Answer:
155.5 rev/min
Explanation:
First, we will calculate the initial moment of inertia I_o. We will consider the ice skater as a rod rotating around its axis. Then, we calculate the final moment of inertia I_f. In this occasion we consider the arms as a rod of length L that is horizontally positioned, L so that the length of an arm is L/2. We will call M_1 the mass that remains close to the rotation axis (90 percent) and M_2 the mass located at the arms (10 percent). Finally, we write the equation for the conservation of angular momentum and we solve for ω_f.
I_o=MR^2/2
=(45)(0.15)^2/2
=0.5 kgm^2
M_1=(0.9)(45)
=40.5 kg
M_2=(0.1)(45)
=4.5 kg
I_f=M_1*R^2/2+M_2*L^2/12
=1.1 kg m^2
I_f*ω_f=I_o*ω_o
ω_f = I_o*ω_o/ I_f
=155.5 rev/min
Answer: The block of 0.4 Kg travel the same distance that the block of
0.2 Kg
Explanation: Considering the second newton law, we have the following
F= m*a
F= P*sin (θ) where θ is the angle for the incline
so mg sin (θ)= m*a
a=g sin(θ)
both block have the same acceleration in the inclined plane so travel the same distance independent of its mass.
Answer:
h = P₁ / 9800
Explanation:
This is a fluid mechanics problem, let's write the Bernoulli equation at two points, the subscript 1 for the lowest point and the subscript of 2 for the point with the highest height.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
at the highest point P₂ = 0 and v₂ = 0,
P1 + ½ ρ v12 = ρ g (y₂ -y₁)
we use the continuity equation for the velocity at the lowest point
A₁ v₁ = A₂ v₂
Since the velocity at the highest point is zero, this implies from the equation that the velocity at the lowest point is also zero. In the no-flow condition
P₁ = ρ g (y₂ -y₁)
h = y₂-y₁
h = P₁ /ρ g
the density of water is ρ = 1000 kg / m³ and g = 9.8 m/s², we substitute
h = P₁ / 9800
Let's do a calculation, suppose that P₁ = 1 10⁵ Pa
h = 1 10⁵ / 9800
h = 10.2 m
Answer:
lift force = 0.213 N
Drag force =
Explanation:
Given: velocity v = 10 m/s
w = 100 rev/sec
diameter d = 3cm
density D = 1.2kg/m3
lift force =
Substituting the values into the equation, we obtain
lift force = 0.213 N
Drag force = C*D*A*v/2
where C = 0.5
substituting the values into the equation again, we have
Drag force =
We'll use the formula Solubility = K * P where K is the Henry's constant and P is the pressure in atm.
If 1 bar = 0.9869 atm.
Then 4.5 bar = X atm
X = 4.4410
So Solubility = 58 * 4.4410 = 257.574 M