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3 years ago

Frequency of the wave below?

Physics

1 answer:

agasfer [191]3 years ago

6 0

It's hard to tell exactly what's happening in that 110 cm that you marked over the wave. What is under the ends of the long arrow ? How many complete waves ? I counted 4.5 complete waves ... maybe ?

If there are 4.5 complete waves in 110cm, then the length of 1 wave is (110/4.5)=24.44cm.

Frequency = speed/wavelength

Frequency = 2m/s /0.2444m

Frequency = 8.18 Hz

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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

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

2 years ago

2 How does Descartes' "quality of motion" differ from the modern momentum?

Rzqust [24]

Answer: Descartes was more of speed which defers from modern day velocity.

Explanation:

Descartes law if conservation referred or defined “motion” rather than “momentum” as what is obtainable in today's world as ”speed” the rate at which something moves rather than “velocity” which is a product of speed and direction. So in conclusion Descartes was more of speed which defers from modern day velocity.

6 0

2 years ago

If m=120kg and a=15m/s2, what is the force

Scorpion4ik [409]

Answer:

F= 1800N

Explanation:

the equation for force is F= ma

so plug in the numbers: F= (120)(15)

solve this to get F= 1800N

tip: don't forget to add the units when writing your answer :)

6 0

2 years ago

A motorboat heads due east at 16 m/s across a river that flows due south at 9.0 m/s. a.) What is the resultant velocity of the b

alukav5142 [94]

Answer:

a)V=18.35 m/s (South -East)

b) t =7.41 m/s

c)D= 66.70 m

Explanation:

Given that

Velocity of boat in east direction = 16 m/s

Velocity of river = 9 m/s

a)The resultant velocity V

$V=\sqrt{16^2+9^2}\ m/s$

V=18.35 m/s (South -East)

We know that

Distance = Velocity x time

Lets t time takes to cross the river

136 = 18.35 x t

t =7.41 m/s

The distance covered downstream

We know that

Distance = Velocity x time

t= 7.41 s

D= 7.41 x 9 m

D= 66.70 m

3 0

2 years ago

How is the electromagnetic spectrum organized

Vlad1618 [11]

The electromagnetic spectrum is the range of all possible frequencies of electromagnetic radiation. The "electromagnetic spectrum" of an object is the characteristic distribution of electromagnetic radiation emitted or absorbed by that particular object. Hope this helped.

7 0

3 years ago