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kipiarov [429]
3 years ago
9

Two point charges of +2.50 x 10^-5 C and -2.50 x 10^-5 C are separated by 0.50m. Which of the following describes the force betw

een them?
A): 90 N, repulsive
B): 90 N, attractive
C): 23 N, attractive
D): 45 N, attractive

If one could answer this ASAP, it will be greatly appreciated
Physics
1 answer:
mr Goodwill [35]3 years ago
5 0

Answer:

Q1 = +2.50 x 10^-5C and Q2 = -2.50 x 10^-5C, r = 0.50m, F=?

Using Coulomb's law:

F = 1/(4πE) x Q1 x Q2/ r^2

Where

k= 1/(4πE) = 9 x 10^9Nm2/C2

Therefore,

F = 9x 10^9 x 2.50 x 10^-5 x2.50 x

10^-5/. ( 0.5)^2

F= 5.625/ 0.25

F= 22.5N approximately

F= 23N.

To find the direction of the force: since Q1 is positive and Q2 is negative, the force along Q1 and Q2 is force of attraction.

Hence To = 23N, attractive. C ans.

Thanks.

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Using formula of torque

\tau=Force\times lever\ arm

So, Here,

\tau_{2}=5 \tau_{1}

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

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Where, F_{1}=First force

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Answer:

a) m = 1.174 grams

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Explanation:

Given:

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- Pressure of air P = 101.3 KPa

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- Universal Gas constant R = 8.314 J/kgK

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Solution:

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- Use Ideal gas law equation and compute number of moles of air n:

                               P*V = n*R*T

                                n = P*V / R*T

                                n = 101.3*10^3 * 0.001 / 8.314*300

                                n = 0.04061 moles

- Compute the mass of the gas:

                                m = n*M

                                m = 0.04061*28.9

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- The gravitational force exerted on the mass of gas is due to its weight:

                                F_g = m*g

                                F_g = 1.174*9.81*10^-3

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                                F_c = P*A

                                F_c = (101.3*10^3)*(0.1)^2

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