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2 years ago

What is the meaning of the reference point in electric potential?.

Physics

1 answer:

mrs_skeptik [129]2 years ago

4 0

Answer:

<h3><u>ELECTRIC POTENTIAL</u></h3>

• the amount of work needed to move a unit charge from a reference point to a specific point against an electric field.

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Whose contributions to astronomy explained how planets were held in their orbits?

ankoles [38]

Answer:

D newton

Explanation:

he did extensive research on gravity, and gravity is what holds planets in orbits.

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3 years ago

What condition is necessary for an object to make a good reference point?

777dan777 [17]

The object is fixed relative to the motion you are trying to describe.

8 0

3 years ago

A rock is thrown downward from an unknown height above the ground with an initial speed of 6.1 m/s. It strikes the ground 1.7 s

insens350 [35]

Answer:

24.531 m

Explanation:

t = Time taken = 1.7 s

u = Initial velocity = 6.1 m/s

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=6.1\times 1.7+\dfrac{1}{2}\times 9.8\times 1.7^2\\\Rightarrow s=24.531\ m$

The initial height of the rock above the ground is 24.531 m

7 0

3 years ago

If the light wave has a wavelength of 10m what would be its velocity

s2008m [1.1K]

If this case could ever happen, the speed would follow from this formula:

$v = f \cdot \lambda$

with f the frequency and lambda the wavelength. We are give a wavelength of 10m. The frequencies of the visible light can range between 400 to about 790 Terahertz, so let us pick a middle point of 600 THz ("green-ish") as a "representative."

$v = 600THz\cdot 10m = 6\cdot 10^{14} \frac{1}{s}\cdot 10 m = 6\cdot10^{15}\frac{m}{s}$

The speed of such a wave would have to be 6e+15 m/s (which would be 7 orders of magnitude higher than the universal speed of light constant)

7 0

3 years ago

Read 2 more answers

Potassium ions (K+) move across a 7.0 -mm- thick cell membrane from the inside to the outside. The potential inside the cell is

Reil [10]

Explanation:

Relation between potential energy and charge is as follows.

U = qV

or, $\Delta U = q \times \Delta V$

= $1.6 \times 10^{-19} \times 70 \times 10^{-3}$

= $112 \times 10^{-22}$ J

or, = $1.12 \times 10^{20} J$

Therefore, we can conclude that change in the electrical potential energy $\Delta U$ is $1.12 \times 10^{20} J$ .

7 0

3 years ago