Answer:
A body travels 10 meters during the first 5 seconds of its travel,and a total of 30 meters over the first 10 seconds of its travel
20miles / 5sec = 4miles /sec would be the average speed for the last 20 m
Explanation:
The answer is 4 m/s.
In the first 5 seconds, a body travelled 10 meters. In the first 10 seconds of the travel, the body travelled a total of 30 meters, which means that in the last 5 seconds, it travelled 20 meters (30m + 10m).
The relation of speed (v), distance (d), and time (t) can be expressed as:
v = d/t
We need to calculate the speed of the second 5 seconds of the travel:
d = 20 m (total 30 meters - first 10 meters)
t = 5 s (time from t = 5 seconds to t = 10 seconds)
Thus:
v = 20m / 5s = 4 m/s
PLEASE GIVE BRAINIEST!! HOPE THIS HELPS
Um you should putting it in a object that it can fill then go from there
The difference in electric potential energy between the two points is
![\Delta U = q \Delta V](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20q%20%5CDelta%20V)
where q is the magnitude of the charge and
![\Delta V](https://tex.z-dn.net/?f=%5CDelta%20V)
is the electric potential difference.
But for energy conservation, the difference in electric potential energy
![\Delta U](https://tex.z-dn.net/?f=%5CDelta%20U)
between the two points is equal to the work done to move the charge between A and B:
![W=\Delta U](https://tex.z-dn.net/?f=W%3D%5CDelta%20U)
so we have
![W=q \Delta V](https://tex.z-dn.net/?f=W%3Dq%20%5CDelta%20V)
and by substituting the numbers of the problem, we find the value of
![\Delta V](https://tex.z-dn.net/?f=%5CDelta%20V)
: