cylindrical water tank is 20.0 m tall, open to the atmosphere at the top, and is filled to the top. It is noticed that a small hole has occurred in the side at a point 16.5 m below the water level and that water is flowing out at the volume flow rate of 2.90 10-3 m3/min. Determine the following. (a) the speed in m/s at which water is ejected from the hole

Answer:

Explanation:

weight of the person = 688 N

a) reading while in the elevator using the scale was  726 N

since when the elevator is going upward the floor of the elevator and the the scale pushes against the person leading to the person experiences a normal force greater than the weight as a result of the acceleration of the elevator and also

F = ma  and W, weight = mg

mass of the body = weight / g = 688 / 9.8 = 70.20 kg

net force on the body = force of normal - weight of the body = 726 - 688 = 38 N

ma = 38 N

70.20 kg × a ( acceleration) = 38 N

a = 38 / 70.20 = 0.54 m/s² and the elevator is moving upward

b) net force, ma = force of normal - weight of the body = 598 - 688 = -90 N

a = -90 / 70.20 = -1.282 m/s² and the elevator is coming downward

Answer:

A sparrow flew faster. the sparrow flew 10 meters per second. The sparow flew 13.(3) meters per second

Explanation:

A sparrow flew faster than the hawk as it completes more distance in 60 seconds than that of hawk which is about 1200 meters. Speed is the distance travelled per unit time.

Speed is the measure of the distance travelled by an object per unit time taken. Speed is a vector quantity. It has both magnitude and direction.

Speed of an object can be calculated as: Distance travelled divided by time taken.

Speed of hawk is 600 meters/ 60 seconds

Speed of hawk = 10 m/s

Speed of sparrow is 400 meters/ 30 seconds

Speed of Sparrow = 13.33 m/s

Learn more about Speed here:

https://brainly.com/question/28224010

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Complete Question

A thin uniform film of refractive index 1.750 is placed on a sheet of glass with a refractive index 1.50. At room temperature ( 18.8 ∘C), this film is just thick enough for light with a wavelength 580.9 nm reflected off the top of the film to be canceled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to 170 ∘C, you find that the film cancels reflected light with a wavelength 588.2 nm .

What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.) Express your answer using two significant figures.

Answer:

the coefficient of linear expansion of the film is [tex]\alpha = 7.93 *10^{-5} / ^oC[/tex]

Explanation:

  From the question we are told that

     The refractive index of the film is  [tex]n_f = 1.750[/tex]

      The refractive index of the glass is [tex]n_g = 1. 50[/tex]

       The wavelength of light reflected at 18°C is [tex]\lambda _r = 580.9nm = 580.9*10^{-9}m[/tex]

      The wavelength of light reflected at 170°C is [tex]\lambda_h = 588.2 nm = 588.2 * 10^{-9}m[/tex]

For destructive interference the condition is  

         [tex]2t = \frac{m \lambda }{n_f}[/tex]

  Where m is the order of interference

              t is the thickness

For the smallest thickness is  when m= 1 and this is represented as

             [tex]t = \frac{\lambda }{2n_f }[/tex]

At 18°C  the thickness would be

              [tex]t_{r} = \frac{580.9 *10^{-9}}{2 * 1.750}[/tex]

              [tex]t_{r} = 166nm[/tex]\

At 170° the  thickness is  

               [tex]t_h = \frac{588.2 *10^{-9}}{2 * 1.750}[/tex]

               [tex]t_h = 168 nm[/tex]

The coefficient of linear expansion f the film is mathematically represented as

              [tex]\alpha = \frac{t_h - t_r}{t_r \Delta T}[/tex]

Substituting value

                 [tex]\alpha = \frac{168 *10^{-9} - 166 *10^{-9} }{166*10^{-9} * (170 -18)}[/tex]

                 [tex]\alpha = 7.93 *10^{-5} / ^oC[/tex]

Answer:

α=8.37 x [tex]10^{-5}[/tex] °[tex]C^{-1}[/tex]

Explanation:

Complete question : What is the coefficient of linear expansion of the film?

SOLUTION:

There is a net ( λ/2 ) phase change due to reflection for this film, therefore, destructive interference is given by

2t = m( λ/n)   where n=1.750

for smallest non-zero thickness

t= λ/2n

At 18.8°C, [tex]t_{o[/tex]=580.9 x [tex]10^{-9}[/tex]/(2 x 1.750)

[tex]t_{o[/tex]= 165.9nm

At 170°C, t= 588.2x [tex]10^{-9}[/tex]/(2x1.750)

t=168nm

t=[tex]t_{o[/tex](1 + αΔT)

=>α= (t-[tex]t_{o[/tex])/ ([tex]t_{o[/tex]ΔT)   [ΔT= 170-18.8 =151.2°C]

α= (168 x [tex]10^{-9}[/tex] - 165.9 x [tex]10^{-9}[/tex])/ (165.9 x [tex]10^{-9}[/tex] x 151.2)

α= 2.1 x [tex]10^{-9}[/tex]/ 2.508 x [tex]10^{-5}[/tex]

α=8.37 x [tex]10^{-5}[/tex] °[tex]C^{-1}[/tex]

Therefore, the coefficient of linear expansion of the film is 8.37 x [tex]10^{-5}[/tex] °[tex]C^{-1}[/tex]

Answer:

29.61 rpm.

