Answer : The partial pressure of is, 67.009 atm
Solution : Given,
Partial pressure of at equilibrium = 30.6 atm
Partial pressure of at equilibrium = 13.9 atm
Equilibrium constant =
The given balanced equilibrium reaction is,
The expression of will be,
Now put all the values of partial pressure, we get
Therefore, the partial pressure of is, 67.009 atm
Answer:
4 m/s
Explanation:
Momentum is defined as:
where
m is the mass of the object
v is its velocity
For the object in this problem, we know:
p = 200 kg m/s is the momentum
m = 50 kg is the mass
Solving for the velocity, we find:
Answer:
Angular velocity is same as frequency of oscillation in this case.
ω = x
Explanation:
- write the equation F(r) = -K with angular momentum <em>L</em>
- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.
- Write the energy of the orbit in relative to r = 0, and solve for "E".
- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.
- Solve for effective potential
- ω = x
