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7nadin3 [17]
3 years ago
12

Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric

force on sphere B due to sphere A is F . F. The magnitude of the electric force on sphere A due to sphere B must be:
Physics
1 answer:
Goshia [24]3 years ago
3 0

Complete Question:

Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric force on sphere B due to sphere A is F . F. The magnitude of the electric force on sphere A due to sphere B must be:

A. 2F

B. F/4

C. F/2

D. F

E. 4F

Answer:

D.

Explanation:

If both spheres can be treated as point charges, they must obey the Coulomb's law, that can be written as follows (in magnitude):

F =\frac{kQ*2Q}{r^{2} }

As it can be seen, this force is proportional to the product of the charges, so it must be the same for both charges.

As this force obeys also the Newton's 3rd Law, we conclude that the magnitude of the electric force on sphere A due to sphere B, must be equal to the the magnitude of the force on the sphere B due to the sphere A, i.e., just F.

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Answer:

T_2 = 0.592

Explanation:

Given

T_1 = 1.48s

See attachment for connection

Required

Determine the time constant in (b)

First, we calculate the total capacitance (C1) in (a):

The upper two connections are connected serially:

So, we have:

\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}

Take LCM

\frac{1}{C_{up}} = \frac{1+1}{C}

\frac{1}{C_{up}}= \frac{2}{C}

Cross Multiply

C_{up} * 2 = C * 1

C_{up} * 2 = C

Make C_{up} the subject

C_{up} = \frac{1}{2}C

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

C_1 = 2 * C_{up}

C_1 = 2 * \frac{1}{2}C

C_1 = C

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

C_p = C+C

C_p = 2C

So, the total capacitance (C2) is:

\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}

\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}

Take LCM

\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}

\frac{1}{C_2} = \frac{5}{2C}

Inverse both sides

C_2 = \frac{2}{5}C

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

T\ \alpha\ C

Convert to equation

T\ =kC

Make k the subject

k = \frac{T}{C}

k = \frac{T_1}{C_1} = \frac{T_2}{C_2}

\frac{T_1}{C_1} = \frac{T_2}{C_2}

Make T2 the subject

T_2 = \frac{T_1 * C_2}{C_1}

Substitute values for T1, C1 and C2

T_2 = \frac{1.48 * \frac{2}{5}C}{C}

T_2 = \frac{1.48 * \frac{2}{5}}{1}

T_2 = \frac{0.592}{1}

T_2 = 0.592

Hence, the time constance of (b) is 0.592 s

8 0
2 years ago
A person notices a mild shock if the current along a path through the thumb and index finger exceeds 82 µA. Find the maximum all
Drupady [299]

Answer:

V_{voltage}=19.68V

Explanation:

Given data

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Resistance R=2.4×10⁵Ω

to find

Voltage

Solution

From Ohms law we know that:

V_{voltage}=I_{current}*R_{resistance}\\V_{voltage}=82*10^{-6}*2.4*10^{5}   \\V_{voltage}=19.68V

4 0
3 years ago
The force of gravity is an inverse square law. this means that, if you double the distance between two large masses, the gravita
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F~1/r²
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3 years ago
A tube is sealed at both ends and contains a 0.0100-m long portion of liquid. The length of the tube is large compared to 0.0100
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Answer:

31.321 rad/s

Explanation:

L = Tube length

A = Area of tube

\rho = Density of fluid

v = Fluid velocity

m = Mass = \rho Al

Centripetal force is given by

F=\dfrac{mv^2}{L}\\ F=\dfrac{m(\omega L)^2}{L}\\ F=m\omega^2\\ F= 0.01A\rho\omega^2L

Pressure is given by

P=\dfrac{F}{A}=\rho gL\\\Rightarrow \dfrac{0.01A\rho\omega^2L}{A}=\rho gL\\\Rightarrow 0.01\omega^2=g\\\Rightarrow \omega^2=\dfrac{g}{0.01}\\\Rightarrow \omega=\sqrt{\dfrac{g}{0.01}}\\\Rightarrow \omega=\sqrt{\dfrac{9.81}{0.01}}\\\Rightarrow \omega=31.321\ rad/s

The angular speed of the tube is 31.321 rad/s

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6. Two light bulbs are designed for use at 120 V and are rated at 75 W and 150 W. Which light bulb has the greater filament resi
Nonamiya [84]

\\ \bull\tt\longrightarrow P=\dfrac{V^2}{R}

  • P is power
  • R is resistance

\\ \bull\tt\longrightarrow R=\dfrac{V^2}{P}

Hence

\\ \bull\tt\longrightarrow R\propto V

\\ \bull\tt\longrightarrow R\propto \dfrac{1}{P}

  • Therefore if power is low then resistance will be high.

The first bulb has less power hence it has greater filament resistance.

5 0
3 years ago
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