Explanation:

Given,

student arm length, l = 67 cm

distance of the bucket, r = 35 m

Minimum angular speed of the bucket so, the water not fall can be calculated by equating centrifugal force with weight.

Now,

[tex]mg = m r \omega^2[/tex]

[tex]\omega = \sqrt{\dfrac{g}{R}}[/tex]

R = 67 + 35 = 102 cm = 1.02 m

[tex]\omega = \sqrt{\dfrac{9.81}{1.02}}[/tex]

[tex]\omega = 3.101\ rad/s[/tex]

[tex]\omega = \dfrac{3.101}{2\pi} = 0.494\ rev/s[/tex]

[tex]\omega = 0.494 \times 60 = 29.61\ rpm[/tex]

minimum angular velocity is equal to 29.61 rpm.

Answer:

29.6 rpm

Explanation:

length of arm = 67 cm

distance of handle to the bottom = 35 cm

radius of rotation, R = 67 + 35 = 102 cm = 1.02 m

The centripetal force acting on the bucket is balanced by the weight of the bucket.

mRω² = mg

R x ω² = g

[tex]\omega = \sqrt\frac{g}{R}[/tex]

[tex]\omega = \sqrt\frac{9.8}{1.02}[/tex]

ω = 3.1 rad/s

Let f is the frequency in rps

ω = 2 x 3.14 x f

3.1 = 2 x 3.14 xf

f = 0.495 rps

f = 29.6 rpm

Answer:

A)P = 1.2 × 10⁵Pa

B)F = 3.2 × 10⁷N

C) P = 1.2 × 10⁵Pa

Explanation:

Part A)

The relative pressure at the bottom of a column of fluid is given by

[tex]p_r = \rho g h[/tex]

where

[tex]\rho[/tex] is the fluid density

g is the gravitational acceleration  

h is the height of the column of fluid

At the bottom of the swimming pool, h=1.9 m, and the water density is  

[tex]\rho[/tex] = 1000 kg/m^3, therefore the relative pressure is

[tex]p_r = (1000 kg/m^3)(9.81 m/s^2)(1.9 m)=1.86 \cdot 10^4 Pa[/tex]

To find the absolute pressure, we must add to this the atmospheric pressure, [tex]p_a[/tex] :

[tex]p= p_r + p_a\\= 1.86 \cdot 10^4 Pa + 1.01 \cdot 10^5 Pa \\=1.2 \times 10^5 Pa[/tex]

part B

Total force acting on the bottom

force = pressure * area

area of pool = 30.0 m × 8.9 m

= 267m²

Force F =

1.2 × 10⁵ * 267m² N

= 32040000 N

F = 3.2 × 10⁷N

Part C

The pressure acting on the side wall will be

now the pressure at the side of the pool at the bottom is simply equal to absolute pressure as they are at same level

P = 1.2 × 10⁵

Answer:

(a) 11.8692‬ ohm

(b) 12.447 A

(c) 17.6 A

Explanation:

a)  inductive reactance Z = L Ω

    = L x 2π x F

    = 45.0 x 10⁻³ x 2(3.14) x 42

    = 11.8692‬ ohm

b) rms current

    = 100 / 8.034

    = 12.447 A

c) maximum current in the circuit

    = I eff x rac2

    = 12.447 x 1.414

    = 17.6 A

Answer:

The angle that the wave would be [tex]\theta = sin ^{-1}\frac{2 \lambda}{D}[/tex]

Explanation:

From the question we are told that  the  opening to  the  harbor acts just like a single-slit so a boat in the harbor that at angle equal to the second diffraction minimum would be safe and the  on at angle greater than the diffraction first minimum would be slightly affected

  The minimum is as a result of destructive interference

       And for single-slit this is mathematically represented as

               [tex]D sin \ \theta =m \lambda[/tex]

where D is the slit with

          [tex]\theta[/tex] is the angle relative to the original direction of the wave

         m is the order of the minimum j

        [tex]\lambda[/tex] is the wavelength

Now since in the question we are told to obtain the largest angle at which the boat would be safe

      And the both is safe at the angle equal to the second minimum then

    The the angle is evaluated as

           [tex]\theta = sin ^{-1}[\frac{m\lambda}{D} ][/tex]

Since for second minimum m= 2

The  equation becomes

               [tex]\theta = \frac{2 \lambda}{D}[/tex]

Answer:

Explanation:

Expression for times period of a satellite can be given as follows

Time period T = 1.8 x 60 x 60

= 6480

T² = [tex]\frac{4\times \pi^2\times r^3}{GM}[/tex] where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.

