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11111nata11111 [884]

3 years ago

11

If the current through a resistor is 12 A and the voltage drop is 12 V, what is the power absorbed by the resistor? Answer to th

e nearest 1 W. Assume the power equation is positive.

1 answer:

klasskru [66]3 years ago

7 0

Answer:

144 W

Explanation:

The power absorbed by a resistor is the product of the current passing through it and the voltage drop across it. Hence

$P=IV$

$P=12\text{ A}\times12\text{ V}=144 \text{ W}$

Other variations of this formula are derived using Ohm's law, $V=IR$

$P=I^2R=\dfrac{V^2}{R}$

$R$ is the resistance of the resistor. It is what causes the voltage drop.

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You have been training and can run a mile in 6.5 minutes. How long would it take to run a lap around the outside track, which is

Paha777 [63]

Answer:

1 minute 36.85 seconds

Explanation:

First we need to convert the miles into meters, as the demanded result should be in meters.

1 mile = 1,609.34 meters

Also, 6.5 minutes should be converted into seconds.

1 minute = 60 seconds

6.5 x 60 = 390 seconds

Now we need to divide the miles with the seconds to see how much meters have been run in a second.

1,609.34 / 390 = 4.13 meters

The suggested meters now should be divided with the distance run in one second.

400 / 4.13 = 96.85 seconds

So we get a result of 96.85 seconds, or 1 minute 36.85 seconds.

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3 years ago

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galina1969 [7]

Answer:

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Explanation:

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3 years ago

Thandy is looking at two cells under the microscope.One is a human cheek cell and the other is a leaf mesophyll cell from a plan

Vanyuwa [196]

Cell are small and us human can't see them with our own eyes. it is in possible to see cell without a microscope

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3 years ago

Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

7 0

3 years ago

A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite

sladkih [1.3K]

Answer:

The kinetic energy is $KE = 7.59 *10^{10} \ J$

Explanation:

From the question we are told that

The radius of the orbit is $r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m$

The gravitational force is $F_g = 6600 \ N$

The kinetic energy of the satellite is mathematically represented as

$KE = \frac{1}{2} * mv^2$

where v is the speed of the satellite which is mathematically represented as

$v = \sqrt{\frac{G M}{r^2} }$

=> $v^2 = \frac{GM }{r}$

substituting this into the equation

$KE = \frac{ 1}{2} *\frac{GMm}{r}$

Now the gravitational force of the planet is mathematically represented as

$F_g = \frac{GMm}{r^2}$

Where M is the mass of the planet and m is the mass of the satellite

Now looking at the formula for KE we see that we can represent it as

$KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

=> $KE = \frac{ 1}{2} *F_g * r$

substituting values

$KE = \frac{ 1}{2} *6600 * 2.3*10^{7}$

$KE = 7.59 *10^{10} \ J$

7 0

3 years ago