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3 years ago

9

Calculate using coulombs: an object has 6 protons and 8 electrons. Calculate the magnitude of the charge of the objects. Thank y

ou.

1 answer:

SVEN [57.7K]3 years ago

8 0

Answer:

-3.2 ×1.6 ×10^-19C

Explanation:

Number of proton = 6

Number of electron = 8

To determine the quantity of the charge, we would first find the difference between the number of

protons and electrons. Then multiply the difference by charge 1.6 x 10 -19 C to determine the charge on the object.

Difference between proton and electron is elementary charge

Elementary charge = proton - electron

= 6 - 8

Elementary charge = -2e

Recall, charge (q) = 1.6 ×10^-19C

1e = 1.6 ×10^-19C

Magnitude of the charge of object = -2× (1.6 ×10^-19C)

Magnitude of the charge of object = -3.2 ×1.6 ×10^-19C

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nadezda [96]

Answer:

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Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

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If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

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$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the block, measured in meters per second.

$\Sigma F$ - Horizontal net force, measured in newtons.

$\Delta t$ - Impact time, measured in seconds.

Now we clear the final speed in (2):

$v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}$

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$v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}$

$v_{f} = 9.8\,\frac{m}{s}$

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Answer:

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L = (1 m)(1/2)

<u>L = 0.5 m</u>

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