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german
3 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]3 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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A student rides a bicycle for 15 miles in 3 hours. What is the student's speed? What else would you need to know for the velocit
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Answer:

5 miles per hour

Explanation:

if you divide 15 by 3 you get 5, therefore the student is going 5 miles per hour.

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3 years ago
Find the resultant force of the following forces :
bezimeni [28]

The resultant of the given forces is; 6√2 N

<h3>How to find the resultant of forces</h3>

We are given the forces as;

10 N along the x-axis which is +10 N in the x-direction

6 N along the y-axis which is +6N in the y-direction

4 N along the negative x-axis which is -4N

Thus;

Resultant force in the x-direction is; 10 - 4 = 6N

Resultant force in the y-direction is; 6N

Thus;

Total resultant force = √(6² + 6²)

Total resultant force = 6√2 N

Read more about finding resultant of a force at; brainly.com/question/14626208

4 0
3 years ago
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A 1500 kg car travelling at 25 m/s collides with a 2500 kg van which had
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4 years ago
Which of the following is located in the temperature climate zone?
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5 0
4 years ago
The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff
stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet  m = 45 kg

coefficient of static friction μ =  0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

\sum f_y =0

The upward force and the downward force is :

N_A+N_B = mg

where;

\mathbf { N_A  \ and  \ N_B} are the normal contact force at center point A and B respectively .

N_A+N_B = 45*9.8

N_A+N_B = 441    ------- equation (1)

Considering the forces on the horizontal axis:

\sum f_x = 0

F_A +F_B  = P

where ;

\mathbf{ F_A \ and \ F_B } are the static friction at center point A and B respectively.

which can be written also as:

\mu_s N_A + \mu_s N_B  = P

\mu_s( N_A +  N_B)  = P

replacing our value from equation (1)

P = 0.30 ( 441)    

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for  the cabinet is not to tip over is calculated by determining the limiting condition  of the unbalanced torque whose effect is canceled by the normal reaction at N_A and it is shifted to N_B:  

Then:

\sum M _B = 0

P*h = mg*0.24

h =\frac{45*9.8*0.24}{132.3}

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

6 0
3 years ago
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