6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM

GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²

= 3.96 X 10¹⁴

Expression for acceleration due to gravity

g = GM / R² where R is radius of satellite

20 = 3.96 X 10¹⁴ / R²

R² = 3.96 X 10¹⁴ / 20

= 1.98 x 10¹³ m

R= 4.45 x 10⁶ m

Answer:

see explanation

Explanation:

Given that,

velocity of 1.50 km/s = 1.50 × 10³m/s

acceleration of 2.00 ✕ 1012 m/s2

electric field has a magnitude of strength of 18.0 N/C

[tex]\bar F= q[\bar E + \bar V \times \bar B]\\\\\bar F = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]\\\\\\m \bar a = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )][/tex]

[tex]9.1 \times 10^-^3^1 \times 2\times 10^1^2 \hat k=-1.6\times10^-^1^9 \hat k [18\hat k+ 1.5\times 10^3 \hat i \times (B_x \hat i +B_y \hat j +B_z \hat k)][/tex][tex]42.2 \times 10^-^1^9 \hat k = -2.4 \times 10^1^6B_y \hat k + 2.4 \times 10 ^1^6 \hat j B_z\\[/tex]

[tex]B_x = undetermined[/tex]

[tex]B_y = \frac{42.2 \times 10^-^1^9}{-2.4 \times 10^-^1^6} \\\\= - 0.0176 T[/tex]

[tex]B_z = 0T[/tex]

Answer:

[tex]1.7\cdot 10^{15} V m^{-1} s^{-1}[/tex]

Explanation:

The electric field between the plates of a parallel-plate capacitor is given by

[tex]E=\frac{V}{d}[/tex] (1)

where

V is the potential difference across the capacitor

d is the separation between the plates

The potential difference can be written as

[tex]V=\frac{Q}{C}[/tex]

where

Q is the charge stored on the plates of the capacitor

C is the capacitance

So eq(1) becomes

[tex]E=\frac{Q}{Cd}[/tex] (2)

Also, the capacitance of a parallel-plate capacitor is

[tex]C=\frac{\epsilon_0 A}{d}[/tex]

where

[tex]\epsilon_0[/tex] is the vacuum permittivity

A is the area of the plates

Substituting into (2) we get

[tex]E=\frac{Q}{\epsilon_0 A}[/tex] (3)

Here we want to find the rate of change of the electric field inside the capacitor, so

[tex]\frac{dE}{dt}[/tex]

If we calculate the derivative of expression (3), we get

[tex]\frac{dE}{dt}=\frac{1}{\epsilon_0 A}\frac{dQ}{dt}[/tex]

However, [tex]\frac{dQ}{dt}[/tex] corresponds to the definition of current,

[tex]I=\frac{dQ}{dt}[/tex]

So we have

[tex]\frac{dE}{dt}=\frac{I}{\epsilon_0 A}[/tex]

In this problem we have

I = 3.9 A is the current

[tex]A=(0.0160 m)\cdot (0.0160 m)=2.56\cdot 10^{-4} m^2[/tex] is the area of the plates

Substituting,

[tex]\frac{dE}{dt}=\frac{3.9}{(8.85\cdot 10^{-12})(2.56\cdot 10^{-4})}=1.7\cdot 10^{15} V m^{-1} s^{-1}[/tex]

The rate at which the electric field is changing in the parallel-plate capacitor is approximately 1.7 x 10¹⁴ V/m·s.

1. The area (A) of the plates:

2. The displacement current (I_D) is given by:

where

3. The electric flux (Φ) is related to the electric field (E) by:

Since:

4. Substitute this into the displacement current equation:

Rearranging for dE/dt:

5. Plugging in the values:

Thus, the rate at which the electric field is changing is approximately 1.7 x 10¹⁴ V/m·s.

Answer:

[tex]f=9.5\ KHz[/tex]

Explanation:

AC Circuit

When connected to an AC circuit, the capacitor acts as an impedance of module

[tex]\displaystyle Z=\frac{1}{wC}[/tex]

Where w is the angular frequency of the power source and C is the capacitance.

If the capacitor is the only element connected to a circuit, then the Ohm's law establishes that

[tex]V=Z.I[/tex]

Where V and I are the rms voltage and current respectively. Replacing the value of Z, we have

[tex]\displaystyle V=\frac{I}{wC}[/tex]

Solving for w

[tex]\displaystyle w=\frac{I}{VC}[/tex]

The question provides us the following values

[tex]C=63\ \mu F=63\cdot 10^{-6}\ F[/tex]

[tex]V=4\ Volt[/tex]

[tex]I=15\ A[/tex]

Plugging in the values

[tex]\displaystyle w=\frac{15}{4\cdot 63\cdot 10^{-6}}[/tex]

[tex]w=59523.81\ rad/s[/tex]

Since

[tex]w=2\pi f[/tex]

Then

[tex]\displaystyle f=\frac{59523.81}{2\pi}=9473.51\ Hz[/tex]

[tex]f=9.5\ KHz[/tex]

Final answer:

The fuse in the AC circuit with a 63.0 µF capacitor will burn out when the generator frequency is increased to approximately 1,001 Hz.

