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ozzi
4 years ago
13

Two charges that are 1 meter apart repel each other with a force of 2 N. If the distance between the charges is increased to 2 m

eters, the force of repulsion will be:
a) 1 N b) 0.5 N c) 8 N d) 4 N
Physics
1 answer:
Savatey [412]4 years ago
5 0

Answer:

b) 0.5 N

Explanation:

From coulomb's law,

F = kq'q/r².................... Equation 1

Where F =force of repulsion between the charges, q' = first charge, q = second charge, r = distance between the charges, k = proportionality constant.

q'q = Fr²/k........................... Equation 2

Given: F = 2 N, r = 1 m, k = 9.0×10⁹ Nm²/C²

Substituting into equation 2

q'q = 2(1)²/(9.0×10⁹)

q'q =  2/9.0×10⁹ C².

If the distance between the charges is increased to 2 meters,

r = 2 m, q'q = 2/9.0×10⁹ C².

Substitute into equation 1

F = 9.0×10⁹(2/9.0×10⁹)/2²

F = 2/4

F = 1/2 = 0.5 N.

The right option is b) 0.5 N

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Answer:

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Explanation:

Given that;

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first we determine the volume of the ball using the following equation;

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we substitute

V = 4/3×π(0.25)³

V =  0.06544 mm³

Now form table 1.1 "Grain sizes" a metal with grain size number of 12 has about 4,200,000 grains/mm³

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Number of grains N = 0.06544 × 4,200,000

N = 274,848 grains

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3 years ago
A total of 25.6 kJ of heat energy is added to a 5.46 L sample of helium at 0.991 atm. The gas is allowed to expand against a fix
bagirrra123 [75]

Answer:

(a) W = 1329.5 J = 1.33 KJ

(b) ΔU = 24.27 KJ

Explanation:

(a)

Work done by the gas can be found by the following formula:

W = P\Delta V

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W = Work = ?

P = constant pressure = (0.991 atm)(\frac{101325\ Pa}{1\ atm}) = 100413 Pa

ΔV = Change in Volume = 18.7 L - 5.46 L = (13.24 L)(\frac{0.001\ m^3}{1\ L}) = 0.01324 m³

Therefore,

W = (100413 Pa)(0.01324 m³)

<u>W = 1329.5 J = 1.33 KJ</u>

<u></u>

(b)

Using the first law of thermodynamics:

ΔU = ΔQ - W (negative W for the work done by the system)

where,

ΔU = change in internal energy of the gas = ?

ΔQ = heat added to the system = 25.6 KJ

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ΔU = 25.6 KJ - 1.33 KJ

<u>ΔU = 24.27 KJ</u>

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Explanation:

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4 years ago
Read 2 more answers
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Delicious77 [7]

Answer: A) Deceleration of the car is -6.6667 m/s² while it came to stop.

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Explanation:

Given Data

Initial velocity of the car ($$v_{i}$$) =   20.0 m/s

Final velocity of the car (v_{f}) = 0 m/s

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Time (in rest) =3 s

To find - A) car's deceleration while it came to a stop

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Deceleration = (( final velocity - initial velocity ) ÷ Time)        (m/s²)

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Distance traveled by the car is equals to the product of the speed and time

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