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sattari [20]
3 years ago
10

A uniform electric field is directed parallel to the +y axis. If a positive test charge begins at the origin and moves upward al

ong the y axis, how does the electric potential vary, if at all?
1. The electric potential will decrease with increasing y.
2. The electric potential will increase with increasing y.
3. The electric potential will remain constant with increasing y.
4. Too little information is given to answer this question.
Physics
1 answer:
lukranit [14]3 years ago
3 0

Answer: option 1 : the electric potential will decrease with an increase in y

Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below

V = kq/y

Where k = 1/4πε0

Where V = electric potential,

k = electric constant = 9×10^9,

y = distance of potential relative to a reference point, ε0 = permittivity of free space

q = magnitude of electronic charge = 1.609×10^-19 c

From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.

We have that

V = k/y

We see the potential(V) is inversely proportional to distance (y).

This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.

This fact makes option 1 the correct answer

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Using examples, explain why the first and second Newton laws of motion are significant for living organisms.
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Answer:

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Explanation:

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6 0
3 years ago
The volume occupied by a sample of gas is 480 mL when the pressure is 115 kPa.What pressure must be applied to the gas to make i
balandron [24]

Answer:

The answer is

<h2>84.9 kPa</h2>

Explanation:

Using Boyle's law to find the final pressure

That's

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the final pressure

P_2 =  \frac{P_1V_1}{V_2}

From the question

P1 = 115 kPa

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So we have

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We have the final answer as

<h3>84.9 kPa</h3>

Hope this helps you

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