Answer:
He probably tripped the wiring, when metal hits the electricity it creates a reaction that burns the wiring.
Explanation:
Decompose the signal (1+0.1 cos5t) cos100t into a linear combination of sinusoidal functions, and find the amplitude, frequency, and phase of each component. Hint: use the identity for cosacosb.
Answer:
amplitudes : 1 , 0.05, 0.05
frequencies : 50/[tex]\pi[/tex], 105/[tex]2\pi[/tex], 95/2[tex]\pi[/tex]
phases : [tex]\pi /2 , \pi /2 , \pi /2[/tex]
Explanation:
signal s(t) = ( 1 + 0.1 cos 5t )cos 100t
signal s(t) = cos100t + 0.1cos100tcos5t . using the identity for cosacosb
s(t) = cos100t + [tex]\frac{0.1}{2}[/tex] [cos(100+5)t + cos (100-5)t]
s(t) = cos 100t + 0.05cos ( 100+5)t + 0.05cos (100-5)t
= cos100t + 0.05cos(105)t + 0.05cos 95t
= cos 2 [tex](\frac{50}{\pi } )t + 0.05cos2 (\frac{105}{2\pi } )t + 0.05cos2 (\frac{95}{2\pi } )t[/tex] [ ∵cos (∅) = sin(/2 +∅ ]
= sin ( 2 [tex](\frac{50}{\pi } ) t[/tex] + /2 ) + 0.05sin ( 2 [tex](\frac{105}{2\pi } ) t + /2 )[/tex] + 0.05sin ( 2 [tex](\frac{95}{2\pi } )t + /2[/tex] )
attached is the remaining part of the solution
A tubular reactor has been sized to obtain 98% conversion and to process 0.03 m^3/s. The reaction is a first-order irreversible isomerization. The reactor is 3 m long, with a cross- sectional area of 25 dm^2. After being built, a pulse tracer test on the reactor gave the following data: tm = 10 s and σ2 = 65 s2. What conversion can be expected in the real reactor?
Answer:
The conversion in the real reactor is = 88%
Explanation:
conversion = 98% = 0.98
process rate = 0.03 m^3/s
length of reactor = 3 m
cross sectional area of reactor = 25 dm^2
pulse tracer test results on the reactor :
mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2
note: space time (t) =
t = [tex]\frac{A*L}{Vo}[/tex] Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor
therefore (t) = [tex]\frac{25*3*10^{-2} }{0.03}[/tex] = 25 s
since the reaction is in first order
X = 1 - [tex]e^{-kt}[/tex]
[tex]e^{-kt}[/tex] = 1 - X
kt = In [tex]\frac{1}{1-X}[/tex]
k = In [tex]\frac{1}{1-X}[/tex] / t
X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then
K = 0.156 [tex]s^{-1}[/tex]
Calculating Da for a closed vessel
; Da = tk
= 25 * 0.156 = 3.9
calculate Peclet number Per using this equation
0.65 = [tex]\frac{2}{Per} - \frac{2}{Per^2} ( 1 - e^{-per})[/tex]
therefore
[tex]\frac{2}{Per} - \frac{2}{Per^2} (1 - e^{-per}) - 0.65 = 0[/tex]
solving the Non-linear equation above( Per = 1.5 )
Attached is the Remaining part of the solution
The velocity field of a flow is given by V = 2x2 ti +[4y(t - 1) + 2x2 t]j m/s, where x and y are in meters and t is in seconds. For fluid particles on the x-axis, determine the speed and direction of flow
Answer:
Explanation:
The value of a will be zero as it is provided that the particle is on the x-axis.
Calculate the velocity of particles along x-axis.
[tex]{\bf{V}} = 2{x^2}t{\bf{\hat i}} + [4y(t - 1) + 2{x^2}t]{\bf{\hat j}}{\rm{ m/s}}[/tex]
Substitute 0 for y.
[tex]\begin{array}{c}\\{\bf{V}} = 2{x^2}t{\bf{\hat i}} + \left( {4\left( 0 \right)\left( {t - 1} \right) + 2{x^2}t} \right){\bf{\hat j}}{\rm{ m/s}}\\\\ = 2{x^2}t{\bf{\hat i}} + 2{x^2}t{\bf{\hat j}}{\rm{ m/s}}\\\end{array}[/tex]
Here,
[tex]A = 2{x^2}t \ \ and\ \ B = 2{x^2}t[/tex]
Calculate the magnitude of vector V .
[tex].\left| {\bf{V}} \right| = \sqrt {{A^2} + {B^2}}[/tex]
Substitute
[tex]2{x^2}t \ \ for\ A\ and\ 2{x^2}t \ \ for \ B.[/tex]
[tex]\begin{array}{c}\\\left| {\bf{V}} \right| = \sqrt {{{\left( {2{x^2}t} \right)}^2} + {{\left( {2{x^2}t} \right)}^2}} \\\\ = \left( {2\sqrt 2 } \right){x^2}t\\\end{array}[/tex]
The velocity of the fluid particles on the x-axis is [tex]\left( {2\sqrt 2 } \right){x^2}t{\rm{ m/s}}[/tex]
Calculate the direction of flow.
[tex]\theta = {\tan ^{ - 1}}\left( {\frac{B}{A}} )[/tex]
Here, θ is the flow from positive x-axis in a counterclockwise direction.
Substitute [tex]2{x^2}t[/tex] as A and [tex]2{x^2}t[/tex] as B.
