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Rina8888 [55]
3 years ago
15

A plastic rod is charged up by rubbing a wool cloth, and brought to an initially neutral metallic sphere that is insulated from

ground. It is allowed to touch the sphere for a few seconds, and then is separated from the sphere by a small distance. After the rod is separated, the rod:A) is repelled by the sphere.B) is attracted to the sphere.C) feels no force due to the sphere.
Physics
1 answer:
Pavlova-9 [17]3 years ago
4 0

Answer:

A) is repelled by the sphere.

Explanation:

On rubbing the plastic rod with the wool cloth the rod gains some electrons from the surface of the wool and becomes electrostatically negative in charge.

When this rod is brought near to a neutral metallic sphere then the electrons of the sphere get repelled from the nearest portion of the metallic sphere as it a conductor and the electrons accumulate on the farthest opposite side of the rod.

But when the rod is brought into contact for some time then the from the portion of the rod which is in contact to the sphere loses the electrons from that region to the sphere since plastic is not an electrical conductor so not all the charges travel to the sphere.

Then when the rod is separated, the charges on the sphere spread uniformly and the similar charged rod faces repulsion.

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An object moves along a straight line path from P to Q under the action of a force (4 3 3 ) N. I j k − + If the coordinates of P
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1 Nm

Explanation:

Given;

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Position 2 = Q = (2, -1, 4)m

The object moves along a straight line path from P to Q, therefore, the distance vector (d) is given by;

d = Q - P

d = (3, 3, -1) - (2, -1, 4)

d = (1, 4, -5)m

Now the work done (W) by the force (F) to move through the distance (d) is the dot product of the two vectors: F and d. i.e

W = F . d

For clarity, let's write vectors F and d in vector unit notation as follows;

F = 4 i + 3 j + 3 k

d = 1 i + 4 j - 5k

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W = (4 i + 3 j + 3 k ) . (1 i + 4 j - 5k)

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6 0
3 years ago
A 3 ϕ, 11 kV, 60 Hz, 25 MVA, Y-connected cylindrical-rotor synchronous machine generator has Ra =0:45 Ω per phase and Xs =4:5 Ω
Vedmedyk [2.9K]

Answer:

Excitation Voltage = Ef = 11.012 kV / phase

Power Angle = δ = + 25.29

Explanation:

Given Data:

Line Voltage: VL = 11 kV

Power Factor: Pf = 0.85 lagging

Synchronous Reactance: Xs = 4.5 ohm / phase

Armature Resistance: Ra = 0.45 ohm / phase

Required:

Excitation Voltage: Ef to be determined

Solution:

The excitation voltage for cylindrical/round rotor synchronous generator can be calculated using formula:  

EF = Vp + Ia Ra + j Xs Ia

In phasor form, the equation can be written as:

Ef ∠δ = Vp ∠0° + (Ia∠-θ)(Ra) + j( Xs)( Ia∠-θ)-------------------- (1)

Where:

Ef ∠δ  = Excitation voltage (δ is the angle between EF and Vp)

Vp = Phase Voltage (Vp is taken as reference phase that’s why angle is 0°)

Ia∠-θ = Armature Current (θ is angle between Vp and Ia, minus sign is taken because the pf is lagging)

<u> </u>

<u>Calculating Vp: </u>

Since, for Y-Connection,

Vp = VL / Sqrt (3)

Vp = 11000 / Sqrt (3)

Vp = 6350.85 V / phase

<u>Calculating Ia :</u>

Since, Power output from the generator is given as:

P = Sqrt (3) x VL x IL x Cos θ

Since, phase current is equal to line current in Y-Connection, therefore, IL = Ip = Ia. Therefore, rearranging the above equation to find Ia:

Ia = P / Sqrt (3) x VL x Cos θ

Since, Apparent Power (S) = Active Power (P) / Power Factor (Cos @) = 25 MVA,  

Therefore,

Ia = S / Sqrt (3) x VL

Ia = 25 MVA / Sqrt (3) x (11 kV)

Ia = 1312.16 A

Since, Cos @ = 0.85

@ = Cos-1 (0.85)

@ = 31.79 degrees

Therefore, Ia = 1312.16 ∠- 31.79 (minus sign is taken because the power factor is lagging)

Calculating Excitation Voltage (Ef):  

Substituting all values in equation (1), we get,

Ef ∠δ  = 6350.85 <0 + (1312.16 ∠-31.79)(0.45) + j(4.5)( 1312.16∠-31.79)

Ef ∠δ  = 9963.4 + j4707.86

Ef ∠δ  = 11019.7 ∠25.29

Excitation Voltage = Ef = 11.012 kV / phase

Power Angle = δ = + 25.29

7 0
3 years ago
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