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FrozenT [24]
3 years ago
9

A proton moves from a location where V = 87 V to a spot where V = -40 V. (a) What is the change in the proton's kinetic energy?

J
(b) If we replace the proton with an electron, what is the change in kinetic energy? J
Physics
1 answer:
Art [367]3 years ago
6 0

Answer: a) 127 eV; b) there is no change of kinetic energy.

Explanation: In order to explain this problem we have to use the change of potentail energy ( conservative field) is equal to changes in kinetic energy. So for the proton ther move to lower potential then they gain kinetic energy from the electric field.  This means the electric force do work in this trayectory and then the protons increased changes its speed.

If we replace the proton by a electron we have a very different situaction, the electrons are located in a lower potental then  they can not move to higher potential  if any  external force does work on the system.

In resumem, the electrons do not move from a point with V=87 to other point with V=-40 V. The electric force point to high potential so the electrons  can not move to lower potential region (V=-40V).

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The brakes in a car increase in temperature by ΔT when bringing the car to rest from a speed v. How much greater would ΔT be if
Wewaii [24]

Answer:

T would be two times the temperature That it was before the double so if the initial temperature was 20 it would now be 40

Explanation:

4 0
3 years ago
According to the nebular theory of solar system formation, what key difference in their early formation explains why the jovian
svp [43]

Answer:

The Jovian planets formed beyond the Frostline while the terrestrial planets formed in the Frostline in the solar nebular

Explanation:

The Jovian planets are the large planets namely Saturn, Jupiter, Uranus, and Neptune. The terrestrial planets include the Earth, Mercury, Mars, and Venus. According to the nebular theory of solar system formation, the terrestrial planets were formed from silicates and metals. They also had high boiling points which made it possible for them to be located very close to the sun.

The Jovian planets formed beyond the Frostline. This is an area that can support the planets that were made up of icy elements. The large size of the Jovian planets is as a result of the fact that the icy elements were more in number than the metal components of the terrestrial planets.

3 0
3 years ago
A wooden rod of negligible mass and length 80.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0
algol13

Answer:

The speed of the animals is 1.64m/s.

Explanation:

Let us work with variables and call the mass of the two rats m_1 and m_2,  and the length of the rod L.

Using the law of conservation of energy, which says the potential energies of the rats must equal their kinetic energies, we know that when the rod swings to the vertical position,

$m_1\frac{L}{2}g -m_2\frac{L}{2}g = \frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2$

$(m_1 -m_2)g\frac{L}{2} = \frac{1}{2}(m_1+m_2)v^2$,

solving for v, we get:

$\boxed{v = \sqrt{\frac{(m_1 -m_2)gL}{(m_1+m_2)}} }$

Putting in the values for m_1, m_2, g, and L we get:

$v = \sqrt{\frac{(0.450kg -0.220kg)(9.8m/s^2)(0.8m)}{(0.450kg+0.220kg)}} $

\boxed{ v= 1.64m/s}

Therefore, as the rod swings through the vertical position , the speed of the rats is 1.64 m/s.

7 0
3 years ago
How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to the "end of the universe"?
meriva
Moon was rotation Earth or sun
it's own orbit . universe is not end
6 0
4 years ago
Read 2 more answers
An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e
timama [110]

Answer:

K = 80.75 MeV    

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}

<em>where E_{ph}: is the photon energy, E_{0p} and E_{0ap}: are the rest energies of the proton and the antiproton, respectively, equals to m₀c², K_{p} and K_{ap}: are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>        

Therefore the kinetic energy of the antiproton is:    

K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}

<u>The proton mass is equal to the antiproton mass, so</u>:

K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}  

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV

K_{ap} = 80.75 MeV              

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!

3 0
3 years ago
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