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Mars2501 [29]
3 years ago
14

Which situation shows potential energy transforming into kinetic energy 1.A car stopped at a stoplight 2.A flowerpot falling fro

m a balcony 3.A cow standing in a field 4.A bicycle slowing down
Physics
1 answer:
Troyanec [42]3 years ago
7 0

Answer:

2.A flowerpot falling from a balcony

Explanation:

Potential energy is the internal energy within a body at rest or at a height above the ground. In this case of the falling flowerpot, the flowerpot initially has a potential energy due to its position on the balcony above the surface of the earth. As the flowerpot falls, the work done against gravity in taking the flowerpot to this height on the balcony (equivalent to the potential energy) is transformed into the kinetic energy of motion as the flowerpot falls down through the height.

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Three metal spoons are on a table. They are each at room temperature. If the three spoons touch.
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________ states that all current entering a branch point in a circuit must exit.
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Answer:

The current through a branch is also called the branch current. The current supplied by the battery in a parallel circuit splits at one or more branch points. All of the current entering a branch point must exit again. This rule is known as Kirchhoff's current law

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3 years ago
I need help on #5-8.
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I've found a link that should assist you or answer your question.
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4 years ago
What is the shortest-wavelength x-ray radiation in m that can be generated in an x-ray tube with an applied voltage of 93.3 kV?
VikaD [51]

(a) 1.33\cdot 10^{-11} m

The x-rays in the tube are emitted as a result of the collisions of electrons (accelerated through the potential difference applied) on the metal target. Therefore, all the energy of the accelerated electron is converted into energy of the emitted photon:

e \Delta V = \frac{hc}{\lambda}

where the term on the left is the electric potential energy given by the electron, and the term on the right is the energy of the emitted photon, and where:

e=1.6\cdot 10^{-19}C is the electron's charge

\Delta V = 93.3 kV = 93300 V is the potential difference

h=6.63\cdot 10^{-34} Js is the Planck constant

c=3.00\cdot 10^8 m/s is the speed of light

\lambda is the wavelength of the emitted photon

Solving the formula for \lambda, we find:

\lambda=\frac{hc}{e\Delta V}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{(1.6\cdot 10^{-19})(93300)}=1.33\cdot 10^{-11} m

(b) 93300 eV (93.3 keV)

The energy of the emitted photon is given by:

E=\frac{hc}{\lambda}

where

h is Planck constant

c is the speed of light

\lambda=1.33\cdot 10^{-11} m is the wavelength of the photon, calculated previously

Substituting,

E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.33\cdot 10^{-11}}=1.50\cdot 10^{-14} J

Now if we want to convert into electronvolts, we have to divide by the charge of the electron:

E=\frac{1.50\cdot 10^{-14} J}{1.6\cdot 10^{-19} J/eV}=93300 eV

(c) The following statements are correct:

The maximum photon energy is just the applied voltage times the electron charge. (1)

The value of the voltage in volts equals the value of the maximum photon energy in electron volts.

In fact, we see that statement (1) corresponds to the equation that we wrote in part (a):

e \Delta V = \frac{hc}{\lambda}

While statement (2) is also true, since in part (b) we found that the photon energy is 93.3 keV, while the voltage was 93.3 kV.

3 0
3 years ago
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