<span>at maximum height the final velocity will be 0
using v=u+at and resolving vertically we get
v=0.6+(-9.81)t
v=0.6-9.81t
0=0.6-9.81t
9.81t=0.6
t=0.6/9.81
t=0.061 to 3sf
Now we need to resolve horizontally to find the horizontal distance
using s=ut+1/2at^2
However we now need the total time taken for the projectile travel and return to the ground. We can assume the time taken for the projectile to reach its maximum height and return to the ground is the same therefore
the total time is 2 x 0.061=0.122seconds. They'll be now horizontal acceleration in this case scenario therefore
Hence s=ut+1/2at^2
since a=0
s=ut
s=0.6 x 0.122
s=0.073m
</span>
R=U/I so
U=RxI
U= 10 x 42
U= 420 volts
Answer:
The object displacement is 0 meters.
Explanation:
Because the expression for the object position as function of time is:
At t= 3.0=t2 seconds the x-position of the object is:
and at t=1.0=t1 seconds the x-position of the objects is:
The total displacement () of the object between t2 and t1 is the difference between x(3.0s) and x(1.0s):
The displacement of the object is zero!!!
Answer:
The input force is the force that when you push down and the output force is the stable going into the paper.
Explanation:
I really hope this helps ;)