[tex]\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{2{x^2}t}}{{2{x^2}t}}} \right)\\\\ = {\tan ^{ - 1}}\left( 1 \right)\\\\ = 45^\circ \\\end{array}[/tex]
The direction of flow is [tex]45^\circ[/tex] from the positive x-axis.
Mathematical modeling aids in technological design by simulating how.
1. A solution should be designed
2. A proposed system might behave
3. Physical models should be built
4. Designs should be used
Mathematical modeling aids in technological design by simulating how proposed system might behave. The correct option is 2.
What is mathematical modelling?Mathematical modelling describes a real world problem in mathematical terms or in the form of equations. This makes an engineer to discover new features about the problem and designer to alter his design for better function and output.
Mathematical models allow engineers and designers to understand how the proposed model and actual prototype will be produced.
Thus, the correct option is 2.
Learn more about mathematical modelling
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Steam is contained in a closed rigid container which has a volume of 2 initially the the pressure and the temperature is the remeraturedrops as a result of heat transfer to the surroundings. Determine
a) the temperature at which condensation first occurs, in °C,
b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.
c) What is the volume, in m3, occupied by saturated liquid at the final state?
The given question is incomplete. The complete question is as follows.
Steam is contained in a closed rigid container with a volume of 1 m3. Initially, the pressure and temperature of the steam are 10 bar and 500°C, respectively. The temperature drops as a result of heat transfer to the surroundings. Determine
(a) the temperature at which condensation first occurs, in [tex]^{o}C[/tex],
(b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.
(c) What is the volume, in [tex]m^{3}[/tex], occupied by saturated liquid at the final state?
Explanation:
Using the property tables
[tex]T_{1} = 500^{o}C[/tex], [tex]P_{1}[/tex] = 10 bar
[tex]v_{1} = 0.354 m^{3}/kg[/tex]
(a) During the process, specific volume remains constant.
[tex]v_{g} = v_{1} = 0.354 m^{3}/kg[/tex]
T = [tex](150 - 160)^{o}C[/tex]
Using inter-polation we get,
T = [tex]154.71^{o}C[/tex]
The temperature at which condensation first occurs is [tex]154.71^{o}C[/tex].
(b) When the system will reach at state 3 according to the table at 0.5 bar then
[tex]v_{f} = 1.030 \times 10^{-3} m^{3}/kg[/tex]
[tex]v_{g} = 3.24 m^{3} kg[/tex]
Let us assume "x" be the gravity if stream
[tex]v_{1} = v_{f} + x_{3}(v_{g} - v_{f})[/tex]
[tex]x_{3} = \frac{v_{1} - v_{f}}{v_{g} - v_{f}}[/tex]
= [tex]\frac{0.3540 - 0.00103}{3.240 - 0.00103}[/tex]
= 0.109
At state 3, the fraction of total mass condensed is as follows.
[tex](1 - x_{5})[/tex] = 1 - 0.109
= 0.891
The fraction of the total mass that has condensed when the pressure reaches 0.5 bar is 0.891.
(c) Hence, total mass of the system is calculated as follows.
m = [tex]\frac{v}{v_{1}}[/tex]
= [tex]\frac{1}{0.354}[/tex]
= 2.825 kg
Therefore, at final state the total volume occupied by saturated liquid is as follows.
[tex]v_{ws} = m \times v_{f}[/tex]
= [tex]2.825 \times 0.00103[/tex]
= [tex]2.9 \times 10^{-3} m^{3}[/tex]
The volume occupied by saturated liquid at the final state is [tex]2.9 \times 10^{-3} m^{3}[/tex].
A nozzle receives an ideal gas flow with a velocity of 25 m/s, and the exit at 100 kPa, 300 K velocity is 250 m/s. Determine the inlet temperature if the gas is argon, helium, or nitrogen.
Given Information:
Inlet velocity = Vin = 25 m/s
Exit velocity = Vout = 250 m/s
Exit Temperature = Tout = 300K
Exit Pressure = Pout = 100 kPa
Required Information:
Inlet Temperature of argon = ?
Inlet Temperature of helium = ?
Inlet Temperature of nitrogen = ?
Answer:
Inlet Temperature of argon = 360K
Inlet Temperature of helium = 306K
Inlet Temperature of nitrogen = 330K
Explanation:
Recall that the energy equation is given by
[tex]$ C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2) $[/tex]
Where Cp is the specific heat constant of the gas.
Re-arranging the equation for inlet temperature
[tex]$ T_{in} = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p} + T_{out}$[/tex]
For Argon Gas:
The specific heat constant of argon is given by (from ideal gas properties table)
[tex]C_p = 520 \:\: J/kg.K[/tex]
So, the inlet temperature of argon is
[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520} + 300$[/tex]
[tex]$ T_{in} = \frac{1}{2} \times 119 + 300$[/tex]
[tex]$ T_{in} = 360K $[/tex]
For Helium Gas:
The specific heat constant of helium is given by (from ideal gas properties table)
[tex]C_p = 5193 \:\: J/kg.K[/tex]
So, the inlet temperature of helium is
[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193} + 300$[/tex]
[tex]$ T_{in} = \frac{1}{2} \times 12 + 300$[/tex]
[tex]$ T_{in} = 306K $[/tex]
For Nitrogen Gas:
The specific heat constant of nitrogen is given by (from ideal gas properties table)
[tex]C_p = 1039 \:\: J/kg.K[/tex]
So, the inlet temperature of nitrogen is
[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039} + 300$[/tex]
[tex]$ T_{in} = \frac{1}{2} \times 60 + 300$[/tex]
[tex]$ T_{in} = 330K $[/tex]
Note: Answers are rounded to the nearest whole numbers.
