Suppose a particle is accelerated through space by a constant 10-N force. Suddenly the particle encounters a second force of 10-
1 answer:
Answer:
The particle will continue moving at constant velocity
Explanation:
When the particle encounters the second force of 10 N, the net force acting on the particle becomes zero, because the two forces are equal in magnitude but opposite in direction:
For Newton's second Law, the acceleration of the particle is proportional to the net force acting on it:
Therefore, since , the acceleration is zero (a=0) and so the particle will keep a constant velocity.
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Answer:
99.63 kg
Explanation:
From the force diagram
N = normal force on the worker from the surface of the roof
f = static frictional force = 560 N
θ = angle of the slope = 35
m = mass of the worker
W = weight of the worker = mg
W Cosθ = Component of the weight of worker perpendicular to the surface of roof
W Sinθ = Component of the weight of worker parallel to the surface of roof
From the force diagram, for the worker not to slip, force equation must be
W Sinθ = f
mg Sinθ = f
m (9.8) Sin35 = 560
m = 99.63 kg
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.