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:
The direction of the contact forces acting on a body is not necessarily perpendicular to the contact surface. The resolution of contact forces in two components i.e. perpendicular to contact surface and along surface. Perpendicular component is normal force and parallel component is friction.
Explanation:
Answer: D) It should be handled in a fume hood, away from open flames.
c) only from warmer areas to colder areas.
The second principle of thermodynamics states that heat can only flow from a hotter body to a cooler one. Specifically, Clausius statement says that is not possible for heat to move by itself from a lower temperature body to a higher temperature body.
Answer:
Hence, work done= 287.54 J
Explanation:
Given data:
angle of ramp with the ground θ =20°
force applied = 76 N
work done on the crate to slide down 4 m down the ramp
W= F×d cosθ ( only the cos component of the force will slide the crate down)
W= 76×4×cos20= 287.54 J