An aggregate blend is composed of 65% coarse aggregate by weight (Sp. Cr. 2.635), 36% fine aggregate (Sp. Gr. 2.710), and 5% filler (Sp. Cr. 2.748). The compacted specimen contains 6% asphalt binder (Sp. Gr. 1.088) by weight of total mix, and has a bulk density of 143.9 lb/ft^3 Ignoring absorption, compute the percent voids in total mix, percent voids in mineral aggregate, and the percent voids filled with asphalt.
Answer:
Explanation:
From the given information:
Addition of all the materials = 65+ 36+ 5 +6 = 112 which is higher than 100 percentage; SO we need to find;
The actual percentage of each material which can be determined as follows:
Percentage of the coarse aggregate will be = 65 × 112/100
= 72.80%
Percentage of the Fine aggregate will be = 36 × 112/100
= 40.32%
Percentage of the filler will be = 5 × 112/100
= 5.6%
Percentage of the asphalt binder will be = 6 × 112/100
= 6.72 %
So; the theoretical specific gravity (Gt) of the mixture can be calculated as follows:
Gt = 100/( 72.80/2.635 + 40.32/2.710 + 5.6/2.748 + 6.72/1.088)
Gt = 100/( 27.628 + 14.878 + 2.039 + 6.177)
Gt = 100/ (50.722)
Gt =1.972
Also;Given that the bulk density = 143.9 lb/ft³
LIke-wsie ; as we know that unit weight of water is =62.43lb/cu.ft
Hence, the bulk specific gravity of the mix (Gm) = 143.9/62.43
=2.305
The percentage of air void = (Gt -Gm )× 100/ Gt
= (1.972 - 2.305) × 100/ 1.972
= -16.89%
The percentage of the asphalt binder is =(6.72/1.088*100)/(72.80+40.32+5.6+6.72)/2.305)
= 617.647/54.42
= 11.35%
Thus; the percentage voids in mineral aggregate = -16.89% + 11.35%
the percentage voids in mineral aggregate = -5.45%
The percent voids filled with asphalt. = 100 × 11.35/-5.45
The percent voids filled with asphalt = - 208.26 %
A certain heat pump produces 200 kW of heating for a 293 K heated zone while only using 75 kW of power and a heat source at 273 K. Calculate the COP of this device as well as the theoretical maximum COP
Answer:
COP(heat pump) = 2.66
COP(Theoretical maximum) = 14.65
Explanation:
Given:
Q(h) = 200 KW
W = 75 KW
Temperature (T1) = 293 K
Temperature (T2) = 273 K
Find:
COP(heat pump)
COP(Theoretical maximum)
Computation:
COP(heat pump) = Q(h) / W
COP(heat pump) = 200 / 75
COP(heat pump) = 2.66
COP(Theoretical maximum) = T1 / (T1 - T2)
COP(Theoretical maximum) = 293 / (293 - 273)
COP(Theoretical maximum) = 293 / 20
COP(Theoretical maximum) = 14.65
A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the rotor at synchronous speed. (b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses. (c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.
Given Information:
Frequency = f = 60 Hz
Complex rated power = G = 100 MVA
Intertia constant = H = 8 MJ/MVA
Mechanical power = Pmech = 80 MW
Electrical power = Pelec = 50 MW
Number of poles = P = 4
No. of cycles = 10
Required Information:
(a) stored energy = ?
(b) rotor acceleration = ?
(c) change in torque angle = ?
(c) rotor speed = ?
Answer:
(a) stored energy = 800 Mj
(b) rotor acceleration = 337.46 elec deg/s²
(c) change in torque angle (in elec deg) = 6.75 elec deg
(c) change in torque angle (in rmp/s) = 28.12 rpm/s
(c) rotor speed = 1505.62 rpm
Explanation:
(a) Find the stored energy in the rotor at synchronous speed.
The stored energy is given by
[tex]E = G \times H[/tex]
Where G represents complex rated power and H is the inertia constant of turbo-generator.
[tex]E = 100 \times 8 \\\\E = 800 \: MJ[/tex]
(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.
The rotor acceleration is given by
[tex]$ P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2} $[/tex]
Where M is given by
[tex]$ M = \frac{E}{180 \times f} $[/tex]
[tex]$ M = \frac{800}{180 \times 50} $[/tex]
[tex]M = 0.0889 \: MJ \cdot s/ elec \: \: deg[/tex]
So, the rotor acceleration is
[tex]$ P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2} $[/tex]
[tex]$ 30 = 0.0889 \frac{d^2 \delta}{dt^2} $[/tex]
[tex]$ \frac{d^2 \delta}{dt^2} = \frac{30}{0.0889} $[/tex]
[tex]$ \frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2 $[/tex]
(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.
The change in torque angle is given by
[tex]$ \Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2 $[/tex]
Where t is given by
[tex]1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2 \\\\t = 0.2 \: sec[/tex]
So,
[tex]$ \Delta \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2 $[/tex]
[tex]$ \Delta \delta = 6.75 \: elec \: deg[/tex]
The change in torque in rpm/s is given by
[tex]$ \Delta \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ } $[/tex]
[tex]$ \Delta \delta =28.12 \: \: rpm/s $[/tex]
The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by
[tex]$ Rotor \: speed = \frac{120 \cdot f}{P} + (\Delta \delta)\cdot t $[/tex]
Where P is the number of poles of the turbo-generator.
