Answer:
w = 132.57 rad / s
Explanation:
To solve this exercise we must define a system formed by the two discs, therefore the angular momentum is conserved
initial
L₀ = I₀ w₀
final
L_f = I w
how the moment is preserved
L₀ = L_f
I₀ w₀ = I w
the moment of inertia of disk 1 is
I₀ = ½ m₁ r₁²
The moment of inertia of the set is
I = I₀ + I₂
I = ½ m₁ r₁² + ½ m₂ r₂²
we substitute
½ m₁ r₁² w₀ = (½ m₁ r₁² + ½ m₂ r₂²) w
w = 
let's reduce the magnitudes to the SI system
w₀ = 30.0 rev / s (2π rad / 1 rev) = 188.5 rad / s
let's calculate
w = ( 
) 188.5
w = ( 
) 188.5
w = 132.57 rad / s
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:
<h2>e. 7.1 MN approx.</h2>
Explanation:
Step one:
given data
density of water= 1000kg/m^3
the dimension of the barge
width= 10m
length= 60m
depth of the boat in the water= 1.2m
Hence the volume occupied by the boat is
volume=10*60*1.2
volume= 720m^2
Step two:
Required is the weight of the barge
we can first find the mass using the relation
density = mass/volume
mass= density*volume
mass= 1000*720
mass= 720000kg
Step three:
Weight =mg
g=9.81m/s^2
W=720000*9.81
W=7063200N
divided by 10^6
W=7.06MN
W=7.1MN approx.
Gravity increases as mass increases.
Gravity decreases when distance decreases
I hope I help.
