A piston–cylinder device contains 0.85 kg of refrigerant- 134a at 2108C. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 158C. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a.

Answer:

  9.29 N . . . . weight of 0.948 kg sphere

Explanation:

The sum of torques on the link BOA is zero, so we have ...

  (right force at A)(OA) = (up force at B)(OB·sin(135°))

Solving for the force at B, we have ...

  up force at B = (10.53 N)(253 mm)/((553 mm)/√2) ≈ 6.81301 N

This force is due to the difference between the buoyant force on the float sphere and the weight of the float sphere. Dividing by the acceleration due to gravity, it translates to the difference in mass between the water displaced and the mass of the sphere.

  ∆mass = (6.81301 N)/(9.8 m/s^2) = 0.695205 kg

__

The center of the sphere of diameter 0.1553 m is below the waterline by ...

  (553 mm)cos(45°) -(353 mm) = 38.0300 mm

The volume of the spherical cap of radius 155.3/2 = 77.65 mm and height 77.65+38.0300 = 115.680 mm can be found from the formula ...

  V = (π/3)h^3(3r -h) = (π/3)(115.680^2)(3·77.65 -115.68) mm^3 ≈ 1.64336 L

So the mass of water contributing to the buoyant force is 1.64336 kg. For the net upward force to correspond to a mass of 0.695305 kg, the mass of the float sphere must be ...

  1.64336 kg -0.695205 kg ≈ 0.948 kg

The weight of the float sphere is then (9.8 m/s^2)·(0.948 kg) = 9.29 N

The weight of the 0.948 kg float sphere is about 9.29 N.

Answer:

[tex]W_{net} = - 1223 kJ[/tex]

Explanation:

State 1:

[tex]P_1 = 20 bar\\T_1 = 360^{0}C\\ h_1 = 3159.3 kJ/kg\\S_1 = 6.9917 kJ/kg[/tex]

State 2:

[tex]P_2 = 20 bar\\x_2 = 1 \\ h_2 = 2799.5 kJ/kg\\u_2 = 2600.3 kJ/kg\\v_2 = 0.09963m^3/kg[/tex]

State 3:

[tex]P_2 = 5 bar\\v_2 = v_3 \\v_3 = v_f + x_3 (v_g - v_f)\\0.09963 = (1.0926 * 10^{-3}) +x_3 (0.3749 - (1.0926 * 10^{-3}))\\x_3 = 0.263[/tex]

[tex]u_{3} = u_f + x_3 ( u_g - u_f)\\u_{3} = 639.68 + 0.263 (2561.2 - 639.68)\\u_{3} = 1146.2 kJ/kg[/tex]

State 4:

[tex]P_{4} = 5 bar\\T_4 = 360^0 C\\h_4 = 3188.4 kJ/kg\\S_4 = 7.660 kJ/kg-K\\Q_{12} = h_2 - h_1 = 2799.5-3159.3 = -359 kJ/kg\\Q_{23} = u_3 - h_2 =1146.2-2006.3 = -1454.1 kJ/kg\\Q_{34} = h_4 - h_3 = 3188.4-1196.04 = 1992.36 kJ/kg\\Q_{41} = T(S_1 - S_4) = (360 + 273) (6.9917 - 7.660) = -423.04 kJ/kg[/tex]

Calculate the network done for the cycle

[tex]W_{net} = m( Q_{12} + Q_{23} + Q_{34} + Q_{41})\\W_{net} = 5( -359.8 - 1454.1 + 1992.36 - 423.04)\\W_{net} = -1223 kJ[/tex]

Find the given attachments for complete explanation

Answer:

A. True

Explanation:

When an electromagnetic field wave strikes a conductor, say a wire, it induces an alternating current that is proportional to the wave in the conductor. This is a reversal of generating electromagnetic wave from accelerating a charged particle. This phenomenon is used in radio antena for receiving radio wave signals and also use in medicine for body scanning.

Answer:

The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].

Explanation:

An ideal gas is represented by the following model:

[tex]P\cdot V = \frac{m}{M}\cdot R_{u} \cdot T[/tex]

Where:

[tex]P[/tex] - Pressure, measured in kilopascals.

[tex]V[/tex] - Volume, measured in cubic meters.

