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Vaselesa [24]
3 years ago
10

A negative charge is moved from point A to point B along an equipotential surface. Which of the following statements must be tru

e for this case? A) The negative charge performs work in moving from point A to point B. B) Work is required to move the negative charge from point A to point B. C) No work is required to move the negative charge from point A to point B. D) The work done on the charge depends on the distance between A and B. E) Work is done in moving the negative charge from point A to point B.
Physics
1 answer:
elena-s [515]3 years ago
7 0

Answer:

C) No work is required to move the negative charge from point A to point B.

Explanation:

An equipotential surface is defined as a surface connecting all the points at the same potential.

Therefore, when a charge moves along an equipotential surface, it moves between points at same potential.

The work done when moving a charge is given by

W=q\Delta V

where

q is the charge

\Delta V is the potential difference between the initial and final point of motion of the charge

However, the charge in this problem moves along an equipotential surface: this means that the potential does not change, so

\Delta V=0

And so, the work done is also zero.

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A 16.0 Ω, 13.0 Ω, and 7.00 Ω resistor are connected in parallel to an emf source. A current of 6.00 A is in the 13.0 Ω resistor.
Darya [45]

Answer:3.54ohms

Explanation: connection in parallel

1/Rt= 1/R1+1/R2+1/R3

1/Rt= 1/16+1/13+1/7

1/Rt= 91+112+208/1456

1/Rt= 411/1456

411Rt= 1456

Rt= 1456/411

Rt= 3.54ohms

3 0
3 years ago
Read 2 more answers
Look at the diagram. Emily needs to complete the circuit. What two points should be connected to complete the circuit and make t
pogonyaev

Answer:

The points 2 and 4 should be connected.

Explanation:

To complete the circuit, we need to connect the two points which when connected, encompass the battery and the bulb in the circuit. The points 2 and 4 do the job, since they connect the terminal of the battery and the terminal of the bulb, and thus complete the circuit.

Therefore,  the choice C is correct.

4 0
3 years ago
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Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441
Tasya [4]
The period of a simple pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
g= \frac{4 \pi^2}{T^2}L (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz
And its period is the reciprocal of its frequency:
T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s

So now we can use eq.(1) to find the gravitational acceleration of the planet:
g= \frac{4 \pi^2}{T^2}L =  \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2
3 0
3 years ago
Which type of force can act through empty space?
katrin2010 [14]

D. Magnetism this is shown by the planet's magnetic pull on moons and space debris.

8 0
3 years ago
The driver of a car travels at 90 km / h, observes some children playing on the road 50 m away, and applies the brakes, managing
Orlov [11]

Answer:

13,750 N

Yes

Explanation:

Given:

v₀ = 90 km/h = 25 m/s

v = 0 m/s

t = 4 s

Find: a and Δx

a = Δv / Δt

a = (0 m/s − 25 m/s) / (4 s)

a = -6.25 m/s²

F = ma

F = (2200 kg) (-6.25 m/s²)

F = -13,750 N

Δx = ½ (v + v₀) t

Δx = ½ (0 m/s + 25 m/s) (4 s)

Δx = 50 m

6 0
3 years ago
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