I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
Answer:
Explanation:
After the collision velocity of the particle is (4î - 3ĵ)m/s . ... A particle of mass 1 kg moving with a velocity of (4i^−3j^)m/s collides with a fixed surface. ... Perfectly inelastic. D ... The common velocity of the blocks after collision is: ... A ball falls from a height of 5 m and strikes the roof of a lift. ... Stay upto date with our Newsletter! i know this is not right but just here for points see ya loser
Answer:
If the temperature of the air in the balloon is less than the temperature of the air surrounding the balloon then the balloon will appear slightly deflated because of the difference in temperature.
As the temperature of the air in the balloon reaches the surrounding air temperature, then the balloon will appear to be fully inflated because the temperature of the air in the balloon is the same as the surrounding air temperature.
They are falling under the sole influence of gravity all objects<span> will </span>fall<span> with the </span>same<span> rate of </span><span>acceleration needless of there size</span>
The truck would of went 150 miles