Explanation:

To find at what frequency the fuse will burn out in the circuit with a 63.0 µF capacitor and a 4.00 V rms voltage generator, we will need to calculate the capacitive reactance (XC) and then use the relationship between current, voltage, and reactance for an AC circuit.

The capacitive reactance is given by the formula XC = 1 / (2πfC), where f is the frequency in hertz (Hz), and C is the capacitance in farads. The rms current I in the circuit is given by the rms voltage V divided by XC:
I = V / XC

Setting I to the maximum allowable current of 15.0 A, we have:
15.0 A = 4.00 V / (1 / (2πf × 63.0 × 10⁻⁶ F))

Solving for f, we get:

f = 1 / (2π × 63.0 × 10⁻⁶ F × (4.00 V / 15.0 A))
f ≈ 1,001 Hz

Therefore, the fuse will burn out when the generator frequency is increased to approximately 1,001 Hz.

Answer:

a. Air fuel Ratio = 19.76 kg air/kg fuel

b. % Theoretical air used = 131%

c. Amount of H2O that condenses as the products are cooled to 25°C at 100kPa = 6.59 kmol

Explanation:

Answer:

Tension, T = 105.09 N

Explanation:

Given that,

Length of the string, l = 0.33 m

Mass of the string, [tex]m=0.3\times 10^{-3}\ kg[/tex]

Fundamental frequency, f = 440 Hz

The expression for the speed in terms of tension is given by :

[tex]v=\sqrt{\dfrac{T}{(m/l)}}[/tex]

[tex]v^2=\dfrac{Tl}{m}\\\\T=\dfrac{v^2m}{l}\\\\T=\dfrac{(340)^2\times 0.3\times 10^{-3}}{0.33}\\\\T=105.09\ N[/tex]

So, the tension in the string is 105.09 N.

Answer:

the correct one is C

Explanation:

For this exercise we must use the work definition

    W = F. s

Where the bold characters indicate vectors and the point is the scalar producer

    W = F s cos θ

Where θ is the angles between force and displacement.

Let us support this in our case. The cable creates an upward tension and with the elevator going down the angle between them is 180º so the work of the cable on the elevator is negative.

The evade has a downward force, its weight so the force goes down and the displacement goes down, as both are in the same direction the work is positive

When examining the statements the correct one is C

Final answer:

The correct statement for an elevator being lowered at constant speed is that the steel cable does negative work on the elevator, and the elevator does negative work on the cable, illustrating the principle that work can be negative if force and displacement are in opposite directions.

Explanation:

The question pertains to the work done by an elevator cable while lowering an elevator at constant speed. According to the principles of work and energy in physics, work done is defined as the force applied in the direction of motion times the distance moved. If an elevator is being lowered at a constant speed, the steel cable exerts an upward force to counteract gravity but the elevator moves downward. Therefore, the displacement of the elevator is in the opposite direction to the force exerted by the cable, resulting in the cable doing negative work on the elevator. Conversely, because the elevator is moving downwards (in the direction opposite to the force exerted by the cable), we can interpret this as the elevator doing negative work on the cable as well, due to the concept that positive work adds energy to a system while negative work removes it.

Thus, the correct statement is: D. The cable does negative work on the elevator, and the elevator does negative work on the cable. This illustrates the application of the definition of work in physics, particularly in scenarios involving opposite directions of force and motion.

Answer:

v = 29.35 m /s

Explanation:

potential energy at the height of 62m

= m g h , m is mass , g is acceleration due to gravity and h is height

= 60 x 9.8 x 62

= 36456 J

negative work done by friction = -10600 J

energy at the bottom = 36456 - 10600 = 25856 J

This energy will be in the form of kinetic energy . If v be velocity at the bottom

1/2 m v² = 25856

1/2 x 60 x v² = 25856

v = 29.35 m /s

Answer:

There is no magnetic field.

Explanation:

Since th capacitor is charged and isolated, magnetic field doesn't exist.

For magnetic field to exist, there must be flow of charge.

Maria's change in velocity while riding her bike is 8 meters per second to the north.

Since the motion is along the same direction (to the north), we do not need to consider the direction as negative or positive. Here's the calculation:

Change in velocity = Final velocity - Initial velocity

= 11 m/s - 3 m/s

= 8 m/s.

So, Maria's change in velocity is 8 meters per second to the north.