[tex]$ Rotor \: speed = \frac{120 \cdot 50}{4} + (28.12)\cdot 0.2 $[/tex]
[tex]$ Rotor \: speed = 1500 + 5.62 $[/tex]
[tex]$ Rotor \: speed = 1505.62 \:\: rpm[/tex]
You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19 . The density of lead is 11.36 . How many atoms of lead are required
Answer:
To answer this question we assumed that the area units and the thickness units are given in inches.
The number of atoms of lead required is 1.73x10²³.
Explanation:
To find the number of atoms of lead we need to find first the volume of the plate:
[tex] V = A*t [/tex]
Where:
A: is the surface area = 160
t: is the thickness = 0.002
Assuming that the units given above are in inches we proceed to calculate the volume:
[tex]V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}[/tex]
Now, using the density we can find the mass:
[tex] m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g [/tex]
Finally, with the Avogadros number ([tex]N_{A}[/tex]) and with the atomic mass (A) we can find the number of atoms (N):
[tex] N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms [/tex]
Hence, the number of atoms of lead required is 1.73x10²³.
I hope it helps you!
A very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one viscosity is twice the other. When the plate is pulled at a velocity of 0.3 m/s, the resulting force on one square meter of plate due to the viscous shear on both sides is 29 N. Assuming viscous flow and neglecting all end effects calculate the viscosities of the oils.
Answer:
The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s
Explanation:
Assuming the two oils are Newtonian fluids.
From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.
τ = μ (∂v/∂y)
There are oils above and below the plate, so we can write this expression for the both cases.
τ₁ = μ₁ (∂v/∂y)
τ₂ = μ₂ (∂v/∂y)
dv = 0.3 m/s
dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)
τ₁ = μ₁ (0.3/0.03) = 10μ₁
τ₂ = μ₂ (0.3/0.03) = 10μ₂
But the shear stress on the plate is given as 29 N per square meter.
τ = 29 N/m²
But this stress is a sum of stress due to both shear stress above and below the plate
τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29
But it is also given that one viscosity is twice the other
μ₁ = 2μ₂
10μ₁ + 10μ₂ = 29
10(2μ₂) + 10μ₂ = 29
30μ₂ = 29
μ₂ = (29/30) = 0.967 Pa.s
μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s
Hope this Helps!!!
A horizontal turbine takes in steam with an enthalpy of h = 2.80 MJ/kg at 45 m/s. A steam-water mixture exits the turbine with an enthalpy of h = 1.55 MJ/kg at 20 m/s. If the heat loss to the surroundings from the turbine is 300 J/s, determine the power the fluid supplies to the turbine. The mass flow rate is 0.85 kg/s.
Answer:
The power that fluid supplies to the turbine is 1752.825 kilowatts.
Explanation:
A turbine is a device that works usually at steady state. Given that heat losses exists and changes in kinetic energy are not negligible, the following expression allows us to determine the power supplied by the fluid to the turbine by the First Law of Thermodynamics:
[tex]-\dot Q_{loss} - \dot W_{out} + \dot m \cdot \left[(h_{in}-h_{out}) + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right] = 0[/tex]
Output power is cleared:
[tex]\dot W_{out} = -\dot Q_{loss} + \dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right][/tex]
If [tex]\dot Q_{loss} = 0.3\,kW[/tex], [tex]\dot m = 0.85\,\frac{kg}{s}[/tex], [tex]h_{in} = 2800\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 1550\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 45\,\frac{m}{s}[/tex] and [tex]v_{out} = 20\,\frac{m}{s}[/tex], then:
[tex]\dot W_{out} = -0.3\,kW + \left(0.85\,\frac{kg}{s} \right)\cdot \left\{\left(2800\,\frac{kJ}{kg}-1550\,\frac{kJ}{kg} \right)+\frac{1}{2}\cdot \left[\left(45\,\frac{m}{s} \right)^{2}-\left(20\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]
[tex]\dot W_{out} = 1752.825\,kW[/tex]
The power that fluid supplies to the turbine is 1752.825 kilowatts.
An airplane flies from San Francisco to Washington DC at an air speed of 800 km/hr. Assume Washington is due east of San Francisco at a distance of 6000 km. Use a Cartesian system of coordinates centered at San Francisco with Washington in the positive x-direction. At cruising altitude, there is a cross wind blowing from north to south of 100 km/hr.
Required:
a. What must be the direction of flight for the plane to actually arrive in Washington?
b. What is the speed in the San Francisco to Washington direction?
c. How long does it take to cover this distance?
d. What is the time difference compared to no crosswind?
Answer:
A.) 7.13 degree north east
B.) 806.23 km/h
C.) 7.44 hours
D.) 0.06 hours
Explanation:
Assume Washington is due east of San Francisco and Francisco with Washington in the positive x-direction
Also, the cross wind is blowing from north to south of 100 km/hr in y coordinate direction.
A.) Using Cartesian system of coordinates, the direction of flight for the plane to actually arrive in Washington can be calculated by using the formula
Tan Ø = y/x
Substitute y = 100 km/h and x = 800km/h
Tan Ø = 100/800
Tan Ø = 0.125
Ø = Tan^-1(0. 125)
Ø = 7.13 degrees north east.
