Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
Answer:
Gamma rays
Explanation:
Gamma rays is at the end of the electromagnetic spectrum, and has the highest energy. It propagates through space at 3x10^8 m/s and has the smallest wavelength and the highest frequency. It is given off by atoms of element as they undergo nuclear disintegration.
There are 2 electrons generated from the oxidation of one water molecule.
<h3>Describe photooxidation.</h3>
The process of a substance interacting with oxygen or losing electrons from chemical species under the influence of light is known as photooxidation. Photooxidation happens in plants when there is environmental stress. It is called photooxidative stress as a result. Reactive oxygen species are produced by the absorption of excess excitation energy in plant tissues. Chloroplasts are harmed by the accumulation of these reactive oxygen species, which is a damaging process in plants. High-intensity light and little 
are the two conditions that cause this photooxidative stress to occur most frequently. It is a procedure that requires light. Photorespiration in 
plants guards against photooxidation.
To know more about Photooxidation visit:
brainly.com/question/14788790
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A: The battery is a store of internal energy (shown as chemical energy). The energy is transferred through the wires to the lamp, which then transfers the energy to the surroundings as light. These are the useful energy transfers - we use electric lamps to light up our rooms.
B: In the case of the light bulb the 95J of energy transferred as heat is wasted energy as it is not useful because the purpose of the device is to produce light.
SORRY I ONLY HAVE ANSWERS FOR A AND B