[tex]m[/tex] - Mass of the ideal gas, measured in kilograms.

[tex]M[/tex] - Molar mass, measured in kilograms per kilomole.

[tex]T[/tex] - Temperature, measured in Kelvin.

[tex]R_{u}[/tex] - Universal constant of ideal gases, equal to [tex]8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}[/tex]

As tank is rigid and insulated, it means that no volume deformations in tank, heat and mass interactions with surroundings occur during expansion process. Hence, final pressure is less that initial one, volume is doubled (due to equal partitioning) and temperature remains constant. Hence, the following relationship can be derived from model for ideal gases:

[tex]\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}[/tex]

Now, final pressure is cleared:

[tex]P_{2} = P_{1}\cdot \frac{T_{2}}{T_{1}}\cdot \frac{V_{1}}{V_{2}}[/tex]

[tex]P_{2} = (750\,kPa)\cdot 1 \cdot \frac{1}{2}[/tex]

[tex]P_{2} = 375\,kPa[/tex]

The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].

Answer:

positive sides:

negative sides:

Answer:

What is the probability of selecting the 4 of spade or black diamond from a deck of 52 playing cards?

a) 2/52

b) 4/52

c) 3/52

d) 1/5

Explanation:

Answer:

Both technician A and technician B are correct

Explanation:

A planetary gearbox consists of a gearbox with the input shaft and the output shaft that is aligned to each other. It is used to transfer the largest torque in the compact form. A planetary gearbox has a compact size and low weight and it has high power density.

One planetary gear set can provide gear reduction, overdrive, and reverse. Also, most transmissions today use compound (multiple) planetary gears set.

So, both technician A and technician B are correct.

Answer:

Explanation:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

[tex]100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\[/tex]

Carrier frequency 100kHz

Upper side band

[tex]100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz[/tex]

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

[tex]100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz[/tex]

At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

[tex]100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz[/tex]

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

[tex]100k, 99.9k, 99.8k\ \ and \ \99.6kHz[/tex]

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

[tex]100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz[/tex]

A) The frequency Components of the Detector Output are;

80 Hz, 120 Hz and 380 Hz

B) The frequency Components if only the lower sideband is transmitted are; 120 Hz, 220 Hz and 420 Hz

A) We are given the frequency components in the message signal as;

We are told that the carrier signal has a frequency; fc = 100kHz

Thus, the frequency components generated are;

Lower side band:

Upper side band:

We are told that the local oscillator now supplies a sinusoidal wave of frequency 100.02 kHz.

Thus, the output frequencies are;

100.02 kHz - 100.1 kHz = 80 Hz

100.02 kHz - 100.2 kHz = 180 Hz

100.02 kHz - 100.4 kHz = 380 Hz

B) Repeating the analysis assuming only the lower sideband is repeated gives us the frequencies as;

100.02 kHz - 99.9 kHz = 120 Hz

100.02 kHz - 99.8 kHz = 220 Hz

100.02 kHz - 99.6 kHz = 420 Hz

Read more about Message Signals at; https://brainly.com/question/25904079

Answer:

Explanation:

Given that:

V = 12.5m/s

L= 2.70m

b= 0.65m

[tex]T_{ \infty} = 27^0C= 273+27 = 300K[/tex]

[tex]T_s= 127^0C = (127+273)= 400K[/tex]

P = 1atm

Film temperature

[tex]T_f = \frac{T_s + T_{\infty}}{2} \\\\=\frac{400+300}{2} \\\\=350K[/tex]

dynamic viscosity =

[tex]\mu =20.9096\times 10^{-6} m^2/sec[/tex]

density = 0.9946kg/m³

Pr = 0.708564

K= 229.7984 * 10⁻³w/mk

Reynolds number,

[tex]Re = \frac{SUD}{\mu} =\frac{\ SUl}{\mu}[/tex]

[tex]=\frac{0.9946 \times 12.5\times 2.7}{20.9096\times 10^-^6} \\\\Re=1605375.043[/tex]

we have,

[tex]Nu=\frac{hL}{k} =0.037Re^{4/5}Pr^{1/3}\\\\\frac{h\times2.7}{29.79\times 10^-63} =0.037(1605375.043)^{4/5}(0.7085)^{1/3}\\\\h=33.53w/m^2k[/tex]

we have,

heat transfer rate from top plate

[tex]\theta _1 =hA(T_s-T_{\infty})\\\\A=Lb\\\\=2.7*0.655\\\\ \theta_1=33.53*2.7*0.65(127/27)\\\\ \theta_1=5884.51w[/tex]

