To Find :
The torque of the stud.
Solution :
We know, torque can be calculated by force and distance between the center and force applied.
So,
[tex]\tau = F\times R\\\\\tau = 40 \times 0.5 \ N .\ m\\\\\tau = 20 \ N .\ m[/tex]
Therefore, the torque on the stud is 20 N m .
Find at the terminals of the circuit
Answer:
ok i will on day
Explanation:
7. The binary addition 1 + 1 + 1 gives
11 [2-bit]
011 [3-bit]
0011 [4-bit]
________
1 + 1 + 1 = 3
________
3 = 2 + 1
2¹ 2⁰
3 = (.. × 0) + (2¹ × 1) + (2⁰ × 1)
3 = ..011
Since 2³, 2⁴, 2⁵, .. are not used, they are represented as 0.
[ 2⁷ 2⁶ 2⁵ 2⁴ 2³ 2² 2¹ 2⁰ ]
[ 128 64 32 16 8 4 2 1 ]
Explain the advantages of using register indirect addressing mode over direct addressing mode with an 8051 assembly code. Also provide an example where it is inefficient to code using register indirect addressing mode.
vertical gate in an irrigation canal holds back 12.2 m of water. Find the average force on the gate if its width is 3.60 m. Report your answer with proper units and 3 sig figs.
Answer:
The right solution is "2625 kN".
Explanation:
According to the question,
The average pressure will be:
= [tex]density\times g\times \frac{h}{2}[/tex]
By putting values, we get
= [tex]1000\times 9.8\times \frac{12.2}{2}[/tex]
= [tex]1000\times 9.8\times 6.1[/tex]
= [tex]59780[/tex]
hence,
The average force will be:
= [tex]Pressure\times Area[/tex]
= [tex]59780\times 3.6\times 12.2[/tex]
= [tex]2625537 \ N[/tex]
Or,
= [tex]2625 \ kN[/tex]
The inputs of two registers R0 and R1 are controlled by a 2-to-1 multiplexer. The multiplexer select line and the register load enable inputs are controlled by inputs C0 and C1. Only one of the control inputs may be equal to 1 at a time. The required transfers are:
Answer: Hello your question is incomplete attached below is the complete question
answer :
Attached below
Explanation:
Given data:
The inputs of two registers are controlled by a 2-to-1 multiplexer.
The multiplexer select line and the register load enable inputs are controlled by inputs Co, C1, and C2.
Using the required transfers in the question to complete the detailed logic diagrams ( attached below )
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This is a test. Please ignore
Explanation:
Test
Consider the same piping system. this time, the same pipe is buried underground. assuming that there is a constant heat flux of 100w/m^2 from the outer surfance of the pipe to the soil determine the exit temperature of the water.
a. 129.1
b. 111.1
c. 82.1
d. 68.1
Complete Question
Complete Question is attached below
Answer:
Option A
Explanation:
From the question we are told that:
inner Diameter of pipe [tex]d_i=100^c[/tex]
Thickness [tex]t=50mm[/tex]
Outer diameter of pipe [tex]d_o=1.1m[/tex]
Length [tex]l=5m[/tex]
Temperature [tex]T_i=130^oC[/tex]
Generally the equation for Heat Balance is mathematically given by
[tex]q*\pi d_oL=mC_p(T_i-T_o)[/tex]
Therefore
[tex]T_o=T_i+\frac{q*\pi d_oL}{mC_p}[/tex]
[tex]T_o=130+\frac{100*3.142 *1.1*5}{0.5*4000}[/tex]
[tex]T_o=129.136^oC[/tex]
Therefore the exit temperature of the water.is [tex]T_o=129.136^oC[/tex]
Option A
Derive the expression for electrical-loading nonlinearity error (percentage) in a rotatory potentiometer in terms of the angular displacement, maximum displacement (stroke), potentiometer element resistance, and load resistance. Plot the percentage error as a function of the fractional displacement for the three cases: RL/RC = 0.1, 1.0, and 10.0
Answer:
The plot for percentage error as a function of fractional displacement ( [tex]\frac{R_{L} }{R_{C} }[/tex]) for the values of 0.1,1.0,10.0 is shown in image attached below.
