Answer:
1.63 N
Explanation:
F = GMm/r^2
= (6.67x10^-11)(10x10^5)(3x10^5) / 3.5^2
= 1.63 N ( 3 sig. fig.)
If you could please give me a already given speed I could estimate it. since there is no speed shown you wouldn't be able to estimate the speed of the moving train.
Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.
v is the speed of cat,
So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
The object that goes through chemical change, changes completely to where you can not change it back to its original form. Physical change you can undo
I included things
Hope this helps!