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Tomtit [17]
2 years ago
10

The power in an electrical current is given by the equation

Physics
2 answers:
Tcecarenko [31]2 years ago
6 0

Answer:

P = VI

Explanation:

the power is equal to the current × voltage

PtichkaEL [24]2 years ago
6 0

Answer:

P = V • I

Explanation:

Power = Voltage • Current

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<span>You have just demonstrated an insight learning. Internal insight occurs if one learned a new way without the help of environmental factors. In here, what the person initially learned that if one saw a broken light bulb from a lamp, he can be cut through the jagged glass if one does not wear a pair of gloves. And maybe because at the moment, the person could not find one, he felt using a cut potato to pick up the pieces of the broken lamp. This is a demonstration of insight learning. The person found a way to pick up the pieces of the broken lamp by using his instinct at the moment. The person is not influenced by an outside source to tell him to use a potato. </span>
5 0
3 years ago
Under certain circumstances, potassium ions (K+) in a cell will move across the cell membrane from the inside to the outside. Th
choli [55]

Answer:

1.368\times 10^{-20}\ J

Explanation:

q = Charge in the potassium ion = 19e-18e

e = Charge of electron = 1.6\times 10^{-19}\ C

V_2-V_1 = Change in potential = 0-(-85.5\times 10^{-3})

Change in electric potential is given by

E=q(V_2-V_1)\\\Rightarrow E=(19e-18e)(0-(-85.5\times 10^{-3})\\\Rightarrow E=1.6\times 10^{-19}\times 85.5\times 10^{-3}\\\Rightarrow E=1.368\times 10^{-20}\ J

The energy is 1.368\times 10^{-20}\ J

3 0
3 years ago
A pitcher hurls a 0.25kg ball around a vertical circular path of radius 0.6 m, applying a tangential force of 30N, before releas
Sever21 [200]
Thank you for posting your question here at brainly. Below is the solution:

Ke up top = 1/2*.25 *225 
<span>gain of Pot energy = .25*9.81*1.2 </span>
<span>work input = (1/4)(2 pi *.6)*30 </span>

<span>so </span>
<span>sum of those 3 energies = </span>
<span>(1/2)(.25)v^2</span>
3 0
3 years ago
Hi, I don't really know I just guessed.
lapo4ka [179]

Answer:

direction

Explanation:

7 0
3 years ago
A resistor R, inductor L, and capacitor C are connected in series to an AC source of rms voltage ΔV and variable frequency. If t
Alexandra [31]

Answer:E=\frac{\pi R\left ( \Delta V\right )^2\sqrt{LC}}{\left ( R^2+\frac{9}{4}\left (\frac{L}{C} \right )\right )}

Explanation:

We know resonant frequency is given by

\omega_0=\frac{1}{\sqrt{LC}}

and the operating frequency is given by

\omega =2\omega_0=\frac{2}{\sqrt{LC}}

The capacitance reactance is given by

X_c=\frac{1}{\omega C}=\frac{\sqrt{LC}}{2C}=\frac{1}{2}\sqrt{\frac{L}{C}}

inductive reactance is given by

X_L=\omega L=\left ( \frac{2}{\sqrt{LC}}\right )L=2\sqrt{\frac{L}{C}}

Thus impedance is

Z=\left ( R^2+\left (X_L-X_C \right )^2 \right )^\frac{1}{2}

Z=\left ( R^2+\left (2\sqrt{\frac{L}{C}}-\frac{1}{2}\sqrt{\frac{L}{C}} \right )^2 \right )^\frac{1}{2}

Z=\left ( R^2+\frac{9}{4}\left ( \frac{L}{C} \right ) \right )^\frac{1}{2}

The average power delivered is

P_{avg.}=\frac{\Delta V^2}{Z}cos\phi =\frac{\left ( \Delta V\right )^2}{Z}\left (\frac{R}{Z} \right )

P_{avg.}=\frac{\left (\Delta V \right )^2R}{\left ( R^2+\frac{9}{4}\left (\frac{L}{C} \right )\right )}

Energy Delivered in one cycle is given by

E=P_{avg}T

E=\frac{\left (\Delta V \right )^2R}{\left ( R^2+\frac{9}{4}\left (\frac{L}{C} \right )\right )}\left ( \frac{2\pi }{\frac{2}{\sqrt{LC}}}\right )

E=\frac{\pi R\left ( \Delta V\right )^2\sqrt{LC}}{\left ( R^2+\frac{9}{4}\left (\frac{L}{C} \right )\right )}              

3 0
3 years ago
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