Explanation:
Electrical loading non linearity error (percentage) is shown below.
[tex]E=\frac{(\frac{v_{o} }{v_{r} }-\frac{Q}{Q_{max} } )}{\frac{Q}{Q_{max} } }[/tex]×[tex]100[/tex]
where Q= displacement of the slider arm
[tex]Q_{max}=[/tex] maximum displacement of a stroke
[tex]\frac{v_{o} }{ v_{r} } =[/tex][tex]\frac{(\frac{Q}{Q_{max} }(\frac{R_{L} }{R_{C} } ) )}{(\frac{R_{L} }{R_{C} } ) +(\frac{Q}{Q_{max} })-(\frac{Q}{Q_{max} })^{2} }[/tex]
here [tex]R_{L}=load resistance[/tex]
[tex]R_{C}=[/tex]total resistance of potentiometer.
Now the nonlinearity error in percentage is
[tex]E=\frac{(\frac{(\frac{Q}{Q_{max} }(\frac{R_{L} }{R_{C} } ) )}{(\frac{R_{L} }{R_{C} } ) +(\frac{Q}{Q_{max} })-(\frac{Q}{Q_{max} })^{2} }-\frac{Q}{Q_{max} } )}{\frac{Q}{Q_{max} } }[/tex]×[tex]100[/tex]
The following attached file shows nonlinear error in percentage as a function of [tex]\frac{R_{L} }{R_{C} }[/tex] displacement with given values 0.1, 1.0, 10.0. The plot is drawn using MATLAB.
The MATLAB code is given below.
clear all ;
clc ;
ratio=0.1 ;
i=0 ;
for zratio=0:0.01:1 ;
i=i+1 ;
tratioa (1,i)=zratio ;
E1(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zrtio)*100 ;
end
ratio=1.0 :
i=0 ;
for zratio=0:0.01:1 ;
i=i+1 ;
tratiob (1,i)=zratio ;
E2(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zratio)*100 ;
end
ratio=10.0 :
i=0 ;
for zratio=0:0.01:1 ;
i=i+1 ;
tratioc (1,i)=zratio ;
E3(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zrtio)*100 ;
end
k=plot(tratioa,E1,tratiob,E2,tratioc,E3)
grid
title({non linear error in % as a function of R_L/R_C})
k(1). line width = 2;
k(1).marker='*'
k(1).color='red'
k(2).linewidth=1;
k(2).marker='d';
k(2).color='m';
k(3).linewidth=0.5;
k(3).marker='h';
k(3).color='b'
legend ('location', 'south east')
legend('R_L/R_C=0.1','R_L/R_C=1.0','R_L/R_C=10.0')
Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a (111) plane and in a [101]direction and is initiated at an applied tensile stress of 13.9 MPa (2020 psi), compute the critical resolved shear stress
Answer:
[tex]\mathbf{\tau_c =5.675 \ MPa}[/tex]
Explanation:
Given that:
The direction of the applied tensile stress =[001]
direction of the slip plane = [[tex]\bar 1[/tex]01]
normal to the slip plane = [111]
Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:
[tex]cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big][/tex]
where;
[tex][d_1\ e_1 \ f_1][/tex] = directional indices for tensile stress
[tex][d_2 \ e_2 \ f_2][/tex] = slip direction
replacing their values;
i.e [tex]d_1[/tex] = 0 ,[tex]e_1[/tex] = 0 [tex]f_1[/tex] = 1 & [tex]d_2[/tex] = -1 , [tex]e_2[/tex] = 0 , [tex]f_2[/tex] = 1
[tex]cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big][/tex]
[tex]cos \ \lambda = \dfrac{1}{\sqrt{2}}[/tex]
Also, to find the angle [tex]\phi[/tex] between the stress [001] & normal slip plane [111]
Then;
[tex]cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big][/tex]
replacing their values;
i.e [tex]d_1[/tex] = 0 ,[tex]e_1[/tex] = 0 [tex]f_1[/tex] = 1 & [tex]d_3[/tex] = 1 , [tex]e_3[/tex] = 1 , [tex]f_3[/tex] = 1
[tex]cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big][/tex]
[tex]cos \phi= \dfrac{1} {\sqrt{3} }[/tex]
However, the critical resolved SS(shear stress) [tex]\mathbf{\tau_c}[/tex] can be computed using the formula:
[tex]\tau_c = (\sigma )(cos \phi )(cos \lambda)[/tex]
where;
applied tensile stress [tex]\sigma =[/tex] 13.9 MPa
∴
[tex]\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})[/tex]
[tex]\mathbf{\tau_c =5.675 \ MPa}[/tex]
A series RLC circuit is driven by an ac source with a phasor voltage Vs=10∠30° V. If the circuit resonates at 10 3 rad/s and the average power absorbed by the resistor at resonance is 2.5W, determine that values of R, L, and C, given that Q =5.
