List two elements that have the same reactivity
2 answers:
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Answer:
The torque on the pulley, when the system is motionless is approximately 9.81 N·m
Explanation:
The given parameters are;
The mass of the object = 2 kg
The friction between the rope and the pulley = 0
The mass of the rope, = 0.5 kg
The mass of the pulley, = 0.01 kg
The radius of the pulley, r = 0.25 m
The torque on the pulley, τ = I·α = F × D
The torque on the pulley, when the system is motionless, τ = F × D
Where;
F = The force acting on the pulley rope = The weight of the mass ≈ 2 kg × 9.81 m/s² = 19.62 N
D = The diameter of the pulley = 2×r = 2 × 0.25 m = 0.5 m
Therefore;
τ = 19.62 N × 0.5 m = 9.81 N·m
The torque on the pulley, when the system is motionless, τ ≈ 9.81 N·m.
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.
Answer:
4.7ft
Explanation:
Pls see attached file