Answer:

a. T-V and P-V diagram are included

b. State 1: Specific volume = 0.0811753 m³/kg

State 2: Specific volume = 0.0811753 m³/kg

State 3: Specific volume = 0.0804155 m³/kg

c. Z = 51.1

d. Quality for state 1 = 100%

Quality for state 2 = 63.47%

Quality for state 3 = 100%

Explanation:

a. T-V and P-V diagram are included

b. State 1: Water vapor

P₁ = 3.0 MPa = 30 bar

T₁ = 300°C = 573.15

Saturation temperature = 233.86°C Hence the steam is super heated

Specific volume = 0.0811753 m³/kg

State 2:

Constant volume formula is P₁/T₁ = P₂/T₂

Specific volume = 0.0811753 m³/kg

T₂ = 200°C = 473.15

Therefore, P₂ = P₁/T₁ × T₂ = 3×473.15/573.15 = 2.4766 MPa

At T₂ water is mixed water and steam and the [tex]v_f[/tex] = 0.00115651 m³/kg

[tex]v_g[/tex] = 0.127222 m³/kg

State 3:

P₃ = 2.5 MPa

T₃ = 200°C

Isothermal compression P₂V₂ = P₃V₃

V₃ = P₂V₂ ÷ P₃ = 2.4766 × 0.0811753/2.5 = 0.0804155 m³/kg

Specific volume = 0.0804155 m³/kg

2) Compressibility factor is given by the relation;

[tex]Z = \dfrac{PV}{RT} = \dfrac{3\times 10^6 \times 0.0811753 }{8.3145 \times 573.15} = 51.1[/tex]

Z = 51.1

3) Gas quality, x, is given by the relation

[tex]x = \dfrac{Mass_{saturated \, vapor}}{Total \, mass} = \dfrac{v - v_f}{v_g - v_f}[/tex]

Quality at state 1 = Saturated quality = 100%

State 2 Vapor + liquid Quality

Gas quality = (0.0811753 - 0.00115651)/ (0.127222-0.00115651) = 63.47%

State 3: Saturated vapor, quality = 100%.

Answer:

[tex] T_A_C = 6.296 kN [/tex]

[tex] R = 10.06 kN [/tex]

Explanation:

Given:

[tex] T_A_B = 8.7 kN[/tex]

Required:

Find the tension TAC and magnitude R of this downward force.

First calculate [tex] \alpha, \beta, \gamma [/tex]

[tex] \alpha = tan^-^1 =\frac{40}{50} = 38. 36 [/tex]

[tex] \beta = tan^-^1 =\frac{50}{30} = 59.04 [/tex]

[tex] \gamma = 180 - 38.36 - 59.04 = 82.6 [/tex]

To Find tension in AC and magnitude R, use sine rule.

[tex] \frac{sin a}{T_A_C} = \frac{sin b}{T_A_B} = \frac{sin c}{R} [/tex]

Substitute values:

[tex]\frac{sin 38.36}{T_A_C} = \frac{sin 59.04}{8.7} = \frac{82.6}{R}[/tex]

Solve for T_A_C:

[tex] T_A_C = 8.7 * \frac{sin 38.36}{sin 59.04} = [/tex]

[tex] T_A_C = 8.7 * 0.724 = 6.296 kN [/tex]

Solve for R.

[tex] R = 8.7 * \frac{sin 82.6}{sin 59.04} = [/tex]

[tex] R = 8.7 * 1.156 [/tex]

R = 10.06 kN

Tension AC = 6.296kN

Magnitude,R = 10.06 kN

Answer:

The total force resisted by the flange bolts is  163.98 N

Explanation:

Solution

The first step is to find  the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ =  diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of  the pipe at the inlet section

A₂ = π /4 D₂²

D₂ =  diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265  * 5 = 1.963 * V₂

V₂ = 12.8 m/s

Note: Kindly find an attached copy of the part of the solution to the given question below

Answer:

We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...

