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3 years ago

9

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?a. The tube should be held horizontally, parallel to the ground b. The potential difference between the ends of the tube does not depend on the tube's orientation. c. The tube should be held vertically perpendicular to the ground

1 answer:

Hunter-Best [27]3 years ago

6 0

Answer:

b) True. potencial diferencie does not depend on orientation

Explanation:

In this exercise we are asked to show which statements are true.

The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.

It does not refer to the height of the system.

We can now review the claims

a) False. Potential not to be refers to height

b) True. Does not depend on orientation

c) False The potential does not refer to the altitude but to the Earth's charge

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Sarah weighs 392 N. She pushes a 56 kg rock with a force of 65 N. What force does the rock exert on Sarah?

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Answer:

The force the rock exerts on Sarah =#is 65 N

Explanation:

The given parameters are;

Sarah's weight = 392 N

The force with which Sarah pushes the rock = 65 N

The mass of the rock = 56 kg

The weight of the rock = The mass of the rock × Acceleration due to gravity

∴ The weight of the rock = 56 kg ×9.81 m/s² = 549.36 N

Given that the force Sarah applies to push the rock = 65 N, then by Newton's third law of motion which states that action and reaction are equal and opposite, the force that the rock exert on Sarah is equal an opposite to the force Sarah is applying

Therefore, the force the rock exerts on Sarah = 65 N (in the opposite direction).

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A runner circles the track exactly 2 times for distance of 800m. it takes 4.0min. what is her average speed in m/s? what is her

Sindrei [870]

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$\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{0\,\mathrm m}{4.0\,\mathrm{min}\cdot\frac{60\,\mathrm s}{1\,\mathrm{min}}=0\,\dfrac{\mathrm m}{\mathrm s}$

The total distance she traversed, however, was $\Delta d=800\,\mathrm m$ , which gives her an average speed of

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3 years ago

a loaded sack of total mass is 1000 gramme falls down from the floor of a lorry 200cm high, calculate the workdone by the gravit

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Answer:

W = 20 J

Explanation:

Given that,

The mass of a loaded sack, m = 1000 g = 1 kg

It falls down from the floor of a lorry 200 cm high, h = 2 m

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W = mgh

Substitute all the values,

W = 1 × 10 × 2

= 20 J

Hence, the required work done by gravity is equal to 20 J.

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3 years ago

A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that emits isotropically at a certa

Aliun [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

b

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

c

The ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

Explanation:

Mathematically Frequency can be represented as

$F = \frac{v}{\lambda}$

Where $\lambda$ is the wavelength and v is the velocity

Now looking at the diagram we see that

For the first frequency we have

Let the wavelength be $\lambda_1 = \lambda$ , and the frequency $F_1 = F$

For the second frequency

Let the wavelength be $\lambda_2 = 2 \lambda$ , and the frequency $F_2 = \frac{F}{2}$

For the third frequency

Let the wavelength be $\lambda_3 = 2\lambda$ , and the frequency $F_3 = \frac{F}{2}$

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

So,

For the first frequency we have

$v_1 = \lambda_1 F_1 = \lambda F$

For the second frequency

$v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F$

For the third frequency

$v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F$

Hence

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

Mathematically angular speed can be represented as

$w = 2 \pi f$

For the first frequency we have

$w_1 = 2\pi F_1 = 2 \pi F$

For the second frequency

$w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2} = \pi F$

For the third frequency

$w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2} = \pi F$

Hence

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

$v = wr$

=> $r = \frac{v}{w}$

Since the linear velocity is constant we have that

$r \ \alpha \ \frac{1}{w}$

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

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3 years ago