Answer:

d = 4180.3m

wavelengt of sound is 0.251m

Explanation:

Given that

frequency of the sound is 5920 Hz

v=1485m/s

t=5.63s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

[tex]velocity=\frac{distance}{time}\\\\ v=\frac{2d}{t} \\\\vt=2d\\\\d=\frac{vt}{2}[/tex]

[tex]d=\frac{1485*5.63}{2}\\d= 4180.3m[/tex]

wavelengt of sound is [tex]\lambda[/tex] = v/f

= (1485)/(5920)

= 0.251 m

Answer:

Explanation:

According to  energy conservation principle;

[tex]\delta \ K.E = \delta \ P.E[/tex]

[tex]8.00*10^{-19} \ J = (1.6*10^{-19} \ C ) |V_b-V_a|[/tex]

[tex]|V_b-V_a|= \frac {8.00*10^{-19} }{(1.6*10^{-19} ) }[/tex]

[tex]|V_b-V_a|= 5 \ V[/tex]

The potential is positive hence the electric fielf=d is negative along the x-axis.

We can then say that the movement of the particle goes to the left through a potential difference of  [tex]|V_b-V_a|= 5 \ V[/tex].

There will be no significant change in the y-direction of the potential energy when the particle moves from point b to point c in the y-direction.

Answer:

Explanation:

velocity of proton  v = 1.5 x 10³ i  m /s

charge on proton e = 1.6 x 10⁻¹⁹ C

Let the magnetic field be B = Bx i + Bz k

force on charged particle ( proton )

F = e ( v x B )

2.06  x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ [ 1.5 x 10³ i x ( Bx i + Bz k) ]

2.06  x10⁻¹⁶ j = - 1.6 x 10⁻¹⁹ x 1.5 x 10³  Bz j) ]

2.06  x10⁻¹⁶ = - 1.6 x 10⁻¹⁹ x 1.5 x 10³  Bz

Bz = -  .8583  

force on charged particle ( electron )

F = e ( v x B )

8.40  x10⁻¹⁶ j = -1.6 x 10⁻¹⁹ [ - 4.4 x 10³ k  x ( Bx i + Bz k) ]

8.4  x10⁻¹⁶ j =  1.6 x 10⁻¹⁹ x 4.4 x 10³  Bx j ]

- 8.4  x10⁻¹⁶ =  1.6 x 10⁻¹⁹ x 4.4 x 10³  Bx

Bx =  - 1.19

Magnetic field = - 1.19 i - .8583 k

magnitude = √ (1.19² + .8583²)

= 1.467 T

If it is making angle θ with x - axis in x -z plane

Tanθ = (.8583 / 1.19 )

36⁰ .

C )

v = - 3.7 x 10³j m /s

e = - 1.6 x 10⁻¹⁶ C

Force = F = e ( v x B )

= -1.6 x 10⁻¹⁹ [ -3.7 x 10³ j  x ( Bx i + Bz k) ]

=  - 1.6 x 10⁻¹⁹ x 3.7 x 10³  Bx  k -1.6 x 10⁻¹⁹ x 3.7 x 10³Bzi ]

=    5.08 i - 7.04 k

Tanθ = 54 ° .

Answer:

[tex]6.53\times10^-^1^7N[/tex]

Explanation:

The magnet of the magnetic field is 53 cm = 0.53m from wire is

[tex]B = \frac{\mu_0 I}{2\pi d}[/tex]

[tex]= \frac{(4\pi \times 10^-^7)(40)}{2 \pi (0.53)} \\\\= \frac{5.0265\times 10^-^5}{3.33} \\\\= 1.5095 \times 10^-^5[/tex]

the magnetic force exerted by the wire on the electron is

[tex]F = Bqv \sin \theta\\\\= 1.5095 \times 10^-^5 \times1.602\times10^-^1^9\times2.7\times10^7\\\\= 6.53\times10^-^1^7N[/tex]

From the right hand rule the direction of the force is parallel to the current (since the particle is electron)

Answer: f = 6.52*10^-16 N

Explanation:

if we assume that the force is directed at the y positive direction, then

B = μi / 2πr, where

μ = 4π*10^-7

B= (4π*10^-7 * 40) / 2 * π * 5.3*10^-2

B = 5.027*10^-5 / 0.333

B = 1.51*10^-4 T

Since v and B are perpendicular, then,

F = qvB

F = 1.6*10^-19 * 2.7*10^7 * 1.51*10^-4

F = 2.416*10^-23 * 2.7*10^7

F = 6.52*10^-16 N

Therefore, the magnitude of the force is, F = 6.52*10^-16 N and it moves in the i negative direction

The velocity that will be needed for the model and prototype to be similar is 108.97m/s  

Explanation:

length of Torpedo = 3m

diameter, [tex]d_{1} = 0.5m[/tex]

velocity of sea water, [tex]v_{1}[/tex]= 10m/s

dynamic viscosity of sea water, η[tex]_{1}[/tex] = 0.00097 Ns/m²

density of sea water, ρ[tex]_{1}[/tex] =  1023 kg/m³

Scale model = 1:15

[tex]\frac{d_{1} }{d_{2} }[/tex] = [tex]\frac{1}{15}[/tex]

Cross multiplying: d[tex]_{2}[/tex] = [tex]15d_{1} }[/tex] = 15 ×0.5 = 7.5m

Let:

velocity of air, [tex]v_{2}[/tex]

viscosity of air, η[tex]_{2}[/tex] =  0.000186Ns/m²

density of air, ρ[tex]_{2}[/tex] =  1.2 kg/m³

For the model and the prototype groups to be equal, Non-dimensional groups should be equal.

