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3 years ago

What is power, and what is its relationship to voltage and amperage? (4 points)

2 answers:

7 0

Power is the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt, equal to one joule per second. The relationship between amperage, voltage, and power is that power equals the amperage quantity times the amount of voltage.

tigry1 [53]3 years ago

5 0

Answer:

The relationship between amperage, voltage, and power is that power equals the amperage quantity times the amount of voltage.

Explanation: brainliest pls

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NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago

I feel really isolated these days and wish I had more friends. I've chosen to put myself in new social situations to try to make

leva [86]

Behaviorist approach to psychology

3 0

3 years ago

Bob walks 370 m south, then jogs 410 m southwest, then walks 370 m in a direction 28 degrees east of north.

wlad13 [49]

We will measure all angles from West, the negative x-axis and divide the journey into 3 parts:
P1 = 370y
P2 = 410cos(45)x + 410sin(45)y = 290x + 290y
P3 = 370cos(270 - 28)x + 370sin(270 - 28) = -174x - 327y

Overall displacement:
x = 290 - 174 = 116 m
y = 370 + 290 - 327 = 333 m

displacement = √(116² + 333²)
= 353 m

Direction:
tan(∅) = y/x
∅ = tan⁻¹ (333 / 116)
∅ = 70.8° from West.

5 0

3 years ago

Read 2 more answers

How can parents help children to gain friends?

Andre45 [30]

Answer:

You could try finding a familiar peer to join the activity with your child. Or ask your child who their friends are at school, or what they look for in a friend at school.

7 0

3 years ago

Read 2 more answers

An automobile tire having a temperature of

EastWind [94]

Answer:

Psm = 30.66 [Psig]

Explanation:

To solve this problem we will use the ideal gas equation, recall that the ideal gas state equation is always worked with absolute values.

P * v = R * T

where:

P = pressure [Pa]

v = specific volume [m^3/kg]

R = gas constant for air = 0.287 [kJ/kg*K]

T = temperature [K]

<u>For the initial state</u>

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P1 = 24 [Psi] + 14.7 = 165.47[kPa] + 101.325 = 266.8 [kPa] (absolute pressure)

T1 = -2.6 [°C] = - 2.6 + 273 = 270.4 [K] (absolute Temperature)

Therefore we can calculate the specific volume:

v1 = R*T1 / P1

v1 = (0.287 * 270.4) / 266.8

v1 = 0.29 [m^3/kg]

As there are no leaks, the mass and volume are conserved, so the volume in the initial state is equal to the volume in the final state.

V2 = 0.29 [m^3/kg], with this volume and the new temperature, we can calculate the new pressure.

T2 = 43 + 273 = 316 [K]

P2 = R*T2 / V2

P2 = (0.287 * 316) / 0.29

P2 = 312.73 [kPa]

Now calculating the manometric pressure

Psm = 312.73 -101.325 = 211.4 [kPa]

And converting this value to Psig

Psm = 30.66 [Psig]

3 0

3 years ago