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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Kinetic energy of the rock just before it hits the ground=KE=33000 J
Explanation:
Weight= 2200N
mg=2200
m(9.8)=2200
m=224.5 kg
initial velocity=0
final velocity =V
using kinematic equation V²=Vi²+2gh
V²=0+2 (9.8)(15)
V=17.1 m/s
now kinetic energy= 1/2 mV²
KE= 1/2 (224.5)(17.1)²
KE=33000 J
Thus the kinetic energy of the rock just before it hits the ground=33000 J
Answer:
if we double the distance the energy stored will be doubled also
Explanation:
The energy stored in a capacitor is given as
Energy stored =1/2(cv²)
Or
= 1/2(Qv)
Where c = capacitance
Q= charge
But the electric field is expressed as
E= v/d
where v= voltage
d= distance
v=Ed
Substituting into any equation above say
Energy stored =1/2(Qv)
Substituting v=Ev
Energy stored =1/2(QEd)
From the equation above it shows that if we double the distance The energy stored will be doubled also
