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2 years ago

6

A man loads 260.5 kg of black dirt into his pickup. 60,452 g blows out on the ride home. How much black dirt does the man have w

hen he reaches home? (with the correct number of significant figures)

1 answer:

Ivan2 years ago

7 0

Answer:

200.048 kg

Explanation:

Total Black dirt loaded in the pickup (A) = 260.5 kg = 260.500 kg

Quantity of the dirt that blew away (B )= 60452 g = 60.452 kg

Remaining quantity of the black dirt is

$= 260.500 - 60.452$

$= 200.048$

Thus, the amount of black dirt the man has when he reaches home = 200.048 kg.

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Answer: (a) Smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) Smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

Explanation:

(a) When both the spheres have same charge then force is repulsive in nature as like charges tend to repel each other.

Therefore, total charge on the two non-conducting spheres will be calculated as follows.

$Q_{1} + Q_{2} = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

= $9 \times 10^{-5} C$

Therefore, force between the two spheres will be calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

or, $Q_{1}(9 \times 10^{-5} - Q_{1}) = 0.104 \times 10^{-9} C^{2}$

$9 \times 10^{-5}Q_{1} - Q^{2}_{1} = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} + 0.104 \times 10^{-9} = 0$

$Q_{1} = 11.7 \times 10^{-5} C, 2.7 \times 10^{-5} C$

This means that smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) When force is attractive in nature then it means both the charges are of opposite sign.

Hence, total charge on the non-conducting sphere is as follows.

$Q_{1} + (-Q_{2}) = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

$Q_{1} - Q_{2} = 9 \times 10^{-5} C$

Now, force between the two spheres is calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

$Q_{1}(Q_{1} - 9 \times 10^{-5}) = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} = 0.104 \times 10^{-9} C^{2}$

$Q_{1} = -11.4 \times 10^{-5}, 9.1 \times 10^{-5}$

Hence, smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

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