Answer:
R = 20Ω
L = 0.1 H
C = 1 × 10⁻⁵ F
Explanation:
Given the data in the question;
Vs = 10∠30°V { peak value }
V"s[tex]_{rms[/tex] = 10/√2 ∠30° V
resonance freq w₀ = 10³ rad/s
Average Power at resonance Power[tex]_{avg[/tex] = 2.5 W
Q = 5
values of R, L, and C = ?
We know that;
Power[tex]_{avg[/tex] = |V"s[tex]_{rms[/tex]|² / R
{ resonance circuit is purely resistive }
we substitute
2.5 = (10/√2)² × 1/R
2.5 = 50 × 1/R
R = 50 / 2.5
R = 20Ω
We also know that;
Q = w₀L / R
we substitute
5 = ( 10³ × L ) / 20
5 × 20 = 10³ × L
100 = 10³ × L
L = 100 / 10³
L = 0.1 H
Also;
w₀ = 1 / √LC
square both side
w₀² = 1 / LC
w₀²LC = 1
C = 1 / w₀²L
we substitute
C = 1 / [ (10³)² × 0.1 ]
C = 1 / [ 1000000 × 0.1 ]
C = 1 / [ 100000 ]
C = 0.00001 ≈ 1 × 10⁻⁵ F
Therefore;
R = 20Ω
L = 0.1 H
C = 1 × 10⁻⁵ F
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Answer:
Explanation:
do you have any other questions besides "tyuuyiopopiouyttrrtrffrlkl,k;;';'l.l"
The current flowing into the collector lead of a certain bipolar junction transistor (BJT) is measured to be 1 nA. If no charge was transferred in or out of the collector lead prior to t = 0, and the current flows for 1 min. calculate the total charge which crosses into the collector.
Answer:
the total charge which crosses into the collector is 60 nC
Explanation:
Given the data in the question;
current flowing into the collector lead of the bipolar junction transistor (BJT); i = 1 nA = 10⁻⁹ A
no charge was transferred in or out of the collector lead prior to t = 0
the current flow time t = 1 min = 60 sec
Now we write the relation between current, charge, and time;
i = dq / dt
where i is current, q is charge and t is time. { d refers to change }
Now,
[tex]q=\int\limits^t_{t=0} {i(t)} \, dt[/tex]
[tex]q=\int\limits^{t=60}_{t=0} { (10^{-9}) } \, dt[/tex]
[tex]q = ( 10^{-9}) (t)_0^{60[/tex]
[tex]q = ( 10^{-9}) ( 60 - 0 )[/tex]
q = 60 × 10⁻⁹ C
q = 60 nC
Therefore, the total charge which crosses into the collector is 60 nC
The term _______________refers to the science of using fluids to perform work.
Answer:
Hi, there your answer is hydraulics
Explanation:
diffrerentiate y=cos^{4} (3x+1)
Answer:
-6sin(6x+2)cos²(3x+1)dx.