Explanation:

We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...

Answer:

The answer is VN =37.416 m/s

Explanation:

Recall that:

Pressure (atmospheric) = 100 kPa

So. we solve for the maximum velocity (m/s) to which water can be accelerated by the nozzles

Now,

Pabs =Patm + Pgauge = 800 KN/m²

Thus

PT/9.81 + VT²/2g =PN/9.81  + VN²/2g

Here

Acceleration due to gravity = 9.81 m/s

800/9.81 + 0

= 100/9.81 + VN²/19.62

Here,

9.81 * 2= 19.62

Thus,

VN²/19.62 = 700/9.81

So,

VN² =1400

VN =37.416 m/s

Note: (800 - 100) = 700

Answer:

[tex]V2 = 37.417ms^{-1}[/tex]

Explanation:

Given the following data;

Water enters the turbine nozzles (inlet) = 800kPa = 800000pa.

Nozzle outlets = 100kPa = 100000pa.

Density of water = 1000kg/m³.

We would apply, the Bernoulli equation between the inlet and outlet;

[tex]\frac{P_{1} }{d}+\frac{V1^{2} }{2} +gz_{1} = \frac{P_{2} }{d}+\frac{V2^{2} }{2} +gz_{2}[/tex]

Where, V1 is approximately equal to zero(0).

Z[tex]z_{1} = z_{2}[/tex]

Therefore, to find the maximum velocity, V2;

[tex]V2 = \sqrt{2(\frac{P_{1} }{d}-\frac{P_{2} }{d}) }[/tex]

[tex]V2 = \sqrt{2(\frac{800000}{1000}-\frac{100000}{1000}) }[/tex]

[tex]V2 = \sqrt{2(800-100)}[/tex]

[tex]V2 = \sqrt{2(700)}[/tex]

[tex]V2 = \sqrt{1400}[/tex]

[tex]V2 = 37.417ms^{-1}[/tex]

Hence, the maximum velocity, V2 is 37.417m/s

Answer:

South Pole and South Pole or North Pole and North Pole.

Answer:

a. [tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex] which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

c. The work done by the balloon boundaries = 10.81 MJ

The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

[tex]v_1 = \dfrac{4}{3} \times \pi \times r^3[/tex]

[tex]\therefore v_1 = \dfrac{4}{3} \times \pi \times \left (\dfrac{10}{2} \right )^3 = 523.6 \ m^3[/tex]

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

[tex]\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}} \right )^{n} = \left (\dfrac{T_{1}}{T_{2}} \right )^{\dfrac{n}{n-1}}[/tex]

We have;

[tex]\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times \pi \times \left (\dfrac{D_2}{2} \right )^3}{\dfrac{4}{3} \times \pi \times \left (\dfrac{D_1}{2} \right )^3} \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]

[tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_2} }{ {D_1}} \right )^{3\times n} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex]

[tex]\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2 \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]

[tex]log \left (\dfrac{D_{1}}{ D_{2}}\right ) = -3\times n \times log\left (\dfrac{ \left{D_1} }{ {D_2}} \right )[/tex]

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

[tex]\left (\dfrac{628.32 }{523.6} \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{1}{4}}[/tex]

T₂ = 300.15/0.784 = 382.75 K = 109.6°C

c. [tex]W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}[/tex]

[tex]p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}} \right )^{n} } = \dfrac{100\times 10^3}{ \left (1.2) \right ^{-\dfrac{1}{3} } }[/tex]

p₂ =  100000/0.941 = 106.265 kPa

[tex]W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3 \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J[/tex]

The work done by the balloon boundaries = 10.81 MJ

Work done against atmospheric pressure, Pₐ, is given by the relation;

Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J

The work done on the surrounding atmospheric air = 10.6 MJ

Answer:

4.75m^2

Explanation:

Given:-

- Temperature of hot fluid at inlet:  [tex]T_h_i = 300[/tex] °C

- Temperature of cold fluid at outlet: [tex]T_c_o = 120[/tex] °C

- Temperature of cold fluid at inlet: [tex]T_c_i = 35[/tex] °C

- The overall heat transfer coefficient: U = 1500 W / m^2 K

- The flow rate of cold fluid: m_c = 10,00 kg/ h

- The flow rate of hot fluid: m_h = 5,000 kg/h

Solution:-

- We will evaluate water properties at median temperatures of each fluid using table A-4.

