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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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Answer:

23.96 N

Explanation:

From the question given above, the following data were obtained:

Mass of Chihuahua (m) = 3.63 kg

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Time (t) = 0.50 s

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Next, we shall determine the acceleration of the Chihuahua. This can be obtained as follow:

Velocity (v) = 3.3m/s

Time (t) = 0.50 s

Acceleration (a) =?

a = v/t

a = 3.3/0.5

a = 6.6 m/s²

Thus, the acceleration of the Chihuahua is 6.6 m/s².

Finally, we shall determine the force need to stop the Chihuahua as shown below:

Mass of Chihuahua (m) = 3.63 kg

Acceleration (a) = 6.6 m/s².

Force (F) =?

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the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)

Explanation:

assuming ideal gas behaviour of the gas , the equation for ideal gas is

P*V=n*R*T

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P = absolute pressure

V= volume

T= absolute temperature

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P=n*R*T/V

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