Explanation:
[tex]dy=4*cos^3(3x+1)*3*(-sin(3x+1))dx=-6sin(6x+2)cos^2(3x+1)dx.[/tex]
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A manufacturing facility with a wastewater flow of 0.011 m3/sec and a BOD5 of 590 mg/L discharges into the Cattaraugus Creek. The creek has a 10-year, 7-day low flow of 1.7 m3/sec. Upstream of the facility, the BOD5 of the creek is 0.6 mg/L. The BOD rate constants k are 0.115 d-1. Determine: [A] UBOD of wastewater. [B] UBOD of creek. [C] What is the initial ultimate BOD after mixing
This question is incomplete, the complete question is;
A manufacturing facility with a wastewater flow of 0.011 m³/sec and a BOD₅ of 590 mg/L discharges into the Cattaraugus Creek. The creek has a 10-year, 7-day low flow of 1.7 m³/sec. Upstream of the facility, the BOD₅ of the creek is 0.6 mg/L. The BOD rate constants k are 0.115 d⁻¹ for the wastewater and 3.7 d⁻¹ for the creek. The temperature of both the creek and tannery of wastewater is 20°C. Determine: [A] UBOD of wastewater. [B] UBOD of creek. [C] What is the initial ultimate BOD after mixing
Answer:
a) Ultimate BOD of wastewater is 1349.188 mg/L
b) Ultimate BOD of creek is 0.6 mg/L
c) the initial ultimate BOD after mixing is 9.27 mg/L
Explanation:
Given the data in the question;
Q[tex]_{wastewater[/tex] = 0.011 m³/s
BOD[tex]_{wastewater[/tex] = 590 mg/L
Q[tex]_{creek[/tex] = 1.7 m³/sec
BOD[tex]_{creek[/tex] = 0.6 mg/L
time t = 5
rate constants k for wastewater = 0.115 d⁻¹
rate constants k for creek = 3.7 d⁻¹
a) UBOD of wastewater.
The Ultimate BOD of wastewater is;
BOD[tex]_{wastewater[/tex] = L₀[tex]_{wastewater[/tex]( 1 - [tex]e^{-kt[/tex] )
where BOD[tex]_{wastewater[/tex] is the BOD of wastewater after 5 days, L₀[tex]_{wastewater[/tex] is the ultimate BOD of wastewater, k is the rate constant of wastewater and t is the time( days ).
we make L₀[tex]_{wastewater[/tex] the subject of formula
BOD[tex]_{wastewater[/tex] = L₀[tex]_{wastewater[/tex]( 1 - [tex]e^{-kt[/tex] )
L₀[tex]_{wastewater[/tex] = BOD[tex]_{wastewater[/tex] / ( 1 - [tex]e^{-kt[/tex] )
so we substitute
L₀[tex]_{wastewater[/tex] = 590 / ( 1 - [tex]e^{(-0.115*5)[/tex] )
L₀[tex]_{wastewater[/tex] = 590 / ( 1 - [tex]e^{(-0.575)[/tex] )
L₀[tex]_{wastewater[/tex] = 590 / ( 1 - 0.5627 )
L₀[tex]_{wastewater[/tex] = 590 / 0.4373
L₀[tex]_{wastewater[/tex] = 1349.188 mg/L
Therefore, Ultimate BOD of wastewater is 1349.188 mg/L
b) UBOD of creek
The Ultimate BOD of creek is;
BOD[tex]_{creek[/tex] = L₀[tex]_{creek[/tex]( 1 - [tex]e^{-kt[/tex] )
we make L₀[tex]_{creek[/tex] the subject of formula
L₀[tex]_{creek[/tex] = BOD[tex]_{creek[/tex] / (1 - [tex]e^{-kt[/tex] )
we substitute
L₀[tex]_{creek[/tex] = 0.6 / ( 1 - [tex]e^{(-3.7*5)[/tex] )
L₀[tex]_{creek[/tex] = 0.6 / ( 1 - [tex]e^{(-18.5)[/tex] )
L₀[tex]_{creek[/tex] = 0.6 / ( 1 - (9.2374 × 10⁻⁹) )
L₀[tex]_{creek[/tex] = 0.6 / 0.99999
L₀[tex]_{creek[/tex] = 0.6 mg/L
Therefore, Ultimate BOD of creek is 0.6 mg/L
c) the initial ultimate BOD after mixing;
