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erma4kov [3.2K]
3 years ago
7

In the early 1900s, it was proposed that the law of conservation of mass should be simultaneously considered with the law of con

servation of energy to explain particular phenomena. Thus, a theory of conservation of mass-energy was proposed. Which of the following reasons could provide evidence to support the proposed theory?
A. After charged particles travel a complete loop around a circuit, the electric potential energy of the charged particles does not change, but the number of available charged particles that can move through the circuit is reduced. This is because charged particles are used in order for circuit elements to operate correctly.
B. After a photon of light is absorbed by certain metals, electrons are found to be ejected from the metals. This is because the energy contained in the massless photon is used to eject an electron with mass out of the metal.
C. After particles of a hot gas collide with other particles in the gas, the initial combined mass of all particles of the gas immediately before the collisions occur is not equal to the final combined mass of all particles immediately after the collisions. This is because some of the particles in the gas are destroyed in the collisions.
D. After the decay of certain unstable nuclei, the initial mechanical energy of an unstable nucleus is not equal to the final mechanical energy of the resultant particles immediately after the decay process. This is because some of the available mechanical energy is converted into a particle that was originally not accounted for.
Physics
1 answer:
svet-max [94.6K]3 years ago
7 0

Answer:

B. After a photon of light is absorbed by certain metals, electrons are found to be ejected from the metals. This is because the energy contained in the massless photon is used to eject an electron with mass out of the metal.

Explanation:

Before, in the early days, it was proposed to form a combined theory by joining the theory of conservation of mass and the theory of conservation of energy and form a combined theory of conservation of mass-energy. It was done to explain a particular theory of $\text{photoelectric effect}$.

The $\text{photoelectric effect}$ is the emission of the electrons form the surface of a metal when light energy strikes on it. Here, in this phenomenon, both mass and energy is conserved.

When the light strikes a metal surface, electrons gets ejected from the surface. The energy of the photon is used to eject the electron form the metal surface.

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What inductance l would be needed to store energy e=3.0kwh (kilowatt-hours) in a coil carrying current i=300a?
myrzilka [38]

The formula for the energy stored in the magnetic field of an inductor is:

                      E  =  (1/2) (inductance) (current)²  .

In the present situation:

Energy = (3 kilo-watt-hour) x (1,000 / kilo) x (joule/watt-sec) x (3,600 sec/hr)

           =  (3 · 1000 · 3,600)  (kilo·watt·hr·joule·sec / kilo·watt·sec·hr)

           =      1.08 x 10⁷ joules .

Now to find the inductance:  

                   E  =  (1/2) (inductance) (current)² 

       (1.08 x 10⁷ joules) = (1/2) (inductance) (300 Amp)²

           (2.16 x10⁷ joules) =  (inductance) (300 Amp)²

             Inductance =  (2.16 x10⁷ joules) / (300 Amp)²

                              =   2.16 x10⁷ / 90,000        Henrys

                           I get        240 Henrys .

This is a big inductance.  Possibly the size of your house.
To get a big inductance, you want to wind the coil
  with a huge number of turns of very fine wire, in
  a small space.
In this case, however, if you plan on running 300A through
  your coil, it'll have to be wound with a very thick conductor ...
  like maybe 1/4-inch solid copper wire, or even copper tubing,
You have competing requirements.
There are cheaper, easier, better ways to store 3 kWh of energy.
In fact, a quick back-of-the-napkin calculation says that
  3 or 4 car batteries will do the job nicely.
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3 years ago
while performing a sonogram, to minimize risk to the fetus, which imaging mode trade off would be most appiclable.
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Answer:

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Explanation:

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A 40 foot beam that weighs 125 pounds is supported at the two ends by walls. It also supports a 600 pound AC unit 5 feet from th
Dmitrij [34]

Answer:

A) 588 pounds

Explanation:

According to the given conditions, we assume the beam to be simply supported at the ends carrying a uniformly distributed load of 125 pounds per feet and a point load of 600 pounds acting at 5 feet from the right support.

Referring the schematic:

<u>Moment about any point will be zero in equilibrium condition. </u>

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F_r\times 40=125\times 20+35\times 600

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8 0
4 years ago
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IrinaVladis [17]

Answer:

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