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german
2 years ago
15

The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr

ound. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held? Select the best answer from the choices provided. View Available Hint(s) Select the best answer from the choices provided. The potential difference between the ends of the tube does not depend on the tube's orientation. The tube should be held horizontally, parallel to the ground. The tube should be held vertically, perpendicular to the ground.
Physics
1 answer:
Amanda [17]2 years ago
6 0

Answer:

The tube should be held vertically and perpendicular to the ground.

Explanation:

Answer: The tube should be held vertically and perpendicular to the ground. The reason is as follows:

Reasoning:

The power lines are parallel to the ground hence, their electric field will be perpendicular to the ground and equipotential surface will be cylindrical.

Hence, if you will put fluorescent tube parallel to the ground then both the ends of the tube will lie on the same equipotential surface and the potential difference will be zero.

So, to maximize the potential the ends of the tube must be on different equipotential surfaces. The surface which is near to the power line has high potential value and the surface which is farther from the line has lower potential value.

hence, to maximize the potential difference, the tube must be placed perpendicular to the ground.

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One of the harmonics of a string fixed at both ends has a frequency of 52.2 Hz and the next higher harmonic has a frequency of 6
Angelina_Jolie [31]

Solution :

Frequency may be defined as the number of observation or number of waves that is taken in per unit time. The unit of frequency is Hertz or Hz.

It is given that :

Successive harmonic frequencies, f = 52.2 Hz

and f' = 60.9 Hz

Therefore, fundamental frequency, F = f' - f

                                                           F = 60.9 - 52.2

                                                          F = 8.7 Hz

Therefore the string which is fixed at both the ends forms all the harmonics.

6 0
2 years ago
Using the same cost and time estimates, consider the time-effectiveness of each engineer.
sammy [17]

Answer:

Camilla

Explanation:

I got it right on edge. :)

4 0
3 years ago
When you lift a box off the ground gravity opposes your lifting force.
ivolga24 [154]

Answer:

TRUE?

Explanation:

Im not really sure what your question is.

6 0
3 years ago
And object which is 3cm high is placed vertically 10cm in front of a concave mirror. If this object produces an image 40cm from
Stolb23 [73]

Answer:

25cm is the answer I think

8 0
2 years ago
Someone drops a brick on a 2.6 kg cart moving at 28.2 cm/s. After the collision, the dropped brick and cart are moving together
Softa [21]

Answer:

2.087 kg

Explanation:

From the law of conservation of momentum,

Total momentum before collision = total momentum after collision.

MU + mu = V(M+m)............................ Equation 1

Where M = mass of  the cart, m = mass of the brick, U = initial velocity of the cart, u = initial velocity of the brick, V = velocity of both cart and brick after collision.

Note: The initial velocity of the brick is zero Thus mu = 0

making m the subject of the equation,

m =(MU/V)-M ........................................ Equation 2

Given: M = 2.6 kg, U = 28.2 cm/s = 0.283 m/s, V = 15.7 cm/s = 0.157 m/s.

Substituting into equation 2,

m = (2.6×0.283/0.157)-2.6

m = 2.087 kg

Thus the mass of the brick = 2.087 kg

5 0
3 years ago
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