Cold fluid:   Tci = 35°C , Tco = 35°C

                            Tcm = 77.5 °C ≈ 350 K  --- > [tex]C_p_c = 4195 \frac{J}{kg.K}[/tex]

 Hot fluid:     Thi = 300°C , Tho = 150°C ( assumed )

                             Thm = 225 °C ≈ 500 K --- > [tex]C_p_h = 4660 \frac{J}{kg.K}[/tex]

- We will use logarithmic - mean temperature rate equation as follows:

                             [tex]A_s = \frac{q}{U*dT_l_m}[/tex]

Where,

                 A_s : The surface area of heat exchange

                 ΔT_lm: the logarithmic differential mean temperature

                 q: The rate of heat transfer

- Apply the energy balance on cold fluid as follows:

                   [tex]q = m_c * C_p_c * ( T_c_o - T_c_i )\\\\q = \frac{10,000}{3600} * 4195 * ( 120 - 35 )\\\\q = 9.905*10^5 W[/tex]

- Similarly, apply the heat balance on hot fluid and evaluate the outlet temperature ( Tho ) :

                   [tex]T_h_o = T_h_i - \frac{q}{m_h * C_p_h} \\\\T_h_o = 300 - \frac{9.905*10^5}{\frac{5000}{3600} * 4660} \\\\T_h_o = 147 C[/tex]

- We will use the experimental results of counter flow ( unmixed - unmixed ) plotted as figure ( Fig . 11.11 ) of the " The fundamentals to heat transfer" and determine the value of ( P , R , F ).

- So the relations from the figure 11.11 are:

                  [tex]P = \frac{T_c_o - T_c_i}{T_h_i - T_c_i} \\\\P = \frac{120 - 35}{300 - 35} \\\\P = 0.32[/tex]    

                 [tex]R = \frac{T_h_i - T_h_o}{T_c_o - T_c_i} \\\\R = \frac{300 - 147}{120 - 35} \\\\R = 1.8[/tex]

Therefore,         P = 0.32 , R = 1.8 ---- > F ≈ 0.97

- The log-mean temperature ( ΔT_lm - cf ) for counter-flow heat exchange can be determined from the relation:

                        [tex]dT_l_m = \frac{( T_h_i - T_c_o ) - ( T_h_o - T_c_i ) }{Ln ( \frac{( T_h_i - T_c_o )}{( T_h_o - T_c_i )} ) } \\\\dT_l_m = \frac{( 300 - 120 ) - ( 147 - 35 ) }{Ln ( \frac{( 300-120 )}{( 147-35)} ) } \\\\dT_l_m = 143.3 K[/tex]

- The log - mean differential temperature for counter flow is multiplied by the factor of ( F ) to get the standardized value of log - mean differential temperature:

                       [tex]dT_l = F*dT_l_m = 0.97*143.3 = 139 K[/tex]

- The required heat exchange area ( A_s ) can now be calculated:

                     [tex]A_s = \frac{9.905*10^5 }{1500*139} \\\\A_s = 4.75 m^2[/tex]

Answer:

A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt

Explanation:

A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt

attached to the answer is a free plot of the output starting with zero degree for one coil rotating in a uniform magnetic field

B ( wave flux density ) = Bm sinwt  and w = 2[tex]\pi[/tex]f = [tex]\frac{2\pi }{T}[/tex] rad/sec

Answer:

Blindspot Location

Explanation:

Just took the quiz

When you do a vehicle check, you do NOT need to keep an eye on Blind spot locations. The correct option is D.

A blind spot is the area of the road that can't be seen by looking forward through windscreen, or by rear-view and side-view mirrors.

While doing vehicle check, we need to check tire inflation, cleanliness of windows and mirrors along with the functioning indicator lights and headlights.

Blind spot locations does not need to be checked.

Thus, the correct option is D.

Learn more about  Blind spot location

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#SPJ2

Answer:

None of these

Explanation:

There are different types of amplifiers, and each has different characteristics.