Lₐ = [( Q[tex]_{wastewater[/tex] × L₀[tex]_{wastewater[/tex] ) + ( Q[tex]_{creek[/tex] × L₀[tex]_{creek[/tex] )] / [ Q[tex]_{wastewater[/tex] + Q[tex]_{creek[/tex] ]
we substitute
Lₐ = [( 0.011 × 1349.188 ) + ( 1.7 × 0.6 )] / [ 0.011 + 1.7 ]
Lₐ = [ 14.841068 + 1.02 ] / 1.711
Lₐ = 15.861068 / 1.711
La = 9.27 mg/L
Therefore, the initial ultimate BOD after mixing is 9.27 mg/L
You are hired as the investigators to identify the root cause and describe what should have occurred based on the following information. The mass of jet fuel required to travel from Toronto to Edmonton is 22,300 kg. The fuel gage correctly indicated that the plane already had 7,682 L of jet fuel in the tank. The specific gravity of the jet fuel is 0.803. Using this information, the crew added 4,916 L of fuel and took off, only to run out of fuel and crash a short while later. Use your knowledge of dimensions and units to work out what went wrong.
Answer:
The mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.
Explanation:
Since density ρ = m/v where m = mass of fuel and v = volume of fuel, we need to find the mass of each volume of fuel.
So, m = ρv now ρ = specific gravity × density of water = 0.803 × 1000 kg/m³ = 803 kg/m³.
To find the mass of the 7,682 L of fuel, its volume is 7,682 dm³ = 7,682 dm³ × 1 m³/1000 dm³ = 7.682 m³.
It's mass, m = 803 kg/m³ × 7.682 m³ = 6168.646 kg
To find the mass of the extra 4,916 L of fuel added, we have
m' = ρv' where v' = 4,916 L = 4,916 dm³ = 4916 dm³ × 1 m³/1000 dm³ = 4.916 m³
m' = 803 kg/m³ × 4.916 m³ = 3947.548 kg
So, the total mass of the fuel is m" = m + m' = 6168.646 kg + 3947.548 kg = 10116.194 kg ≅ 10,166.2 kg
Since this mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.
The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) True Strain 49300 0.11 61300 0.21 What true stress is necessary to produce a true plastic strain of 0.25
Answer:
64640.92 psi
Explanation:
True stress ( psi ) True strain
49300 0.11
61300 0.21
Determine the true stress necessary to produce a true plastic strain of 0.25
бT1 = 49300
бT2 = 61300
бT3 = ?
∈T1 = 0.11
∈T2 = 0.21
∈T3 = 0.25
note : бTi = k ∈Ti^h
∴ 49300 = k ( 0.11 )^h ----- ( 1 )
61300 = k ( 0.21)^h ------ ( 2 )
solving equations 1 and 2 simultaneously
49300/61300 = ( 0.11 / 0.21 )^h
0.804 = (0.52 )^h
next step : apply logarithm
log ( 0.804 ) = log(0.52)^h
h = log 0.804 / log (0.52)
= 0.33
back to equation 1
49300 = k ( 0.11 )^0.33
k = 49300 / (0.11)^0.33
= 102138
therefore бT3 = K (0.25)^h
= 102138 ( 0.25 )^ 0.33
= 64640.92
A column carries 5400 pounds of load and is supported on a spread footing. The footing rests on coarse sand. Design the smallest square footing (to the next 3-inch increment) that will safely carry the column load. The footing will be 1 ft 9 in. deep and will be constructed of cast-in-place concrete.
Answer:
Following are the responses to the given question:
Explanation:
Some General Motors flex fuel vehicles do not use a fuel sensor to measure the percentage of ethanol in the fuel. These vehicles use ________ as a base fuel to calculate the percentage using oxygen sensor readings.
