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Korolek [52]
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.
Physics
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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A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo
horrorfan [7]

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

s=r_{2}-r_{1}

Where, r_{1} = initial position

r_{2} = final position

Put the value into the formula

s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)

s= -6.80i+6.20j-0.1k

(a). We need to calculate the work done on the object

Using formula of work done

W=F\cdot s

Put the value into the formula

W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)

W=-13.6+55.8-0.53

W=41.67\ J

(b). We need to calculate the average power due to the force during that interval

Using formula of power

P=\dfrac{W}{t}

Where, P = power

W = work

t = time

Put the value into the formula

P=\dfrac{41.67}{5.40}

P=7.71\ Watt

(c). We need to calculate the angle between vectors

Using formula of angle

\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})

Put the value into the formula

\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})

\theta=79.7^{\circ}

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c).  The angle between vectors is 79.7°

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3 years ago
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