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A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and

cricket20 [7]

Answer:

The frequency $f$ = 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 = $\frac{d}{2}$ = $\frac{44}{2}$ = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = $\frac{d}{2}$ = $\frac{3.0}{2}$ = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = $\frac{d}{2}$ = $\frac{2.0}{2}$ = 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:

$P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2$

$pgy_1=\frac{1}{2}pv^2_2 +pgy_2$

$v_2=\sqrt{2g(y_1-y_2)}$

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 = $\sqrt{2*9.8*(0.35-0)}$

v₂ = $\sqrt {6.86}$

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

$v_3 = \frac{v_2r_2^2}{v_3^2}$

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = $(2.62 m/s)*$ $(\frac{1.5mm^2}{1.0mm^2} )$

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;

$f=\frac{v_s}{4(h-v_3t)}$

where;

$v_s$ = velocity of sound

h = height of the fluid

v₃ = velocity of the fluid in the cylinder

$f=\frac{343}{4(0.40-(0.0589)(0.4)}$

$f= \frac{343}{0.6576}$

$f$ = 521.59 Hz

∴ The frequency $f$ = 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:

$\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})$

$\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)$

$\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}$

where;

$v_s$ (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 ;

$\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}$

= 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.

8 0

3 years ago

A 2.5 kg tribble is placed in a bucket and whirled in a 1.4 m radius vertical circle at a constant tangential speed. If the forc

Over [174]

Given that,

Mass of a tribble, m = 2.5 kg

Radius, r = 1.4 m

The force on the tribble from the bucket does not exceed 10 times its weight.

To find,

The maximum tangential speed.

Solution,

The force acting on the tribble is equal to the centripetal force.

F = 10mg

The formula for the centripetal force is given by :

$F=\dfrac{mv^2}{r}$

v is maximum tangential speed

$v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{mgr}{m}} \\\\v=\sqrt{{10gr}} \\\\v=\sqrt{10\times 9.8\times 1.4} \\\\v=11.7\ m/s$

So, the maximum tangential speed is 11.7 m/s.

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2 years ago

A mule pulls a cart of milk 10 meters with a force of 50 Newton's.Calculate the work done by a mule

san4es73 [151]

Work = Force x Distance

Assuming that this work is being done parallel to the displacement that is, but under that assumption:

W = (50)(10)

W = 500 J

8 0

3 years ago

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What is the acceleration of a 600,000 kg freight train, if each of itsthree engines can provide 100,000 n of force?

sergey [27]

The acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

<h3>How to calculate acceleration?</h3>

The acceleration of a freight train can be calculated using the following formula:

Force = mass × acceleration

According to this question, a 600,000kg freight train can produce 100,000N of force. The acceleration is as follows:

100,000 = 600,000 × a

100,000 = 600,000a

a = 0.167m/s²

Therefore, the acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

Learn more about acceleration at: brainly.com/question/12550364

#SPJ1

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1 year ago

If you're studying the science of sound, you're studying____?

timurjin [86]

The study of sound is called sonics and the study of sound waves are acoustics

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