9 \times 10^{9} \times (96000)^{2}  }{8^{2} }

   F = 1.296 \times 10^{18} N

Therefore, the attractive force between them is 1.296 \times 10^{18} N

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Answer:

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F = 5 * 9.81 = 49.05 [N]

W = 49.05 * 1.6 = 78.48 [N]

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3 years ago
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5 0
3 years ago
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Step2247 [10]
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3 years ago
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A block weighing 10 newtons is resting on a plane inclined 30° to the horizontal. What is the magnitude of the normal force acti
aleksley [76]

Magnitude of normal force acting on the block is 7 N

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10N = 1.02kg

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Normal force acting on the block = N

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N = 8.66 N

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