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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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An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, KE_{t} = \frac{3}{2}k_{b}T

k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

\lambda_{e} = \frac{h}{p_{e}}

\lambda_{e} = \frac{h}{m_{e}{v_{e}}              (1)

where

h = Planck's constant = 6.626\times 10^{- 34}m^{2}kg/s

p_{e} = momentum of an electron

v_{e} = velocity of an electron

m_{e} = 9.1\times 10_{- 31} kg = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

v_{e} = 2.86\times 10^{5} m/s                    (2)

Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

where

m_{p} = 1.6726\times 10_{- 27} kg = mass of proton

v_{p} = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T

}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}

v_{p} = 6.674\times 10^{3} m/s                               (4)                    

Using eqn (4) in (3):

\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

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3 years ago
If you cut a piece of wood in half, each half would have *
Anon25 [30]

Answer:

c

Explanation:

let's assume that you are splitting the piece of wood horizontally. This would not change the density of the wood at all, only the length. I rest my case. Ur awesome and have an amazing day.

7 0
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give an example of motion in which displacement of the particle is zero but acceleration is not zero in journey?
BartSMP [9]

Motion of a ball thrown by a person upwards and caught after some time is an example of motion in which displacement of the particle is zero but acceleration is not zero in journey.

The displacement of the ball is zero because the starting and end point of the motion are same, i.e, the person's hands.During its motion, the acceleration of ball is constant and non zero called as acceleration due to gravity, g= -9.8 m/s². The velocity of ball is continuously changing. It first decreases during the upward motion of the ball and then increases during the downward journey.The acceleration remains constant and non zero all the time.

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Identify two fields where physical quantities are used in motion calculations​
larisa86 [58]

The two fields were physical quantities are used in motion calculations are length and mass with time.

The physical quantity in a field is referred as every point in a particular space time.

<h3>How physical quantities are used in motion calculations?</h3>

 If we consider an object, the physical property of the object is considered as physical quantity and to measure that object is known as units. The Physical quantity can be classified as elemental physical quantity and derived physical quantity. Length, mass, time, etc.. are elemental physical quantity, momentum, density, acceleration, etc... are derived physical quantity. Only for charge and temperature the physical quantity will be less than zero.

Length, mass and time  are the physical quantities used in motion calculations.

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Two blocks, stacked one on top of the other, slide on a frictionless horizontal surface. The surface between the two blocks is r
devlian [24]

Answer:

F_{net} = 31.88 N

Explanation:

When top block is just or about to slide on the lower block then we can say that the frictional force on it will be maximum static friction

So we will have

F_{net} = ma

F_s = ma

\mu_s mg = ma

a = \mu_s g

a = (0.50)(9.81)

a = 4.905 m/s^2

now for the Net force on two blocks to move together

F_{net} = (m_1 + m_2) a

F_{net} = (2.3 + 4.2)(4.905)

F_{net} = 31.88 N

4 0
3 years ago
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