Answer:
Explanation:
If the work done on the cart is NET work
Then the work will result in an increase in kinetic energy
KE₀ + W = KE₁
½mv₀² + W = ½mv₁²
½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²
v₁ = 1.626991...
v₁ = 1.6 m/s
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).
D
The exact location of electrons in electron shells of atoms cannot be exactly ascertained. This is why VSPER atomic models represent the position of electrons (s, p, d, & f) using the probability of where they would most be expected to be found.
Explanation:
This is because merely observing electrons changes their behavior. Remember that to observe something one has to shine light on it so it bounces back to the eye. Due to the negligible mass of electrons, mere photons of light will change their direction of movement, spin or other behaviors/properties.
Learn More:
For more on electron clouds check out;
brainly.com/question/12199882
brainly.com/question/11686000
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<span>This is because Helium
has two valence electrons compared to Hydrogen which has only one. Helium has
more energy levels for an electron to jump thus more spectral lines to occur.
The spectral lines relating to each change of energy level would be more
grouped together and hence the greater chance of them falling in the visible
range.</span>