Reynold's number:   (ρ[tex]_{2}[/tex] ×[tex]v_{2}[/tex] ×d[tex]_{2}[/tex])/η[tex]_{2}[/tex] =  (ρ[tex]_{1}[/tex] ×[tex]v_{1}[/tex] ×d[tex]_{1}[/tex])/η[tex]_{1}[/tex]

[tex]v_{2}[/tex] = η[tex]_{2}[/tex]/(ρ[tex]_{2}[/tex] ×d[tex]_{2}[/tex])  ×  (ρ[tex]_{1}[/tex] ×[tex]v_{1}[/tex] ×d[tex]_{1}[/tex])/η[tex]_{1}[/tex]

[tex]v_{2}[/tex] =  [tex]\frac{0.000186}{1.2* 7.5}[/tex]×[tex]\frac{1023 *10*0.5}{0.00097}[/tex] , note: * means multiplication

[tex]v_{2}[/tex] = 108.97m/s  

velocity that will be needed for the model and prototype to be similar = 108.97m/s  

Answer:

The value of the distance is [tex]\bf{14.52~cm}[/tex].

Explanation:

The velocity of a particle(v) executing SHM is

[tex]v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)[/tex]

where, [tex]\omega[/tex] is the angular frequency, [tex]A[/tex] is the amplitude of the oscillation and [tex]x[/tex] is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., [tex]x = 0[/tex].

The maximum velocity([tex]\bf{v_{m}}[/tex]) is

[tex]v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]

Divide equation (1) by equation(2).

[tex]\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)[/tex]

Given, [tex]v = 0.25 v_{m}[/tex] and [tex]A = 15~cm[/tex]. Substitute these values in equation (3).

[tex]&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm[/tex]

The heat rate, efficiency, and volume of a straight fin made from 2024 Aluminum alloy can be calculated using relevant formulas considering its physical dimensions, the transferred heat, and the profile of the fin. Comparisons across different profiles (rectangular, triangular, parabolic) are commonly done using numerical or graphical solution methods.

To compare the heat rate, efficiency, and volume for a straight fin made of 2024 Aluminum alloy with a rectangular, triangular, and parabolic profile, we first need to convert all known variables into SI units. The fin has a base thickness (t) of 3 mm or 0.003 m, a length (L) of 15 mm or 0.015 m, and it's exposed to a fluid at a temperature (T infinity) of 20C. The base temperature (Tb) of the fin is 100oC, and the heat transfer coefficient (h) is 50 W/m2K.

We can estimate the heat transfer rate (Q) by applying the formula Q = hA(Tb - T infinity), where A represents the surface area of the fin which would depend on the fin's profile. For a unit width, the area of a rectangular fin is A = wt = unit width*L, for a triangular profile A = 0.5*wt, and for a parabolic profile A = (2/3)*wt.

Fin efficiency can be calculated by dividing the actual heat transferred by the fin by the maximum possible heat transfer. Since these are dependent on the fin's profile (shape), numerical or graphical solution methods are commonly used for calculations.

Volume can be calculated as the product of the surface area and thickness, which again would depend on the fin's profile.

Understanding these differences among fin profiles is important in heat transfer management and finding ways to increase fin efficiency.

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Compare the fin heat rate using equation: Qfin = √(hPkA) (Tb - T∞), efficiency using equation: η = Qfin / (hAt (Tb - T∞)) , and volumes are found to be bLt, (1/2)bLt and (2/3)bL for rectangular, triangular, and parabolic profiles, respectively.

1.) Heat Transfer Analysis of a Straight Fin

To compare the fin heat rate, efficiency, and volume for different fin profiles (rectangular, triangular, and parabolic), we need to perform the following calculations:

Step 1: Identify Knowns and Convert to SI Units

Given:

Step 2: Determine the Fin Parameter, m

The fin parameter, m, is calculated using the formula:

For a rectangular fin:

We adjust the perimeter and area for parabolic and triangular profiles likewise.

Step 3: Heat Transfer Rate

For each fin profile, the heat transfer rate, Qfin, is given by:

where A is the cross-sectional area, and P is the perimeter of the respective fin.

Step 4: Fin Efficiency

Fin efficiency, η, is determined as:

where At is the total surface area of the fin.

Step 5: Compare Fin Volume

The volume, V, can be found by:

After calculating these values, you can compare the heat rate, efficiency, and volume between the fin profiles to determine the most efficient design.