Therefore, the correct answer is "None of these "

Answer:

Heat rejected to cold body = 3.81 kJ

Explanation:

Temperature of hot thermal reservoir Th = 1600 K

Temperature of cold thermal reservoir Tc = 400 K

efficiency of the Carnot's engine = 1 - [tex]\frac{Tc}{Th}[/tex]

eff. of the Carnot's engine = 1 - [tex]\frac{400}{600}[/tex]

eff = 1 - 0.25 = 0.75

efficiency of the heat engine = 70% of 0.75 = 0.525

work done by heat engine = 2 kJ

eff. of heat engine is gotten as = W/Q

where W = work done by heat engine

Q = heat rejected by heat engine to lower temperature reservoir

from the equation, we can derive that

heat rejected Q = W/eff = 2/0.525 = 3.81 kJ

Answer:

Settling Velocity (Up)= 2.048*10^-5 m/s

Reynolds number Re = 2.159*10^-3

Explanation:

We proceed as follows;

Diameter of Particle = 0.09 mm = 0.09*10^-3 m

Solid Particle Density = 2002 kg/m3

Solid Fraction, θ= 0.45

Temperature = 26.7°C

Viscosity of water = 0.8509*10^-3 kg/ms

Density of water at 26.7 °C = 996.67 kg/m3

The velocity between the interface, i.e between the suspension and clear water is given by,

U = [ ((nf/ρf)/d)D^3] [18+(1/3)D^3)(1/2)]

D = d[(ρp/ρf)-1)g*(ρf/nf)^2]^(1/3)

D = 2.147

U = 0.0003m/s (n = 4.49)

Up = 0.0003 * (1-0.45)^4.49 = 2.048*10^-5 m/s

Re=0.09*10^-3*2.048*10^-5*996.67/0.0008509 = 2.159*10^-3

Answer:

a) 4160 V

b) 12 kW and 81 kVAR

c)  54 kW and 477 kVAR

Explanation:

1) The phase voltage is given as:

[tex]V_p=\frac{3810.5}{\sqrt{3} }=2200 V[/tex]

The complex power S is given as:

[tex]S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA[/tex]

[tex]where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA[/tex]

The line current I is given as:

[tex]I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)} =100\angle -36.87^o\ A[/tex]

The phase voltage at the sending end is:

[tex]V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV[/tex]

The magnitude of the line voltage at the source end of the line ([tex]V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V[/tex]

b) The Total real and reactive power loss in the line is:

[tex]S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000[/tex]

The real power loss is 12000 W = 12 kW

The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

[tex]S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000[/tex]

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

Answer:

s = v√[2(H - h)/g]

This formula describes the distance from the building that Prof Murthy must be when you release the egg

Explanation:

First, we need to find the time required by the egg to reach the head of Professor. For that purpose, we use 1st equation of motion in vertical form:

Vf = Vi + gt

where,

Vf = Velocity of egg at the time of hitting head of the Professor

Vi = initial velocity of egg = 0 m/s  (Since, egg is initially at rest)

g = acceleration due to gravity

t = time taken by egg to come down

Therefore,

Vf = 0 + gt

t = Vf/g   ----- equation (1)

Now, we use 3rd euation of motion for Vf:

2gs = Vf² - Vi²

where,

s = height dropped = H - h

Therefore,

2g(H - h) = Vf²

Vf = √[2g(H - h)]

Therefore, equation (1) becomes:

t = √[2g(H - h)]/g

t = √[2(H - h)/g]

Now, consider the horizontal motion of professor. So, the minimum distance of professor from building can be found out by finding the distance covered by the professor in time t. Since, the professor is running at constant speed of v m/s. Therefore:

s = vt

s = v√[2(H - h)/g]

This formula describes the distance from the building that Prof Murthy must be when you release the egg

Answer:

a = 0.07m or 70mm

b = 0.205m or 205mm

Explanation:

Given the following data;

Modulus of rigidity, G = 14MPa=14000000Pa.

c = 80mm = 0.08m.

P = 46kN=46000N.

Shearing stress (r) in the rubber shouldn't exceed 1.4MPa=1400000Pa.

Deflection (d) of the plate is to be at least 7mm = 0.007m.

