Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around
The farthest position the mouse reaches inside the tunnel is 4 meters into the tunnel.
From the graph,
The positions reached after,
5 s = 4 m
10 s = 2 m
20 s = 2 m
35 s = 3 m
40 s = 0 m
So the farthest position here is 4 m into the tunnel.
The rate of change of positions is displacement. So displacement will be change in initial and final positions divided by change in time.
s = Δx / Δt
Therefore, the farthest position the mouse reaches inside the tunnel is 4 meters into the tunnel.
To knw more about displacement
brainly.com/question/28609499
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Calculate velocity at halfway to the ground.
vfinal = root 2ad
v = root (2*9.81m/s^2*25)
v = 69.367175234 m/s
Kinetic energy = 1/2mv^2
Kinetic energy = 1/2 * (69.367175234 m/s^s^2
Ek = 2405.9025 Joules