Ceres is a dwarf planet located in the main asteroid belt.

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

2 years ago

which force field can accelerate an electron, but never change its speed? a) electric field b) magnetic field c) both of these d

Drupady [299]

Option a; Electric field can accelerate an electron, but never change its speed

An electric field (also known as an E-field) is a physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It can also refer to the physical field of a charged particle system. Electric fields are created by electric charges and time-varying electric currents. Electric and magnetic fields are both aspects of the electromagnetic field, one of nature's four fundamental interactions (also known as forces). Electric fields are significant in many areas of physics and are used in electrical technology. In atomic physics and chemistry, for example, the electric field is the attractive force that holds the atomic nucleus and electrons together in atoms. It is also the driving force behind chemical bonds between atoms.

Learn more about Electric field here:

brainly.com/question/15800304

#SPJ4

4 0

1 year ago

When an object of weight w is suspended from the center of a massless string?

igomit [66]

I attached a picture of the diagram associated with this question.

Now,
When we check the vertical components of the tension in the rope, we will find that we have two equal components acting upwards.
These two components support the weight and each of them has a value of TcosΘ

The net force acting on the body is zero.
Fnet=Force of tension acting upwards-Force due to weight acting downwards
0 = 2TcosΘ -W
W = 2TcosΘ
T = W / 2cosΘ

4 0

3 years ago

Read 2 more answers

A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats

ololo11 [35]

Answer:

Part a)

$f_B = 290 Hz$

Part B)

percentage increase is

$percentage = 1.38$ %

Explanation:

Part a)

As we know that the beat frequency is

$f_A - f_B = 3$

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

$293 - f_B = 3$

$f_B = 290 Hz$

Part B)

percentage increase in the tension of the string will be given as

$f_A - f_B' = 1$

$f_B' = 292 Hz$

now we have

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

so we have

$T_1 = C (290)^2$

$T_2 = C(292)^2$

so we have

$\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}$

percentage increase is

$percentage = 1.38$

4 0

3 years ago

Two people travel from Columbus to Cleveland, one by car at an average velocity of 90km/h, and one by plane at an average veloci

Flura [38]

Answer:

Explanation:

Yes , their displacement may be equal .

Suppose the displacement is AB where A is starting point and B is end point .

The car is covering the distance AB by going from A to B on straight line . On the other hand plane goes from A to C , then from C to D and then from D to B . In this way plane reaches B from A on a different path which is longer than path of the car . In the second case also displacement of plane is AB . In the second case distance covered is longer but displacement is same that is AB .

7 0

2 years ago