From shearing strain;

[

[tex]Modulus Of Elasticity, E = \frac{d}{a} =\frac{r}{G}[/tex]

Making a the subject formula;

[tex]a = \frac{Gd}{r}[/tex]

Substituting into the above formula;

[tex]a = \frac{14000000*0.007}{1400000}[/tex]

[tex]a = \frac{98000}{1400000}[/tex]

[tex]a = 0.07m or 70mm[/tex]

a = 0.07m or 70mm.

Also, shearing stress;

[tex]r = \frac{P}{2bc}[/tex]

Making b the subject formula;

[tex]b = \frac{P}{2cr}[/tex]

Substituting into the above equation;

[tex]b = \frac{46000}{2*0.08*1400000}[/tex]

[tex]b = \frac{46000}{224000}[/tex]

[tex]b = 0.205m or 205mm[/tex]

b = 0.205m or 205mm

Answer:

See explanation

Explanation:

See the document for the complete FBD and the introductory part of the solution.

Static Balance ( Sum of Forces = 0 ) in all three directions

                 ∑[tex]F_G_X = W - G_x = 0[/tex]

                 [tex]G_X = W = 1500 lb[/tex]

                 ∑[tex]F_G_Y = P - G_Y = 0[/tex]

                 [tex]G_Y = P = -10,800 lb[/tex]

                ∑[tex]F_G_Z = - G_Z = 0[/tex]

Where, ( [tex]G_X, G_Y, G_Z[/tex] ) are internal forces at section ( G ) along the defined coordinate axes.

Static Balance ( Sum of Moments about G = 0 ) in all three directions

              [tex]M_G = r_O_G x F_O[/tex]

Where,

              r_OG: The vector from point O to point G

              F_OG: The force vector at point O

- The vector ( r_OG ) and ( F_OG ) can be written as follows:

              [tex]r_O_G = [ -( 3 + \frac{H}{2} ) i + (\frac{r_o}{12})j - ( \frac{r_o}{12} + \frac{L}{2})k ] \\\\r_O_G = [ -( 6 ) i + (0.25)j - (6)k ] \\[/tex]

              [tex]F_O_G = [ ( W ) i + ( P ) k ]\\\\F_O_G = [ (1500) i - ( 10,800 ) k ] lb[/tex]

- Then perform the cross product of the two vectors ( r_OG ) and ( F_OG ):

     [tex]( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = \left[\begin{array}{ccc}i&j&k\\-6&0.25&-6\\1500&-10,800&0\end{array}\right] \\\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = -( 6*10,800 ) i - ( 6*1500 ) j + [ ( 10,800*6) - ( 0.25*1500) ] k\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = - (64,800)i - (9,000)j + (64,425)k[/tex]

- The internal torque ( T ) and shear force ( V ) that act on slice ( G ) are due to pressure force ( P ) as follows:

             [tex]T = P*[\frac{L}{2}] = (10,800)*(6) = 64,800 lb.ft[/tex]

             [tex]V = P = -10,800 lb[/tex]

- For the state of stress at point "C" we need to determine the the normal stress along x direction ( σ_x ) and planar stress ( τ_xy ) as follows:-

             σ_x = [tex]-\frac{G_x}{A} - \frac{M_G_Y. z*}{I_Y_Y} + \frac{M_G_Z. y*}{I_Z_Z}[/tex]

Where,

          A: The area of pipe cross section

          [tex]A = \pi * [ ( \frac{r_o}{12})^2 - ( \frac{r_i}{12})^2 ] = \pi * [ ( \frac{3}{12})^2 - ( \frac{2.75}{12})^2 ] = 0.03136 ft^2[/tex]

          z*: The distance of point "C" along z-direction from central axis ( x )

          [tex]z*= [\frac{r_i}{12} ] = [\frac{2.75}{12} ] = 0.22916 ft[/tex]

         I_YY: The second area moment of pipe along and about "y" axis:

         [tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]

         y*: The distance of point "C" along y-direction from central axis ( x )

         [tex]y* = 0[/tex]

- The normal stress ( σ_x ) becomes:

          σ_x = [tex][-\frac{1500}{0.03136} - \frac{-9,000*0.22916}{0.00090} + \frac{64,425*0}{0.00090} ] * (\frac{1}{12})^2 = 15.5 ksi[/tex]

- The planar stress is ( τ_xy ) is a contribution of torsion ( T ) and shear force ( V ):

           τ_xy = [tex]- \frac{T.c}{J} + \frac{V.Q}{I.t}[/tex]

Where,

           c: The radial distance from central axis ( x ) and point "C".