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Answer:
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Hot combustion gases, modeled as air behaving as an ideal gas, enter a steady-state turbine at 145 psi, 2700oR, and exit at 29 psi and 1974oR. The power output is 74.26 hp, and heat transfer from the turbine to the surroundings occurs at a rate of 14 BTU/s.
Required:
Determine the power output of the turbine, in hp.
Hot combustion gases, modeled as air behaving as an ideal gas, enter a steady-state turbine at 145 psi, 2700oR, and exit at 29 psi and 1974oR. The power output is 74.26 hp, and heat transfer from the turbine to the surroundings occurs at a rate of 14 BTU/s.
Required:
a)Determine the power output of the turbine, in hp.
b) The Flow rate
Answer:
a) [tex]w=74.26hp[/tex]
b) [tex]m=0.22[/tex]
Explanation:
From the question we are told that:
Initial Pressure [tex]p_1= 145 psi[/tex]
Initial Temperature [tex]T_1 =2700oR=>2240.33^oF[/tex]
Final Pressure [tex]p_2= 29 psi[/tex]
Final Temperature [tex]t_2=1974oR=>1514.33^oF[/tex]
Output Power [tex]w=74.26hp[/tex]
Heat transfer Rate [tex]Q=14BTU/s[/tex]
Generally the equation for Steady flow energy is mathematically given by
[tex]Q-w=m(h_2-h_1)[/tex]
Where
[tex]m=Flow\ rate[/tex]
From Steam table
[tex]h_1=704btu/ib (at\ p_1= 145\ psi,\ T_1 =2700oR=>2240.33^oF )[/tex]
[tex]h_2=401btu/ib (at\ p_2= 29psi\ t_2=1974oR=>1514.33^oF )[/tex]
Therefore
[tex]-14-74.26=m(401-704)[/tex]
[tex]m=\frac{-14-74.26}{(401-704)}[/tex]
[tex]m=0.22[/tex]
3. Which of these instruments is used to measure wind speed? A. anemometer C. wind sock B. thermometer D. wind vane It is an instrument that can show both the wind speed and direction. A. Anemometer C. Wind sock B. thermometer D. wind vane 4. 5. At what time does air temperature changes? A. from time to time C. in the evening only B. in the afternoon only D. in the morning only
Answer:
wind vane if it can be used to show wind speed and the other is a
Explanation:
please mark 5 star if im right and brainly when ya can
A centrifugal pump is used to extract water from a reservoir at 14,000 gal/min. The pipe connecting the pump inlet to the reservoir is 12 inches in diameterand is 65 ft long. The average friction factor in this pipe is 0.018. The pump performance curves indicate that, at this flow rate, the head rise across the pump is 320 ft, the efficiency is 81% and the required NPSH is 25 ft. Please estimate:
This question is incomplete, the complete question is;
A centrifugal pump is used to extract water from a reservoir at 14,000 gal/min. The pipe connecting the pump inlet to the reservoir is 12 inches in diameter and is 65 ft long.
The average friction factor in this pipe is 0.018. The pump performance curves indicate that, at this flow rate, the head rise across the pump is 320 ft, the efficiency is 81% and the required NPSH is 25 ft.
Please estimate: The required brake horsepower.
Answer:
The required brake horsepower is 1400.08
Explanation:
Given the data in the question;
Power required to drive the pump can be determined using the formula;
P = r[tex]_w[/tex]QH / η₀(0.745)
given that; centrifugal pump is used to extract water from a reservoir at 14,000 gal/min.
Q = 14,000 gal/min = ( 14,000 × 0.00006309 )m³/sec = 0.883 m³/sec
the head rise across the pump is 320 ft,
H = 320 ft = ( 320 × 0.3048 )m = 97.536 m
the efficiency η₀ = 81% = 0.81
r[tex]_w[/tex] = 9.81 kN/m³
so we substitute our values into the formula
P = [ 9.81 × 0.883 × 97.536 ] / 0.81(0.745)
P = 844.87926528 / 0.60345
P = 1400.08 HP
Therefore, The required brake horsepower is 1400.08
A long power transmission cable is buried at a depth (ground-to-cable-centerline distance) of 2 m. The cable is encased in a thin-walled pipe of 0.1-m diameter, and, to render the cable superconducting (with essentially zero power dissipation), the space between the cable and pipe is filled with liquid nitrogen at 77 K. If the pipe is covered with a super insulator (ki = 0.005 W/mK) of 0.05-m thickness and the surface of the earth (kg = 1.2 W/mK) is at 300 K, what is the cooling load (W/m) that must be maintained by a cryogenic refrigerator per unit pipe length?
