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garri49 [273]
3 years ago
5

Select True or False for the follwing statements about electric field lines.

Physics
1 answer:
aalyn [17]3 years ago
5 0

Answer:

Please see below as the answer is self explanatory.

Explanation:

1) A positive point charge released from rest will initially accelerate along an E-field line : TRUE.

The electric field lines have the same direction that will have a positive test charge, when acted on by the force exerted by the charge distribution that creates the field.

2) E-Field lines make circles around positive charges: FALSE.

E-Field lines start in positive charges and finish in negative charges, except at infinity.

3) E-field lines point inward toward negative charges: TRUE

As the E-field lines follow the trajectory that would have a positive test charge, as negative charges attract the positive charges, the positive test charge will travel towards the negative charges.

4) E-Field line may cross: FALSE

As the E-Field is the gradient of the electric potential (so they have the direction of the maximum change of the potential), if  they may cross, this would mean that the potential could be changing differently at the same point, which it is not possible.

5) E-Field lines point outward from positive charges: TRUE

As explained above, the E-field lines follow the trajectory that it would have a positive test charge, so if the field is created by a positive charge, the test charge would go away from it, so the field lines point outward from positive charges.

6) Where the E-field lines are dense the E-field must be weak : FALSE

By convention, the E-field lines are proportional to the charge that creates the field, so if the E-field lines are dense, this means that the field is strong.

7) E-field lines do not begin or end in a charge-free region except at infinity: TRUE.

Except at infinity, E-field lines originate at positive charges and finish at negative charges, so they can't originate in a charge-free region.

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Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
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Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

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We have: Gravitational Potential Energy (U) = mgh

Here, m = 3 Kg

g = 9.8 m/s²        [ constant value for earth system ]

h = 3 m


Substitute their values into the expression:

U = 3 × 9.8 × 3

U = 88.2 J


In short, Your Answer would be 88.2 Joules


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Answer:

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Starting from the front door of your ranch house, you walk 50.0 m due east to your windmill, and then you turn around and slowly
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Answer:

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Average speed,

S=\frac{D}{t}\\ S=\frac{90}{70}\\ S=1.29\frac{m}{s}

Explanation:

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We have to find the average velocity. We know that velocity is defined as the rate of change of displacement with respect to time.

To find the average velocity we have to find the total displacement.

since displacement along east direction is 50m

and displacement along west=40m

so total displacement,

d=50m-40m\\d=10m

total time,

t=28 s+42 s\\t=70 s

therefore, average velocity

v=\frac{d}{t}\\ v=\frac{10m}{70s}\\v=.1428 \frac{m}{s}

(b)Average Speed:

Average speed is defined as the ratio of total distance to the total time

it means

Average speed= total distance/total time

here total distance,

D= 50m+40m\\D=90m

and total time,

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therefore,

Average speed,

S=\frac{D}{t}\\ S=\frac{90}{70}\\ S=1.29\frac{m}{s}

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