Answer:
20 [N], in the opposite direction of the first force.
Explanation:
We know that newton's second law stipulates that the sum of forces on a body must be equal to the product of mass by acceleration.
![SumF = m*a\\30 + F = 2*5\\F = 30 - (2*5)\\F = - 20 [N]](https://tex.z-dn.net/?f=SumF%20%3D%20m%2Aa%5C%5C30%20%2B%20F%20%3D%202%2A5%5C%5CF%20%3D%2030%20-%20%282%2A5%29%5C%5CF%20%3D%20-%2020%20%5BN%5D)
The negative sign means that the other force acting on the body must be in the opposite direction to the force of 30 [N]
Ernest Rutherford is the answer you are looking for my friend.
Answer:
False
Explanation:
ac = v^2/r
acceleration is not dependent on the mass of the orbiting object.
Answer:
Explanation:
cSep 20, 2010
well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.
Anonymous
Sep 20, 2010
First you need to solve for time by using
d=(1/2)(a)(t^2)+(vi)t
1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s
t=.45 sec
Then you find the horizontal distance traveled by using
v=d/t
1.3m/s=d/.54sec
d=.585m
Then you need to find the time of player B by using
d=(1/2)(a)(t^2)+(vi)t
1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0
t=.61 sec
Finally to find player Bs initial horizontal velocity you use the horizontal equation
v=d/t
v=.585m/.61 sec
so v=.959m/s
-- Put the rod into the freezer for a while. As it cools,
it contracts (gets smaller) slightly.
-- Put the cylinder into hot hot water for a while. As it heats,
it expands (gets bigger) slightly.
-- Bring the rod and the cylinder togther quickly, before the
rod has a chance to warm up or the cylinder has a chance
to cool off.
-- I bet it'll fit now.
-- But be careful . . . get the rod exactly where you want it as fast
as you can. Once both pieces come back to the same temperature,
and the rod expands a little and the cylinder contracts a little, the fit
will be so tight that you'll probably never get them apart again, or even
move the rod.