Answer:
[tex]q=9.9w/m[/tex]
Explanation:
From the question we are told that:
Ground-to-cable-center line Distance [tex]d_g=2m[/tex]
Diameter of Cable case [tex]d=0.1m[/tex]
Temperature of Nitrogen [tex]T_n=77K[/tex]
Insulator [tex]ki = 0.005 W/mK[/tex]
Thickness [tex]t=0.05[/tex]
Mass [tex]kg = 1.2 W/mK[/tex]
Temperature of earth surface [tex]T_e=300K[/tex]
Generally the equation for Heat rate per unit length is mathematically given by
[tex]q=\frac{T_g-T_n}{R_g+R_e}[/tex]
Where
[tex]R_g=[kg(\frac{2\pi}{(in4d_g/d_x)}]^{_1}[/tex]
[tex]R_g=[(1.2)(\frac{2\pi}{(in4(2)/0.2)}]^{-1}[/tex]
[tex]R_g=0.489[/tex]
And
[tex]R_e=\frac{In(\frac{D_0}{D_1})}{2\pi ki}[/tex]
[tex]R_e=\frac{In(2)}{2*3.142 0.005}[/tex]
[tex]R_e=22.1[/tex]
Therefore
[tex]q=\frac{223}{0.489+22.064}[/tex]
[tex]q=9.9w/m[/tex]
How did the development of John Hadley's octant into a sextant enhance its measuring capabilities?
O A.
navigators could use the instrument in stormy conditions
OB.
navigators could measure angles to the nearest minute
O C. navigators could measure angles up to 60°
OD.
navigators could measure angles up to 120°
Answer:
b
Explanation:
A spring having a stiffness k is compressed a distance δ. The stored energy in the spring is used to drive a machine which requires power P. Determine how long the spring can supply energy at the required rate.
Units Used: kN = 103 N
Given: k = 5 kN/m; δ = 400 mm; P = 90 W
Answer:
30w
Explanation:
Technician A says lever action pushes a rod into the brake booster and master cylinder
when the driver pushes on the brake pedal. Technician B says the produces hydraulic
pressure in the master cylinder. Who is correct?
When you double the velocity, you _____ the kinetic energy.
Answer:
you quadruple the kinetic energy
A project has an initial cost of $100,000 and uniform annual benefits of $12,500. At the end of its 8-year useful life, its salvage value is $30,000. At a 10% interest rate, the net present worth of the project is approximately:__________
1- $-19,318
2- $0
3- $+30,000
4- $+100,000
Answer:
[tex]X=-\$19318[/tex]
Explanation:
From the question we are told that:
Initial cost [tex]P= $100,000[/tex]
Annual benefits [tex]A= $12,500[/tex]
Salvage value [tex]S= $30,000[/tex]
Interest rate [tex]I=10\%=>0.10[/tex]
Time [tex]t=8years[/tex]
Generally the equation for Net Project worth X is mathematically given by
[tex]X=-P+A+S[/tex]
Where
Present worth of Annual benefits A is
[tex]A'=A(P/a,0.10,8)[/tex]
[tex]A'=12500*5.3349[/tex]
[tex]A'=\$66686.25[/tex]
Present worth of Salvage Price S is
[tex]S'=S(P/a,0.10,8)[/tex]
[tex]S'=30000*0.46651[/tex]
[tex]S'=\$13995.3[/tex]
Therefore
[tex]X=-P+A'+S'[/tex]
[tex]X=-100000+66686.25+\$13995.3[/tex]
[tex]X=-\$19318[/tex]
