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Burka [1]
3 years ago
8

Which of these is true about kinetic energy but not necessarily true about potential energy

Physics
1 answer:
cestrela7 [59]3 years ago
3 0

Kinetic energy is never negative, but potential energy can be.

Potential energy depends on height above some reference level,
and you can pick any level you want as the reference.  So, if the
object is below the reference level you pick, then its potential
energy relative to your reference level is negative. 

What that means is:  You have to lift it / do work on it / give it more
energy than it has now ... in order to move it to the reference level.

(That's exactly the situation with electrons bound to an atom.  Their
energy is considered negative, because we have to do work and
give them more energy to rip them away from the atom.)
_____________________________________

Regarding the other choices:

-- Kinetic energy is scalar ... Yes.  So is potential energy.

-- Kinetic energy increases with height ...
   No. It doesn't, but potential energy does.

-- Kinetic energy depends on position ...
   No. It doesn't, but potential energy does.

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Properties of most medals include
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Answer:

Metals are lustrous, malleable, ductile, good conductors of heat and electricity. Other properties include: State: Metals are solids at room temperature with the exception of mercury, which is liquid at room temperature (Gallium is liquid on hot days).

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2 years ago
A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of b
nekit [7.7K]
<h2>Answer:</h2>

In circuits, the average power is defined as the average of the instantaneous power  over one period. The instantaneous power can be found as:

p(t)=v(t)i(t)

So the average power is:

P=\frac{1}{T}\intop_{0}^{T}p(t)dt

But:

v(t)=v_{m}cos(\omega t) \\ \\ i(t)=i_{m}cos(\omega t)

So:

P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}

P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}

In terms of RMS values:

V_{RMS}=V=\frac{v_{m}}{\sqrt{2}} \\ \\ I_{RMS}=I=\frac{i_{m}}{\sqrt{2}} \\ \\ Then: \\ \\ P=VI

7 0
2 years ago
Match the element to the best description of its ionization energy
kompoz [17]
Ionization energy, according to <span>chem.libretexts.org,</span><span> is the quantity of </span>energy<span> that an isolated, gaseous atom in the ground electronic state must absorb to discharge an electron, resulting in a cation. This </span>energy<span> is usually expressed in kJ/mol, or the amount of </span>energy<span> it takes for all the atoms in a mole to lose one electron each.</span>
4 0
3 years ago
The Iceberg Kid, infamous for imposing a deep freeze on whoever disagrees with him, discovers that in his advanced teenaged year
Mademuasel [1]

Answer:

7.24 Ω

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8 0
2 years ago
What is an equilibrant
professor190 [17]
The question seems to be what is an equilibrant force.

The answer is "an added force that produces equilibrium.

Here you have more insight:

<span>an object that has no net force acting on it? This object indeed is in equilibrium but the object is not the equilibran force.

the reaction force in an action-reaction pair of forces?

the reaction force is not an equilibrant force. The reaction force exists always but equilibrium is only possible if the net force is cero.

an added force that produces equilibrium? this is the right  answer.</span>
5 0
3 years ago
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