Answer:

There are four factors affecting resistance which are Temperature,Length of wire,Area of the cross section of wire and nature of the material.When there is current in a conductive material,The free electrons move through the material and occasionally collide with atoms.  

Explanation:

Answer: its length, material, temperature, and cross section area which can also be considered as diameter.

Explanation:

a) 1.84 m

b) 1.55 m

Explanation:

a)

In this problem, the only force acting on Zak along the direction of motion (horizontal direction) is the force of friction, which is

[tex]F_f=-\mu mg[/tex]

where

[tex]\mu=0.250[/tex] is the coefficient of friction

m is Zak's mass

[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity

According to Newton's second law of motion, the net force acting on Zak is equal to the product between its mass (m) and its acceleration (a), so we have

[tex]F=ma[/tex]

Here the only force acting is the force of friction, so this is also the net force:

[tex]-\mu mg = ma[/tex]

Therefore we can find Zak's acceleration:

[tex]a=-\mu g=-(0.250)(9.8)=-2.45 m/s^2[/tex]

Since Zak's motion is a uniformly accelerated motion, we can now use the following suvat equation:

[tex]v^2-u^2=2as[/tex]

where

v = 0 is the final velocity (he comes to a stop)

u = 3.00 m/s is the initial velocity

[tex]a=-2.45 m/s^2[/tex] is the acceleration

s is the distance covered before stopping

Solving for s,

[tex]s=\frac{v^2-u^2}{2a}=\frac{0^2-3.0^2}{2(-2.45)}=1.84 m[/tex]

b)

In this second part, Zak gives a push to Greta.

We can find Greta's velocity after the push by using the work-energy theorem, which states that the work done on her is equal to her change in kinetic energy:

[tex](F-F_f)d =\frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]

where

F = 125 N is the force applied by Zak

d = 1.00 m is the distance

[tex]F_f=\mu mg[/tex] is the force of friction, where

[tex]\mu=0.250[/tex]

m = 20.0 kg is Greta's mass

[tex]g=9.8 m/s^2[/tex]

v  is Greta's velocity after the push

u = 0 is Greta's initial velocity

Solving for v, we find:

[tex]v=\sqrt{\frac{2(F-\mu mg)d}{m}}=\sqrt{\frac{2(125-(0.250)(20.0)(9.8))(1.00)}{20.0}}=2.76 m/s[/tex]

After that, Zak stops pushing, so Greta will slide and the only force acting on her will be the force of friction; so the acceleration will be:

[tex]a=-\mu g = -(0.250)(9.8)=-2.45 m/s^2[/tex]

And so using again the suvat equation, we can find the distance she slides after Zak's push ends:

[tex]s=\frac{v'^2-v^2}{2a}[/tex]

where

v = 2.76 m/s is her initial velocity

v' = 0 when she stops

Solving  for s,

[tex]s=\frac{0-(2.76)^2}{2(-2.45)}=1.55 m[/tex]

(a) The distance traveled by Zack before stopping is 1.84 m.

(b) The distance traveled by Greta after Zack's push ends is 1.56 m.

The given parameters;

The distance traveled by Zack before stopping is calculated as follows;

The acceleration of Zack;

[tex]-F_k = ma\\\\-\mu_k mg = ma\\\\-\mu_k g = a\\\\-(0.25 \times 9.8) = a\\\\- 2.45 \ m/s^2 = a[/tex]

The distance traveled by Zack;

[tex]v^2 = u^2 + 2as\\\\when \ Zack \ stops \ v = 0\\\\0 = u^2 + 2as\\\\0 = (3)^2 +2(-2.45)s\\\\0 = 9 - 4.9s\\\\4.9s = 9\\\\\s = \frac{9}{4.9} \\\\s = 1.84 \ m[/tex]

The distance traveled by Greta is calculated as follows;

Apply law of conservation of energy to determine the velocity of Greta after the push.

[tex]Fd - F_kd = \frac{1}{2} mv^2\\\\125\times 1 - (0.25 \times 20 \times 9.8 \times 1) = (0.5 \times 20)v^2\\\\76 = 10v^2\\\\v^2 = \frac{76}{10} \\\\v ^2 = 7.6\\\\v = \sqrt{7.6} \\\\v = 2.76 \ m/s[/tex]

The acceleration of Greta;

[tex]a = -\mu_k g\\\\a = 0.25 \times 9.8\\\\a = -2.45 \ m/s^2[/tex]

The distance traveled by Greta;

[tex]v_f^2 = v^2 + 2as\\\\when \ Greta \ stops \ v_f = 0\\\\0 = v^2 + 2as\\\\-2as = v^2\\\\-2(-2.45)s = (2.76)^2\\\\4.9s = 7.62\\\\s = \frac{7.62}{4.9} \\\\s = 1.56 \ m[/tex]

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Answer:

1. Our ears can sort out the individual sine waves from a mixture of two or more sine waves, so we hear the pure tones that make up a complex tone.

