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
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If one of two interacting charges is doubled, the force between the charges will double.
Explanation:
The force between two charges is given by Coulomb's law

K=constant= 9 x 10⁹ N m²/C²
q1= charge on first particle
q2= charge on second particle
r= distance between the two charges
Now if the first charge is doubled,
we get 
F'= 2 F
Thus the force gets doubled.
The energy carried by the incident light is

where h is the Planck constant and f is the frequency of the light. The threshold frequency is the frequency that corresponds to the minimum energy needed to eject the electrons from the metal, so if we substitute the threshold frequency in the formula, we get the minimum energy the light must have to eject the electrons:
#1). Anthony does the same amount of work as Angel, with <em>more power</em>.
#2). Power = (Work)/(Time) = 41,000 J / 500 s = <em>82 watts .</em>
#3). Power = (Work) / (Time) = 83 J / 3 sec = <em>27.7 watts</em>
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