<span>PV = nRT
moles of H2 = 1/2 = 0.5
moles of He = 1/4 =0.25
T = 273 + 27
partial pressure of H2
Px1 = 0.5x0.083x300
P=12.45 atmospheres
PP of He
px1 = 0.25x0.083x300
P =6 22 atmospheres
Totla pressure = 6.22 + 12.45 = 18.68 atm</span>
A hanging mass of 23 kg will exert a force of:
23 * 9.81 = 225.6 Newtons on the string
Thus, during rotation, the force exerted should remain less than this. Force on an object in circular motion is given by:
F = (m * v^2) / r
225.6 = (2.95 * v^2) / 0.791
v = sqrt(60.491)
v = 7.78 m/s
Thus, the magnitude of the velocity should not exceed 7.78 meters per second.
To solve this problem it is necessary to apply the concepts related to the condition of path difference for destructive interference between the two reflected waves from the top and bottom of a surface.
Mathematically this expression can be described under the equation
Where
n = Refractive index
t = Thickness
In terms of the wavelength the path difference of the reflected waves can be described as
Where
\lambda = Wavelenght
Equation the two equations we have that
Our values are given as
Wavelength of light
Therefore the minimum thickness of the oil for destructive interference to occur is approximately 34.0 nm
The work done on the bullet is force•distance. This is also equivalent to the energy it has leaving the muzzle (work energy theorem). This kinetic energy is (1/2)(0.006kg)(336m/s)^2=338.7J. Then 338.7J=F•0.83m
F=338.7/0.83=408N