Explanation:

A complex tone is a sound wave that consist of two or more forms of audible sound frequencies. Sound wave is a mechanical wave that is longitudinal, and could be represented by a sine wave because of it sinusoidal manner of propagation.

A Fourier analyzer can be used to differentiate individual sine waves from a combination of two or more of it; which is as the same function performed by human ear. To the human ear, a sound wave that consist of more than one sine wave will have perceptible harmonics which would be distorted and turn to a noise.

Thus, the human ear makes it possible to hear the pure tones that make up a complex tone.

Answer:

1. A Fourier analyzer sorts out the individual sine waves from a mixture of two or more sine waves, so we hear the pure tones that make up a complex tone.

Explanation:

Fourier analysis is a technique that is used to determine which sine waves constitute a given signal, i.e. to deconstruct the signal into its individual sine waves.  It is the process of decomposing a periodic function into its constituent sine or cosine waves.

What goes on inside our ears is a mathematical process called a Fourier transform. In the ear, there's a combination of different waves, Fourier analysis identifies contributions at different frequencies, allowing us to reconstruct the individual signals that go into it.

A complex tone perceived by the air is is an individual sine wave that the ear, by acting as a Fourier analyzer, decomposes to serious of sine waves that we hear as pure tones.

Complete Question

The complete Question is shown on the first and second uploaded image

Answer:

The underlined words are the answers

Part A

Moving from boron to carbon, the intensity of the bulb Increases  because Z increases from 5 to 6 , The thickness of the frosting stays the  same because the core electron configuration is the same for both atoms

Part B

Moving from boron to aluminum the intensity of the bulb Increases because Z increases  from 5 to 13 . The thickness of the frosting also increases because Al has the core configuration of Ne, while B has the core configuration of He

Explanation:

Here Z denotes the atomic number

        Ne denoted the element called Neon and its electronic  configuration is

                     [tex]1s^2 \ 2s^2 \ 2p^6[/tex]

    He denoted the element called Helium  and its electronic  configuration is

                     [tex]1s^2[/tex]

     B denoted the element called Boron   and its electronic  configuration is

                [tex]1s^2 \ 2s^2\ 2p^1[/tex]

Looking at its electronic configuration we can see that the core is He

I,e              [tex][He]\ 2s^2 2p^1[/tex]

Al denoted the element called Aluminium  and its electronic configuration is    

             [tex]1s^2 \ 2s^2 \ 2p^2 \ 3s^2 \ 3p^1[/tex]

Looking at its electronic configuration we can see that the core is Ne

I,e              [tex][Ne]\ 3s^2 \ 3p^1[/tex]

Answer:

Pabs = 247150 [Pa]

Explanation:

The pressure in the depth h can be calculated by the following expression.

Pabs = Po + (rho * g * h)

Where:

g = gravity = 9.81[m/s^2]

rho = density = 1000 [kg/m^3]

h = depth = 15 [m]

Po = 100000 [Pa]

Pabs = 100000 + (1000*9.81*15)

Pabs = 247150 [Pa]

Answer:

18.8m

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.390

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know  from work-energy theorem, that the work done by the net force on a body is equal to the change in its kinetic energy.

Hence, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

[tex]f=\mu N[/tex]

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, N=mg

[tex]f=\mu mg\\=0.390\times 90\times 9.8\\=343.98\ N[/tex]

work done by friction is a negative work as friction and displacement are in opposite direction and is given as

[tex]W=-fd=-343.98d[/tex]

Now, change in kinetic energy is given as:

[tex]\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J[/tex]

Therefore, from work-energy theorem,

[tex]W=\Delta K\\\\-343.98d=6480\\\\d=\frac{6480}{343.98}\\\\d=18.8m[/tex]

Hence, the skater covers a distance of 18.8 m before stopping.

Answer:

18.84m

Explanation:

We are given;

Mass; m = 90 Kg

Initial velocity; u = 12 m/s

Coefficient of friction; μ = 0.390

Let,the combined mass of the ice skater and the skate be M

Thus, So,if the coefficient of frictional force is μ,then frictional force acting is; μN. N= Mg. Thus F_friction = μMg

Now, the deceleration due to friction will be, F_friction/M

Thus, deceleration = μMg/M

M will cancel out and we have; μg

Now, deceleration means negative acceleration. Thus acceleration (a) =

-μg

Now, to find the distance, let's use equation of motion which is;

v² = u² + 2as

Putting -μg for a, we have;

v² = u² + 2(-μg)s

v² = u² — 2μgs

We want to know the distance covered before coming to rest. Thus, v = 0m/s

So,

0² = 12² - (2 x 0.39 x 9.8 x s)

0 = 144 - 7.644s

Thus,

7.644s = 144

Thus, s = 144/7.644 = 18.84m