Answer:
0.2
Explanation:
The given parameters are;
The acceleration of the train, a = 0.2·g
The mass of the person standing on the train = m
Let μ represent the coefficient of static friction, we have;
The force acting on the person, F = m × a = m × 0.2·g
The force of friction acting between the feet and the floor, 
= m·g·μ
For the person not to slide we have;
The force acting on the person = The force of friction acting between the feet and the floor
F =
∴ m × 0.2·g = m·g·μ
From which we get;
0.2 = μ
The coefficient of static friction that must exist between the feet and the floor if the person is not to slide, μ = 0.2.
It is like that, except most nails are steel or stainless steel, slowing to rusting process to about 5 years.
Answer:
0.358Kg
Explanation:
The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system
0.5Ke^2 = 0.5Mv^2
0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2
Using conservation of momentum between the bullet and the block
0.0115(265) = (M + 0.0115)v
3.0475 = (M + 0.0115)v
v = 3.0475/(M + 0.0115)
plugging into Energy equation
12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2
12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )
12.56 = 0.5 × 9.2872/ M + 0.0115
12.56 = 4.6436/ M + 0.0115
12.56 ( M + 0.0115 ) = 4.6436
12.56M + 0.1444 = 4.6436
12.56M = 4.6436 - 0.1444
12.56 M = 4.4992
M = 4.4992÷12.56
M = 0.358 Kg
