If n=1 you divide by 2
If n=2 you divide by 4
If n=3 you divide by 8
so any n you divide by 2 to the power n
Answer:
Explanation:
Given that
d= 1.5 in ( 1 in = 0.0254 m)
d= 0.0381 m
P= 75 hp ( 1 hp = 745.7 W)
P= 55927.5 W
N= 1800 rpm
We know that power P is given as
T=Torque
N=Speed
T=296.85 N.m
The maximum shear stress is given as
We know that 1 MPa =0.145 ksi
Answer:
979.6 kg/m³
Explanation:
We know pressure P = hρg where h = height of liquid = 10.5 m, ρ = density of liquid and g = acceleration due to gravity = 9.8 m/s²
So, density ρ = P/hg
Since P = 100.8 kPa = 100.8 × 10³ Pa
substituting the values of the variables into the equation for ρ, we have
ρ = P/hg
= 100.8 × 10³ Pa ÷ (10.5 m × 9.8 m/s²)
= 100.8 × 10³ Pa ÷ 102.9 m²/s²
= 0.9796 × 10³ kg/m³
= 979.6 kg/m³
So, the density of the liquid is 979.6 kg/m³
Power is the work done per unit time. Therefore,
Therefore,
Answer:
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Explanation:
We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is
w² = mg d / I
In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow
d = L / 2
The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated
I = ¼ m r2 + ⅓ m L2
I = m (¼ r2 + ⅓ L2)
now let's use the concept of density to calculate the mass of the system
ρ = m / V
m = ρ V
the volume of a cylinder is
V = π r² L
m = ρ π r² L
let's substitute
w² = m g (L / 2) / m (¼ r² + ⅓ L²)
w² = g L / (½ r² + 2/3 L²)
L >> r
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