           [tex]c = \frac{r_i}{12} = \frac{2.75}{12} = 0.22916 ft[/tex]

          J: The polar moment of inertia of the annular cross section of pipe:

          [tex]J = \frac{\pi }{2}* [ ( \frac{r_o}{12})^4 - ( \frac{r_i}{12})^4 ] = \frac{\pi }{2}* [ ( \frac{3}{12})^4 - ( \frac{2.75}{12})^4 ] = 0.00180 ft^4[/tex]

          Q: The first moment of area for point "C" = semi-circle

          [tex]Q = Y_c*A_c = \frac{4*( r_m)}{3\pi } * \frac{\pi*( r_m)^2 }{2} = \frac{2. ( r_m)^3}{3} \\\\Q = \frac{2. [ ( \frac{r_o}{12})^3 - ( \frac{r_i}{12})^3] }{3} = \frac{2. [ ( \frac{3}{12})^3 - ( \frac{2.75}{12})^3] }{3} = 0.00239ft^3[/tex]

          I: The second area moment of pipe along and about "y" axis:

         [tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]    

         t: The effective thickness of thin walled pipe:

         [tex]t = 2* [ \frac{r_o}{12} - \frac{r_i}{12} ] = 2* [ \frac{3}{12} - \frac{2.75}{12} ] = 0.04166 ft[/tex]

- The planar stress is ( τ_xy ) becomes:

        τ_xy =  [tex][ - \frac{-64,800*0.22916}{0.0018} + \frac{-10,800*0.00239}{0.0009*0.04166} ] * [ \frac{1}{12}]^2 = 52.4 ksi[/tex]

- The principal stresses at point "C" can be determined from the following formula:-

       σ_x = 15.55 ksi,  σ_y = 0 ksi , τ_xy = 52.4 ksi

       σ_1 =[tex]\frac{sigma_x+sigma_y}{2} + \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]

       σ_2 = [tex]\frac{sigma_x+sigma_y}{2} - \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]

        σ_1 = [tex]\frac{15.55+0}{2} + \sqrt{(\frac{15.55+0}{2})^2 + (52.4)^2 } = 60.75 ksi[/tex]

        σ_2 =[tex]-\sqrt{\left(\frac{15.55+0}{2}\right)^2\:+\:\left(52.4\right)^2\:}+\frac{15.55+0}{2} = -45.20 ksi[/tex]

- The angle of maximum plane stress ( θ ):

       θ = [tex]0.5*arctan ( \frac{tow_x_y}{\frac{sigma_x-sigma_y}{2} } )= 0.5*arctan*( \frac{52.4}{7.8} ) = 40.8 deg[/tex]

Note: The plane stresses at point D are evaluated using the following procedure given above. Due to 5,000 character limit at Brainly, i'm unable to post here.

Answer:

Roll force, F = 5.6 MN

Explanation:

Sheet width, b = 0.5 m

Roll contact length, [tex]l_{p} = 40 mm[/tex]

Strength coefficient, [tex]\sigma_{0} = 200 MPa[/tex]

Thickness, h = 11 mm

Since the sheet of metal is cold worked by 40%, the reduction in thickness will be:

Δh = 40% * 11 = 0.4 * 11 = 4.4 mm

Strain, e = (Δh)/h

e = 4.4/11 = 0.4

The roll force is calculated by the formula:

[tex]F = \sigma_{f} l_{p} b[/tex]

[tex]\sigma_{f} = \sigma_{0} (e+1)\\\sigma_{f} = 200 (0.4+1)\\\sigma_{f} = 200 *1.4\\\sigma_{f} = 280 MPa[/tex]

Substituting the value of [tex]\sigma_{f}[/tex], [tex]l_{p}[/tex], and b into the formula for the roll force:

[tex]F = \sigma_{f} l_{p} b\\F = 280 * 0.04 * 0.5\\F = 5.6 MN[/tex]