Answer : The energy of one photon of hydrogen atom is, 
Explanation :
First we have to calculate the wavelength of hydrogen atom.
Using Rydberg's Equation:

Where,
= Wavelength of radiation
= Rydberg's Constant = 10973731.6 m⁻¹
= Higher energy level = 3
= Lower energy level = 2
Putting the values, in above equation, we get:


Now we have to calculate the energy.

where,
h = Planck's constant = 
c = speed of light = 
= wavelength = 
Putting the values, in this formula, we get:


Therefore, the energy of one photon of hydrogen atom is, 
Answer:
7800kg/m³
Explanation:
Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is
Given the density of iron = 7.8 g/cm3.
The SI units must be in kg/m³
7.8g = 7.8/1000 kg
7.8g = 0.0078kg
1cm³ = 0.000001m³
7.8g/cm³
= 0.0078/0.000001 kg/m³
= 7800kg/m³
Hence the density in SI unit is 7800kg/m³
Answer:
Ff = 839.05 N
Explanation:
We can use the equation:
Ff = μ*N
where <em>N</em> can be obtained as follows:
∑ Fc = m*ac ⇒ N - F = m*ac = m*ω²*R ⇒ N = F + m*ω²*R
then if
F = 32 N
m = 133 Kg
R = 0.635 m
ω = 95 rev /min = (95 rev / min)(2π rad / 1 rev)(1 min / 60 s) = 9.9484 rad /s
we get
N = 32 N + (133 Kg)*(9.9484 rad /s)²*(0.635 m) = 8390.53 N
Finally
Ff = μ*N = 0.10*(8390.53 N) = 839.05 N
Answer with Explanation:
We are given that
Diameter of fighter plane=2.3 m
Radius=
a.We have to find the angular velocity in radians per second if it spins=1200 rev/min
Frequency=
1 minute=60 seconds
Angular velocity=
Angular velocity=
b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.

c.Centripetal acceleration=
Centripetal acceleration==