If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
Light will travel more slowly in a material with a higher index of refraction
Answer:
Given that
speed u=4*10^6 m/s
electric field E=4*10^3 N/c
distance b/w the plates d=2 cm
basing on the concept of the electrostatices
now we find the acceleration b/w the plates to find the horizontal distance traveled by the electron when it hits the plate.
acceleration a=qE/m=
=
m/s
now we find the horizontal distance traveled by electrons hit the plates
horizontal distance
![X=u[2y/a]^{1/2}](https://tex.z-dn.net/?f=X%3Du%5B2y%2Fa%5D%5E%7B1%2F2%7D)
=![4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}](https://tex.z-dn.net/?f=4%2A10%5E6%5B2%2A2%2A10%5E%7B-2%7D%2F7%2A10%5E%7B14%7D%5D%5E%7B1%2F2%7D)
=
= 3 cm
<u>We are Given:</u>
Mass of the block (m) = 500 grams or 0.5 Kg
Initial velocity of the block (u) = 0 m/s
Distance travelled by the block (s) = 8 m
Time taken to cover 8 m (t)= 2 seconds
Acceleration of the block (a) = a m/s²
<u>Solving for the acceleration:</u>
From the seconds equation of motion:
s = ut + 1/2* (at²)
<em>replacing the variables</em>
8 = (0)(2) + 1/2(a)(2)²
8 = 2a
a = 4 m/s²
Therefore, the acceleration of the block is 4 m/s²
<span>When a magnet moves near a wire, it's changing field causes the electrons in the wire to flow as electric current.</span>
