An elevator supported by a single cable descends a shaft at a constant speed. The only forces acting on the elevator are the ten
1 answer:
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Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:
or
where <em>K</em> is the block's kinetic energy at the equilibrium point,
Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is
Compute the work performed by friction:
By Newton's second law, the net vertical force on the block is
∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0 ==> <em>n</em> = <em>mg</em>
where <em>n</em> is the magnitude of the normal force from the surface pushing up on the block. Then if <em>f</em> is the magnitude of kinetic friction, we have <em>f</em> = <em>µmg</em>, where <em>µ</em> is the coefficient of kinetic friction.
So we have
Answer:
The frequency = 521.59 Hz
The rate at which the frequency is changing = 186.9 Hz/s
Explanation:
Given that :
Diameter of the tank = 44 cm
Radius of the tank = =
= 22 cm
Diameter of the spigot = 3.0 mm
Radius of the spigot = =
= 1.5 mm
Diameter of the cylinder = 2.0 cm
Radius of the cylinder = =
= 1.0 cm
Height of the cylinder = 40 cm = 0.40 m
The height of the water in the tank from the spigot = 35 cm = 0.35 m
Velocity at the top of the tank = 0 m/s
From the question given, we need to consider that the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.
The expression for Bernoulli's Equation is as follows:
where;
P₁ and P₂ = initial and final pressure.
v₁ and v₂ = initial and final fluid velocity
y₁ and y₂ = initial and final height
p = density
g = acceleration due to gravity
So, from our given parameters; let's replace
v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²
∴ we have:
v₂ =
v₂ =
v₂ = 2.61916
v₂ ≅ 2.62 m/s
Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:
v₂A₂ = v₃A₃
v₂r₂² = v₃r₃²
where;
v₂r₂ = velocity of the fluid and radius at the spigot
v₃r₃ = velocity of the fluid and radius at the cylinder
where;
v₂ = 2.62 m/s
r₂ = 1.5 mm
r₃ = 1.0 cm
we have;
v₃ =
v₃ = 0.0589 m/s
∴ velocity of the fluid in the cylinder = 0.0589 m/s
So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;
where;
= velocity of sound
h = height of the fluid
v₃ = velocity of the fluid in the cylinder
= 521.59 Hz
∴ The frequency = 521.59 Hz
b)
What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?
The rate at which the frequency is changing is related to the function of time (t) and as such:
where;
(velocity of sound) = 343 m/s
v₃ (velocity of the fluid in the cylinder) = 0.0589 m/s
h (height of the cylinder) = 0.40 m
t (time) = 4.0 s
Substituting our values; we have ;
= 186.873
≅ 186.9 Hz/s
∴ The rate at which the frequency is changing = 186.9 Hz/s when the cylinder has been filling for 4.0 s.