Answer:
2.4 m/s
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.08 kg)(0.5 m/s) + (0.05 kg)(0 m/s) = (0.08 kg)(-0.1 m/s) + (0.05 kg) v
0.04 kg m/s = -0.08 kg m/s + (0.05 kg) v
0.12 kg m/s = (0.05 kg) v
v = 2.4 m/s
Answer: 29.50 m
Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:
f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).
f=m.a=μ*N= m*a= μ*m*g= m*a
then
a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2
With this value we can determine the short distance to stop the train
as follows:
x= vo*t- (a/2)* t^2
Vf=0= vo-a*t then t=vo/a
Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m
Answer:
The correct answer is
a) 1, 2, 3
Explanation:
In rolling down an inclined plane, the potential energy is Transferred to both linear and rotational kinetic energy thus
PE = KE or mgh = 1/2×m×v² + 1/2×I×ω²
The transformation equation fom potential to kinetic energy is =
m×g×h = 
= 
= 
=
Therefore the order is with increasing rotational kinetic energy hence
the first is the sphere 1 followed by the disc 2 then the hoop 3
the correct order is a, 1, 2, 3
This is a question that would have literally have taken two seconds to look up on google but the answer is 1.88 years.
