There are several many equations that are available to relate the distance,
speed, and time of a body moving vertically in gravity. Happily, the only one
I can always remember without looking it up happens to be the right one to
use for this question !
Distance = (1/2) x (gravity) x (time)²
3.8 m = (1/2) x (9.8 m/s²) x (time)²
Divide each side
by 4.9 m/s² :
(3.8 m) / (4.9 m/sec²) = (time)²
0.7755 sec² = time²
Square root
of each side:
0.88 second = time
1.244 m per second the person driving will go
Answer:
- Contact 1 with 3
, initial charge of 1.8 C.
- contact 1 with 2 and then 1 with 3
, first body should have 3.6 C
Explanation:
The excess charge on a body is distributed evenly throughout the body.
We can have two different configurations:
- Contact 1 with 3
When the third body was touched with the first, the initial charge was distributed between the two, so that when each one separated, it had half the charge, in this configuration the first body should have an initial charge of 1.8 C.
- contact 1 with 2 and then 1 with 3
Another possible configuration of the exercise is that the first body touches the second and the charge decrease to the half and then touches the third where it again decreases by half, so that the first body only gives it every ¼ of its initial load.
Therefore in this configuration if the third body has a load of 0.9C the first body should have 3.6 C
Elements; 1, 9, 12
Compounds: 3, 7, 11
Mixtures: 2, 4, 5, 6, 8, 10
