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3 years ago

What is the relationship between power and voltage?

Physics

1 answer:

kipiarov [429]3 years ago

3 0

<u>Answer:</u>

The power and voltage are related through Power Rule .

<u> Explanation : </u>

Power Rule states that the current I that flows the element in the circuit with a loss in voltage V, then the amount of power dissipated by that element in the circuit is the multiplicative product of voltage and the current.

Mathematically, power law is

$\boldsymbol{P}=\boldsymbol{I} \times \boldsymbol{V}$

The rate of work done or the rate of energy consumption or production is termed as power.Unit of power is denoted as W(watts).

Potential energy between any two points on a circuit is called as Voltage and is measured in volts (V) .

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A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread nonuniformly throug

Vaselesa [24]

Answer:

$1.57 * 10^{3} Q$

Explanation:

The volume charge density is defined by ρ = $\frac{Q}{V}$ (Equation A), where Q is the charge and V, the volume.

The units in the S.I. are $\frac{Coulombs}{m^{3} }$ , so we have to express the radius in meters:

inner radius = $4 cm * \frac{1 m}{100 cm} = 0.04m$

outer radius = $6 cm * \frac{1m}{100cm} = 0.06m$

Now, we know that the volume of the sphere is calculated by the formula:

$V = \frac{4}{3}\pi r^{3}$ , and as we have an spherical shell, the volume is calculated by the difference between the outher and inner spheres:

V = $\frac{4}{3}\pi (r_{o} ^{3} - r_{i} ^{3})$ , where $r_{o}$ is the outer radius and $r_{i}$ is the inner radius.

Replacing the volume formula in the Equation A:

ρ = $\frac{Q}{\frac{4}{3}\pi(r^{3} _{o}-r_{i} ^{3})}$

ρ = $\frac{3Q}{4\pi (r_{o} ^{3}-r_{i} ^{3} ) }$

Replacing the values of the outer and inner radius whe have:

ρ = $\frac{3Q}{4\pi (1.52 * 10^{-4})}$

ρ = $1.57 * 10^{3} Q$

4 0

3 years ago

An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-

kirill [66]

Answer:0.478 c

Explanation:

Given

mass of lighter Particle $(m_1)=3\times 10^{-28} kg$

mass of heavier Particle $(m_2)=1.51\times 10^{-27} kg$

speed of lighter particle $(v_1)=0.834 c$

Let speed of heavier particle $=v_2$

and Momentum of the particle is given by

$P=\frac{mv}{\sqrt{1-(\frac{v}{c})^2}}$

$P_1=\frac{m_1v_1}{\sqrt{1-(\frac{v_1}{c})^2}}$

$P_1=\frac{3\times 10^{-28}\times 0.834 c}{\sqrt{1-(\frac{0.834 c}{c})^2}}$

$P_1=8.219\times 10^{-28} kg c$

$P_2=\frac{m_2v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

as momentum is conserved therefore $P_1=P_2$

$8.219\times 10^{-28} kg c=\frac{1.51\times 10^{-27}\times v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

$v_2=0.478 c$

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