Therefore, the direction of flight for the plane to actually arrive in Washington is 7.13 degree north east
B.) The speed in the San Francisco to Washington direction can be achieved by using pythagorean theorem
Speed = sqrt ( 800^2 + 100^2)
Speed = sqrt (650000)
Speed = 806.23 km/h
C.) Let us use the speed formula
Speed = distance / time
Substitute the speed and distance into the formula
806.23 = 6000/ time
Make Time the subject of formula
Time = 6000/806.23
Time = 7.44 hours
D.) If there is no cross wind,
Time = 6000/800
Time = 7.5 hour
Time difference = 7.5 - 7.44
Time difference = 0.06 hours
You are installing network cabling and require a cable solution that provides the best resistance to EMI.Which of the following will you choose for this installation?
Answer: b. STP
Explanation:
Twisted Pair Cables are best used for network cabling but are usually prone to EMI (Electromagnetic Interference) which affects the electrical circuit negatively.
The best way to negate this effect is to use Shielding which will help the cable continue to function normally. This is where the Shielded Twisted Pair (STP) cable comes in.
As the name implies, it comes with a shield and that shield is made out of metal which can enable it conduct the Electromagnetic Interference to the ground. The Shielding however makes it more expensive and in need of more care during installation.
9. A Co has 500,000 total shares outstanding and each share is priced at 20$. B Co has 300,000 total shares outstanding and each share is priced at 40$. You have 100 shares in A Co and 200 shares in B Cos. After consolidation how many new shares you will own in consolidated AB Co?
Answer:
In consolidated AB Co 300 shares.
Explanation:
Consolidation is a process in which two different organizations are united. In this question A Co and B Co are consolidated and a new Co names AB Co is formed. The shares of both the companies will be combined and their total share capital will be increased.
The benefit of using the generalized enthalpy departure chart prepared by using PR and TR as the parameters instead of P and T is that the single chart can be used for all gases instead of a single particular gas.
a. True
b. False
A 2.75-kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate (E = 200 GPa, ν = 0.30). Determine the resulting change in (a) the 50-mm gage length, (b) the width of portion AB of the test coupon, (c) the thickness of portion AB, (d) the cross- sectional area of portion AB.
Answer:
I have attached the diagram for this question below. Consult it for better understanding.
Find the cross sectional area AB:
A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m
Forces is given by:
F = 2.75 × 10³ N
Horizontal Stress can be found by:
σ (x) = F/A
σ (x) = 2.75 × 10³ / 19.2 × 10⁻⁶m
σ (x) = 143.23 × 10⁶ Pa
Horizontal Strain can be found by:
ε (x) = σ (x)/ E
ε (x) = 143.23 × 10⁶ / 200 × 10⁹
ε (x) = 716.15 × 10⁻⁶
Find Vertical Strain:
ε (y) = -v · ε (y)
ε (y) = -(0.3)(716.15 × 10⁻⁶)
ε (y) = -214.84 × 10⁻⁶
PART (a)For L = 0.05m
Change (x) = L · ε (x)
Change (x) = 35.808 × 10⁻⁶m
PART (b)
For W = 0.012m
Change (y) = W · ε (y)
Change (y) = -2.5781 × 10⁻⁶m
PART(c)
For t= 0.0016m
Change (z) = t · ε (z)
where
ε (z) = ε (y) ,so
Change (z) = t · ε (y)
Change (z) = -343.74 × 10⁻⁹m
PART (d)
A = A(final) - A(initial)
A = -8.25 × 10⁻⁹m²
(Consult second picture given below for understanding how to calculate area)
The resulting change in the 50-mm gauge length; the width of portion AB of the test coupon; the thickness of portion AB; the cross- sectional area of portion AB are respectively; Δx = 35.808 × 10⁻⁶ m; Δy = -2.5781 × 10⁻⁶m; Δ_z = -343.74 × 10⁻⁹m; A = -8.25 × 10⁻⁹m²
What is the stress and strain in the plate?Let us first find the cross sectional area of AB from the image attached;
A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m
We are given;
Tensile Load; F = 2.75 kN = 2.75 × 10³ N
Horizontal Stress is calculated from the formula;
σₓ = F/A
σₓ = (2.75 × 10³)/(19.2 × 10⁻⁶)m
σₓ = 143.23 × 10⁶ Pa
Horizontal Strain is calculated from;
εₓ = σₓ/E
We are given E = 200 GPa = 200 × 10⁹ Pa
Thus;
εₓ = (143.23 × 10⁶)/(200 × 10⁹)
εₓ = 716.15 × 10⁻⁶
Formula for Vertical Strain is;
ε_y = -ν * εₓ
We are given ν = 0.30. Thus;
ε_y = -(0.3) * (716.15 × 10⁻⁶)
ε_y = -214.84 × 10⁻⁶
A) We are given;
Gauge Length; L = 0.05m
Change in gauge length is gotten from;
Δx = L * εₓ
Δx = 0.05 × 716.15 × 10⁻⁶
Δx = 35.808 × 10⁻⁶ m
B) From the attached diagram, the width is;
W = 0.012m
Change in width is;
Δy = W * ε_y
Δy = 0.012 * -214.84 × 10⁻⁶
Δy = -2.5781 × 10⁻⁶m
C) We are given;
Thickness of plate; t = 1.6 mm = 0.0016m
Change in thickness;
Δ_z = t * ε_z
where;
ε_z = ε_y
Thus;
Δ_z = t * ε_y
Δ_z = 0.0016 * -214.84 × 10⁻⁶
Δ_z = -343.74 × 10⁻⁹m
D) The change in cross sectional area is gotten from;
ΔA = A_final - A_initial
From calculating the areas, we have;
A = -8.25 × 10⁻⁹ m²
Read more about stress and strain in steel plates at; https://brainly.com/question/1591712
Time, budget, and safety are almost always considered to be
1. Efficiency
2. Constraints
3. Trade-offs
4. Criteria
Answer:
The answer is option # 2. (Constraints).
Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a particular location where flow and heat transfer are fully developed, the wall temperature is 370K. Determine:
a. The oil mean temperature.
b. The centerline temperature.
c. The axial gradient of the mean temperature.
d. The heat transfer coefficient.
Answer:
(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K
Explanation:
Solution
Given that:
The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s
Diameter of the tube, D = 1cm =0.01 m
Electrical heat rate, q =76 W/m
Wall Temperature, Ts = 370 K
Now,
From the properties table of engine oil we can deduce as follows:
thermal conductivity, k =0.139 W/m .K
Density, ρ = 854 kg/m³
Specific heat, cp = 2120 J/kg.K
(a) Thus
The wall heat flux is given as follows:
qs = q/πD
=76/π *0.01
= 2419.16 W/m²
Now
The oil mean temperature is given as follows:
Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)
Tb =370 - 11/24 * (2419.16 * 0.005/0.139)
Tb = 330.12 K
(b) The center line temperature is given below:
Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)
Tc =304.73 K
(c) The flow velocity is given as follows:
V = m/ρ (πR²)
Now,
The The axial gradient of the mean temperature is given below:
dTb/dx = 2 *qs/ρ *V*cp * R
=2 *qs/ρ*[m/ρ (πR²) *cp * R
=2 *qs/[m/(πR)*cp
dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120
dTb/dx = 19.81 K/m
(d) The heat transfer coefficient is given below:
h =48/11 (k/D)
=48/11 (0.139/0.01)
h =60.65 W/m². K
We need to design a logic circuit for interchanging two logic signals. The system has three inputs I1I1, I2I2, and SS as well as two outputs O1 and O2. When S is low, we should have O1 = I1 and O2 = I2. On the other hand, when S is high,we should have O1 = I2 and O2 =I1. Thus, S acts as the control input for a reversing switch. Use Karnaugh maps to obtain a minimal SOP(sum ofproduct) design. Draw the circuit.
Explanation:
Inputs and Outputs:
There are 3 inputs = I₁, I₂, and S
There are 2 outputs = O₁ and O₂
The given problem is solved in three major steps:
Step 1: Construct the Truth Table
Step 2: Obtain the logic equations using Karnaugh map
Step 3: Draw the logic circuit
Step 1: Construct the Truth Table
The given logic is
When S = 0 then O₁ = I₁ and O₂ = I₂
When S = 1 then O₁ = I₂ and O₂ = I₁
I₁ | I₂ | S | O₁ | O₂
0 | 0 | 0 | 0 | 0
0 | 0 | 1 | 0 | 0
0 | 1 | 0 | 0 | 1
0 | 1 | 1 | 1 | 0
1 | 0 | 0 | 1 | 0
1 | 0 | 1 | 0 | 1
1 | 1 | 0 | 1 | 1
1 | 1 | 1 | 1 | 1
Step 2: Obtain the logic equations using Karnaugh map
Please refer to the attached diagram where Karnaugh map is set up.
The minimal SOP representation for output O₁
[tex]$ O_1 = I_1 \bar{S} + I_2 S $[/tex]
The minimal SOP representation for output O₂
[tex]$ O_2 = I_2 \bar{S} + I_1 S $[/tex]
Step 3: Draw the logic circuit
Please refer to the attached diagram where the circuit has been drawn.
Given the circuit at the right in which the following values are used: R1 = 20 kΩ, R2 = 12 kΩ, C = 10 µ F, and ε = 25 V. You close the switch at t = 0. Find (a) the current in R1 and R2 at t=0, (b) the voltage across R1 after a long time. (Careful with this one.)
Answer:
a.) I = 7.8 × 10^-4 A
b.) V(20) = 9.3 × 10^-43 V
Explanation:
Given that the
R1 = 20 kΩ,
R2 = 12 kΩ,
C = 10 µ F, and
ε = 25 V.
R1 and R2 are in series with each other.
Let us first find the equivalent resistance R
R = R1 + R2
R = 20 + 12 = 32 kΩ
At t = 0, V = 25v
From ohms law, V = IR
Make current I the subject of formula
I = V/R
I = 25/32 × 10^3
I = 7.8 × 10^-4 A
b.) The voltage across R1 after a long time can be achieved by using the formula
V(t) = Voe^- (t/RC)
V(t) = 25e^- t/20000 × 10×10^-6
V(t) = 25e^- t/0.2
After a very long time. Let assume t = 20s. Then
V(20) = 25e^- 20/0.2
V(20) = 25e^-100
V(20) = 25 × 3.72 × 10^-44
V(20) = 9.3 × 10^-43 V
An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s². The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s². (Hint: The police will not go against the law.) a) Find the total time required for the police car to overtake the automobile. (12 marks) b) Find the total distance travelled by the police car while overtaking the automobile. (2 marks) c) Find the speed of the police car at the time it overtakes the automobile. (2 marks) d) Find the speed of the automobile at the time it was overtaken by the police car. (2 marks)
Answer:
A.) Time = 13.75 seconds
B.) Total distance = 339 m
C.) V = 11.18 m/s
D.) V = 10.2 m/s
Explanation: Given that the automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone.
