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
Area of a circle is
A= pi r^2
so
500m^2 = 3.14 r2
500/pi = r ^2
152.1549...=r^2
square root both sides
r=12.61566...
d=2r
d=25.2
to 3 sig fig
Answer:
1600 kg
Explanation:
use the formula p=mv. p=3200, v=2. Plug in and rearrange.
3200=(m)(2)
m= 3200/2
m=1600
GPE= height x mass x gravitational field strength
5 x 10 x 9,8=490J
Treating the system as a point-like particle allows us to assign a quantity to the object and monitor this quantity throughout any changes. The complexity of the system which includes geometry, appearance, and extensions can complicate the studying of the system.
