Answer:
A u = 0.36c B u = 0.961c
Explanation:
In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains
u ’= (u-v) / (1- uv / c²)
Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory
The data give is u ’= 0.43c and the initial core velocity v = 0.94c
Let's clear the speed with respect to the observer (u)
u’ (1- u v / c²) = u -v
u + u ’uv / c² = v - u’
u (1 + u ’v / c²) = v - u’
u = (v-u ’) / (1+ u’ v / c²)
Let's calculate
u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)
u = 0.51c / (1 + 0.4042)
u = 0.36c
We repeat the calculation for the other piece
In this case u ’= - 0.35c
We calculate
u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)
u = 1.29c / (1- 0.329)
u = 0.961c
If it is a matter of which way you are going you could lean forward. It would help to put all the weight opposite of where you are falling.
Answer
given,
D = 50 mm = 0.05 m
d = 10 mm = 0.01 m
Force to compress the spring
F = 3160 N
stress correction factor from stress correction curve is equal to 1.1
now, calculation of corrected stress
= 442.6 Mpa
The tensile strength of the steel material of ASTM A229 is equal to 1300 Mpa
now,
since corrected stress is less than the 
hence, spring will return to its original shape.
M = 7.0 kg, the mass of the groceries
h = 1.2 m, the elevation of the bag of groceries
The bag of groceries moves a constant velocity over the 2.7-m room.
At constant velocity, there is no applied force, and the kinetic energy remains constant.
At an elevation of 1.2 m, there is an increase in PE (potential energy) given by
V = m*g*h
= (7.0 kg)*(9.8 m/s²)*(1.2 m)
= 82.32 J
The change in PE is equal to the work done.
Answer: 82.3 J
Mechanical advantage = output force/input force
MA = Fo/Fi