Then,
Initial velocity U of the motorist = 15.65m/s
acceleration a = - 3.05 m/s^2
Initial velocity u of the police man = 11.18 m/s
Acceleration a = 1.96 m/s^2
The police will overtake at distance S as the motorist decelerate and come to rest.
Where V = 0 and a = negative
While the police accelerate.
Using 2nd equation of motion for the motorist and the police
S = ut + 1/2at^2
Since the distance S covered will be the same, so
15.65t - 1/2×3.05t^2 = 11.18t +1/2×1.96t^2
Solve for t by collecting the like terms
15.56t - 1.525t^2 = 11.18t + 0.98t^2
15.56t - 11.18t = 0.98t^2 + 1.525t^2
4.38t = 2.505t^2
t = 4.38/2.505
t = 1.75 seconds approximately
But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit.
Therefore, the total time required for the police car to overtake the automobile will be:
12 + 1.75 = 13.75 seconds
B.) Using the same formula
S = ut + 1/2at^2
Where S = total distance travelled
Substitutes t into the formula
S = 11.18(13.75) + 1/2 × 1.96 (13.75)^2
S = 153.725 + 185.28
S = 339 m approximately
C.) The speed of the police car at the time it overtakes the automobile will be constant = 11.18 m/s
D.) Using first equation of motion
V = U - at
Since the motorist is decelerating
V = 15.65 - 3.05 × 1.75
V = 15.65 - 5.338
V = 10.22 m/s
Therefore, the speed of the automobile at the time it was overtaken by the police car is 10.2 m/ s approximately
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm
Complete Question:
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.
Answer:
[tex]T_{min} =[/tex] 26 mins 40 secs
Explanation:
Reduction in depth, Δd = 20 mm
Depth of cut, [tex]d_c = 0.5 mm[/tex]
Number of passes necessary for this reduction, [tex]n = \frac{\triangle d}{d_c}[/tex]
n = 20/0.5
n = 40 passes
Tool width, w = 5 mm
Width of metal plate, W = 200 mm
For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times
Speed of tool, v = 100 mm/s
[tex]Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec[/tex]
minimum time required to reduce the depth of the plate by 20 mm:
[tex]T_{min} =[/tex] number of passes * Time/pass
[tex]T_{min} =[/tex] n * Time/pass
[tex]T_{min} =[/tex] 40 * 40
[tex]T_{min} =[/tex] 1600 = 26 mins 40 secs
Answer:
the minimum time required to reduce the depth of the plate by 20 mm is 26 minutes 40 seconds
Explanation:
From the given information;
Assuming the tool moves 100 mm/sec
The number of passes required to reduce the depth from 30 mm to 20 mm can be calculated as:
Number of passes = [tex]\dfrac{30-20}{0.5}[/tex]
Number of passes = 20
We know that the width of the tool is 5 mm; therefore, to reduce the depth per pass; the tool have to travel 20 times
However; the time per passes is;
Time/pass = [tex]\dfrac{20*L}{velocity \ of \ the \ feed}[/tex]
where;
length L = 400mm
velocity of the feed is assumed as 100
Time/pass [tex]=\dfrac{20*400}{100}[/tex]
Time/pass = 80 sec
Thus; the minimum time required to reduce the depth of the plate by 20 mm can be estimated as:
[tex]T_{min} = Time/pass *number of passes[/tex]
[tex]T_{min} = 20*80[/tex]
[tex]T_{min} = 1600 \ sec[/tex]
[tex]T_{min}[/tex] = 26 minutes 40 seconds
a surveyor is trying to find the height of a hill . he/she takes a sight on the top of the hill and find that the angle of elevation is 40°. he/she move a distance of 150 metres on level ground directly away from the hill and take a second sight. from this point the angl.e of elevation is 22°. find the height of the
hill
Answer:
height ≈ 60.60 m
Explanation:
The surveyor is trying to find the height of the hill . He takes a sight on the top of the hill and finds the angle of elevation is 40°. The distance from the hill where he measured the angle of elevation of 40° is not known.
Now he moves 150 m on level ground directly away from the hill and take a second sight from this point and measures the angle of elevation as 22°. This illustration forms a right angle triangle. The opposite side of the triangle is the height of the hill. The adjacent side of the triangle which is 150 m is the distance on level ground directly away from the hill.
Using tangential ratio,
tan 22° = opposite/adjacent
tan 22° = h/150
h = 150 × tan 22°
h = 150 × 0.40402622583
h = 60.6039338753
height ≈ 60.60 m
Which of the following is normally included in the criteria of a design?
1. Materials
2. Time
3. Budget
4. Efficiency
Answer:
i took the test and its budget
Explanation:
Budget, of the following is normally included in the criteria of a design. Thus, option (c) is correct.
What is design?The term design refers to the visual representation of image and shapes. The image is based on vector raster and bitmap. Who design the graphic is called graphic designer. The design main purpose to design of logo, word art, and advertisement design. The design mostly software as Coral Draw, Photoshop, Canva etc.
The criteria of a design are the budget. The budget was the financial as well as the monetary terminology. The criteria of the budget are the decided to how much they spend, and they save. The design on the country budget and the five-year plan.
As a result, the budget following is normally included in the criteria of a design. Therefore, option (c) is correct.
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When the transportation of natural gas in a pipeline is not feasible for economic reasons, it is first liquefied using nonconventional refrigeration techniques and then transported in super-insulated tanks. In a natural gas liquefaction plant, the liquefied natural gas (LNG) enters a cryogenic turbine at 30 bar and –160°C at a rate of 20 kg/s and leaves at 3 bar. If 120 kW power is produced by the turbine, determine the efficiency of the turbine. Take the density of LNG to be 423.8 kg/m3.
