Answer:OB=58.3m
Explanation:
So here cow wanders 30m in north and turns 22 degrees in right side and moves 40m more, as shown in figure given.
now take the starting point as a origin such that cow moves in x-y co-ordinate axis.
As shown in figure length OA is the length when cow moves in north or y direction. Later she takes 22 degrees turn to right and moves 40m more.
So the final displacement is the length of cow from the origin that is length OB.
now co-ordinates of B are [40cos22°,40sin22°+30] i.e [37.084,44.984]
now displacement of cow= length of OB
= ![\sqrt{[37.084]^{2}+[44.984]^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5B37.084%5D%5E%7B2%7D%2B%5B44.984%5D%5E%7B2%7D%20%20%7D)
=
OB =
The marbles that are 'more energetic' fall out of the tray, in the same way particles have enough energy to escape and turn into a gas.
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
Answer:
Energy = 1.38*10^13 J/mol
Explanation:
Total number of proton in F-19 = 9
Total number of neutron in F-19 = 10
Expected Mass of F-19
= 9*1.007 + 10*1.008 = 19.152 u
Actual mass of F-19 = 18.998 u
Energy of one particle of F-19 = 931.5*Δm = 931.5*(19.152-18.998)
= 143.234 MeV
Energy of one mole of F-19 = 143.234*10^6*1.6*10^-19*6.022*10^23
= 1.38*10^13 J/mol
