Answer:
m₁ / m₂ = 1.3
Explanation:
We can work this problem with the moment, the system is formed by the two particles
The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero
p₀o = m₁ v₁ + m₂ v₂
pf = 0
m₁ v₁ + m₂ v₂ = 0
m₁ / m₂ = -v₂ / v₁
m₁ / m₂= - (-6.2) / 4.7
m₁ / m₂ = 1.3
Another way to solve this exercise is to use the mass center relationship
Xcm = 1/M (m₁ x₁ + m₂ x₂)
We derive from time
Vcm = 1/M (m₁ v₁ + m₂v₂)
As they say the velocity of the center of zero masses
0 = 1/M (m₁ v₁ + m₂v₂)
m₁ v₁ + m₂v₂ = 0
m₁ / m₂ = -v₂ / v₁
m₁ / m₂ = 1.3
What is the constant velocity?
<span>We need that number in order to solve the problem
</span>Or
<span>It has a graph in which it describes the different acceleration</span>
<span>here's a cheap trick
it would take the same time to accelerate from rest to top speed
as it would take to decelerate from top speed to zero
so
instead of
d = Vi t + 1/2 a t^2 where Vi is positive and a is negative
we'll use
Vi = 0 and a is positive
giving
85 = 0 + 1/2 (0.43) t^2 = 0.215 t^2
t^2 = 395.345
t = 19.88s or 20. s to 2 sig figs
or we ccould find Vi from
Vf*2 = Vi^2 + 2 a d
0 = Vi^2 + 2 (0.43) 85
Vi^2 = 71.4
Vi = 8.45m/s
then
85 = 8.45 t + 1/2 (-0.43) t^2
85 = 8.45 t - 0.215 t^2
0.215 t^2 - 8.45 + 85 = 0
t = 19.65s or 20. s to 2 s.f.(minor difference arises from rounding Vi)
or another cheap trick
when a is constant
Vavg = (Vf + Vi) /2 = 8.45/2 = 4.225
and
d = Vavg t
85 = 4.225 t
t = 20.12 or 20. s to 2 s.f. (minor differences from intermidiate roundings)
anyway you choose you get 20. s</span>
The answer is 17500 J for 35*10*50 and the answer is 16500 J for 55*10*30. Thus , the first object has more gravitational potential energy
The correct answer is entropy.
In fact, second law of thermodynamics states that the total entropy of an isolated system can never decrease. It can remain constant if the system is in equilibrium or in case of a reversible process: however, perfectly reversible process do not exist in the real world, so the entropy of a system always increases whenever there is a process that involves energy changes.