Answer:
the isentropic efficiency of turbine is 99.65%
Explanation:
Given that:
Mass flow rate of LNG m = 20 kg/s
The pressure at the inlet [tex]P_1 =30 \ bar[/tex] = 3000 kPa
turbine temperature at the inlet [tex]T_1 = -160^0C[/tex] = ( -160+273)K = 113K
The pressure at the turbine exit [tex]P_2 = 3 bar[/tex] = 300 kPa
Power produced by the turbine W = 120 kW
Density of LNG [tex]\rho = 423.8 \ kg/m^3[/tex]
The formula for the workdone by an ideal turbine can be expressed by:
[tex]W_{ideal} = \int\limits^2_1 {V} \, dP[/tex]
[tex]W_{ideal} ={V} \int\limits^2_1 \, dP[/tex]
[tex]W_{ideal} ={V} [P]^2_{1}[/tex]
[tex]W_{ideal} ={V} [P_1-P_2][/tex]
We all know that density = mass * volume i.e [tex]\rho= m*V[/tex]
Then ;
[tex]V = \dfrac{m}{\rho}[/tex]
replacing it into the above previous derived formula; we have:
[tex]W_{ideal} ={ \dfrac{m}{\rho}} [P_1-P_2][/tex]
[tex]W_{ideal} ={ \dfrac{20}{423.8}} [3000-300][/tex]
[tex]W_{ideal} ={ \dfrac{20}{423.8}} [2700][/tex]
[tex]W_{ideal} =0.04719*[2700][/tex]
[tex]W_{ideal} =127.42 kW[/tex]
However ; the isentropic efficiency of turbine is given by the relation:
[tex]n_{isen} =\dfrac{W}{W_{ideal}}[/tex]
[tex]n_{isen} =\dfrac{120}{120.42}[/tex]
[tex]n_{isen} =0.9965[/tex]
[tex]n_{isen} =[/tex] 99.65%
Therefore, the isentropic efficiency of turbine is 99.65%
what are the besl measures used for data variation or dispersion
Answer:
STANDARD DAVIATION OR SD
Explanation:
BECAUSE it is commonly in measuring of dispersion and it also the most roubst measure of variability
An undersea research chamber is spherical with an external diameter of 3.50 mm . The mass of the chamber, when occupied, is 21700 kg. It is anchored to the sea bottom by a cable. Find the followings
Required:
a. The buoyant force on the chamber.
b. The tension in the cable?
Answer:
a. The buoyant force on the chamber is 220029.6 N
b. The tension in the cable is 7369.6 N
Explanation:
The diameter of the sphere cannot be in millimeter (mm), if the chamber must occupy a big mass as 21700kg
Given;
diameter of the sphere, d = 3.50 m
radius of the sphere, r = 1.75 mm = 1.75 m
mass of the chamber, m = 21700 kg
density of water, ρ = 1000 kg/m³
(a)
Buoyant force is the weight of water displaced, which is calculated as;
Fb = ρvg
where;
v is the volume of sphere, calculated as;
[tex]V = \frac{4}{3} \pi r^3\\\\V = \frac{4}{3} \pi (1.75)^3\\\\V = 22.452 \ m^3[/tex]
Fb = 1000 x 22.452 x 9.8
Fb = 220029.6 N
(b)
The tension in the cable will be calculated as;
T = Fb - mg
T = 220029.6 N - (21700 x 9.8)
T = 220029.6 N - 212660 N
T = 7369.6 N
Using models helps scientists conduct research. How else can research using models save lives?
Answer: find the answer in the explanation.
Explanation:
Some of the methods of science are hypothesis, observations and experiments.
Scientific research and findings cut across many areas and field of life. Field like engineering, astronomy and medicine e.t.c
Using models in research is another great method in which researchers and scientists can master a process of a system and predict how it works.
Wilbur and Orville Wright wouldn't have died if the first airplane built by these two wonderful brothers was first researched by using models and simulations before embarking on a test.
Modelling in science involves calculations by using many mathematical methods and equations and also by using many laws of Physics. In this present age, many of these are computerised.
In space science, it will save lives, time and money to first model and simulate if any new thing is discovered in astronomy rather than people going to the space by risking their lives.
Also in medicine, drugs and many medical equipment cannot be tested by using people. This may lead to chaos and loss of lives.
So research using models in astronomy, medicine, engineering and many aspects of science and technology can indeed save lives.
The drag coefficient for a newly designed hybrid car is predicted to be 0.21. The cross-sectional area of the car is 30 ft2 . Determine the aerodynamic drag on the car when it is driven through still air at 55 mph.
Answer:
D = 1311.94 lb/ft
Explanation:
We are given the velocity as;
V = 55 mph.
First of all, let's convert it to ft/s
V = 55 × (5280/3600) ft/s
V = 80.67 ft/s
The equation for the aerodynamic drag force on the car is given as;
D = C_d•½ρ•V²•A
Where;
C_d is drag coefficient = 0.21
ρ is density of air
V is velocity = 80.67 ft/s
A is area 55 ft²
Now, from tables, the density of air under S.T.P condition is 1.225 kg/m³. Converting to lb/ft³ gives; 0.0624 lb/ft³
Plugging in the relevant values, we have;
D = 0.21•½•0.0624•80.67²•30
D = 1311.94 lb/ft