Answer:
Explanation:
Just start with the trivial. If gravity was a constant then E = mgh and 1/2 m v^2 = E
so v=sqrt(2gh)=sqrt(2*9.8*1590000) = 5582m/s
Now this will be too high as gravity reduces with distance.
However it is still true that 1/2mv^2 = loss of gravitational potential energy
so 1/2 v^2 = loss of gravitational potential ( i.e a field without considering mass )
As g = GM/ Ro^2 and P = - GM/R
the Po = - 9.8 * (6370*10^3)= - 62.4 * 10 ^ 6 J/kg
P1= Po * 6370/(6370+1590) = - 49.93 * 10 ^ 6 J/kg
find the CHANGE and then from that the velocity
ie v = sqrt(2*( P1 - Po)) = 5094 m/s
Note how it is a bit smaller than the first estimate but not by such a margin that they are unrecognizably different.
Arcsin or sin^-1
Hope this helps!
Answer:
a) x_{cm} = m₂/ (m₁ + m₂) d
, b) x_{cm} = 52.97 pm
Explanation:
The expression for the center of mass is
= 1 / M ∑
Where M is the total masses, mI and xi are the mass and position of each element of the system.
Let's fix our reference system on the oxygen atom and the molecule aligned on the x-axis, let's use index 1 for oxygen and index 2 for carbon
x_{cm} = 1 / (m₁ + m₂) (0+ m₂ x₂)
Let's reduce the magnitudes to the SI system
m₁ = 17 u = 17 1,661 10⁻²⁷ kg = 28,237 10⁻²⁷ kg
m₂ = 12 u = 12 1,661 10⁻²⁷ kg = 19,932 10⁻²⁷ kg
d = 128 pm = 128 10⁻¹² m
The equation for the center of mass is
x_{cm} = m₂/ (m₁ + m₂) d
b) let's calculate the value
x_{cm} = 19.932 10⁻²⁷ /(19.932+ 28.237) 10⁻²⁷ 128 10-12
x_{cm} = 52.97 10⁻¹² m
x_{cm} = 52.97 pm