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Natali5045456 [20]
3 years ago
13

1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t

he electric force? The constant k = 9 x 109 N-m2/C2
Physics
1 answer:
tiny-mole [99]3 years ago
6 0

Explanation:

Charges,q_1=8\ \mu C=8\times 10^{-6}\ C

q_2=-5\ \mu C=-5\times 10^{-6}\ C

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.

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6 0
3 years ago
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Hen a gfci receptacle device is installed on a 20-ampere branch circuit (12 awg copper), what is the minimum volume allowance (i
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a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diame
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Answer:

The upper limit on the flow rate = 39.46 ft³/hr

Explanation:

Using Ergun Equation to calculate the pressure drop across packed bed;

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However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

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where;

ΔP = pressure drop

Q = flow rate

Given that:

9.6 = A12 + B12²

Then

12A + 144B = 9.6       --------------   equation (1)

24A + 576B = 24.1    ---------------  equation (2)

Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

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Q² + 35.02Q - 2938.8 = 0

Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

Q = 39.46 ft³/hr

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