Answer:
"192.3 watt" is the right answer.
Explanation:
Given:
Efficient amplifier,
= 65%
or,
= 0.65
Power,
[tex]P_c=250 \ watt[/tex]
As we know,
⇒ [tex]P_t=P_c(1+\frac{\mu^2}{2} )[/tex]
By putting the values, we get
[tex]=P_c(1+\frac{1}{2} )[/tex]
[tex]=1.5 \ P_c[/tex]
Now,
⇒ [tex]P_i=(P_t-P_c)[/tex]
[tex]=1.5 \ P_c-P_c[/tex]
[tex]=\frac{P_c}{2}[/tex]
DC input (0.65) will be equal to "[tex](\frac{P_c}{2} )[/tex]".
hence,
The DC input power will be:
= [tex]\frac{250}{2}\times \frac{1}{0.65}[/tex]
= [tex]\frac{125}{0.65}[/tex]
= [tex]192.3 \ watt[/tex]
to check for ripple voltage from the alternator, connect a digital multimeter and select
Answer:
isn't it summer? sjsushsiansudndd
Even though the content of many alcohol blends doesn’t affect engine drive ability using gasoline with alcohol in warm weather may cause
A 2.0-in-thick slab is 10.0 in wide and 12.0 ft long. Thickness is to be reduced in three steps in a hot rolling operation. Each step will reduce the slab to 75% of its previous thickness. It is expected that for this metal and reduction, the slab will widen by 3% in each step. If the entry speed of the slab in the first step is 40 ft/min, and roll speed is the same for the three steps, determine: (a) length and (b) exit velocity of the slab after the final reduction
Answer:
26.02 ft
86.7690 ft/min
Explanation:
After 3 steps
0.75³(2.0 thickness)
T = 0.84375
W = (1+0.03)³10
= 10.92727 inches
A To get length
2.0 x 10 x 12 x12 = 0.84375 x 10.92727x lf
= 2880 = 9.21988Lf
Lf = 2880/9.21988
= 312.368 inches
Convert to feet
322.368 x 0.0833
= 26.02 ft
B.
= 2 x 10 x 40 = 0.84375 x 10.92727 x vf
800 = 9.21988vf
Vf = 800/9.21988
Vf = 86.7690 ft/min
Things to be done before isolation
Resistance depends on which three properties of a wire?
Color and texture are not directly related to a wire’s resistance.
1. color, thickness, texture
2. thickness, length, temperature
3. length, texture, temperature
4. temperature, color, texture
Answer:
2
Explanation:
From the formula R=(ro)A/l resistance depends on the length of the wire, the area of the wire(thickness) and the resistivity(ro) which depends on the material and temperature.
A particle which moves in two-dimensional curvilinear motion has coordinates in millimeters which vary with time t in seconds according to X=2t^2 +3t–1 and y = 5t - 2. Determine the coordinates of the center of curvature C at time t = 1s.
Answer:
The answer is "22.501,-22.899"
Explanation:
Just as in the previous problems find the angle the velocity makes with the x-axis and radius of curvature.
[tex]x= 2t^2 + 3t — 1\\\\y=5t-2\\\\x=4t+3\\\\y=5\\\\\tan \alpha (t = 1) =\frac{y}{x}=\frac{5}{4+3}=\frac{5}{7} \to alpha=35.54^{\circ}\\\\[/tex]
For the radius of curvature, we can use the expression from the last two problems, but first express the position and derivatives as y(x).
[tex]y(x)=2(\frac{y+2}{5})^2+3(\frac{y+2}{5})-1=\frac{1}{25}(2y^2+23y+13)\\\\y'(x)=\frac{1}{25}(4y+23)\\\\y''(x)=\frac{4}{25}\\\\\rho(t=1)=\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}=\frac{(1+(\frac{35}{25})^2)^{\frac{3}{2}}}{4}25=31.828[/tex]
The position for the center of the radius of curvature [tex]\vec{r}[/tex], (finding this expression is easy and is left as an exercise for the reader.)
[tex]\to \vec{r} = \hat{x}(x + \rho \sin \alpha) + \hat{y}(y- \rho \cos \alpha)\\\\= (4 + 18.501, 3-25.899)\\\\=(22.501, -22.899)[/tex]
Type the correct answer in each box. Spell all words correctly. According to the priority matrix, which tasks should an entrepreneur complete first? According to the priority matrix, entrepreneurs should first complete tasks that are blank and important.
Answer:
Development of creative and develop ideas
Explanation:
First task as an entrepreneur is to be creative and develop ideas. The person must design the product based on which he will develop the business strategy.
The remaining activities such as marketing, fund raising, recruitment etc. comes at a later stage.
Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 oF and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by Cp
Complete Question:
Problem 8 Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 °F and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by
C_p=10.03+0.0184T C_p[=]Btu/lbmole- "F ; T[=] °F C,
Answer:
[tex]P'=0.377hp[/tex]
Explanation:
From the question we are told that:
Initial Temperature T_1=70 F
Final Temperature [tex]T_2=250pisa =114.94F[/tex]
Efficiency [tex]E=75\%=0.75[/tex]
Generally the equation for Work-done is mathematically given by
[tex]W=\int C_pT[/tex]
[tex]W=10.03(114.94-70 )+0.0184((114.94)^2-70^2 )[/tex]
[tex]W=527.21btu/ibmole[/tex]
[tex]W=11.982btu/ibm[/tex]
Generally the equation for Efficiency is mathematically given by
[tex]E=\frac{isotropic Power}{Actual P'}[/tex]
[tex]E=\frac{P}{P'}[/tex]
Since
Isotropic Power
[tex]P=0.0167*11.982btu/ibm[/tex]
[tex]P=0.2btu/s[/tex]
Therefore
[tex]P'=\frac{0.2}{0.75}[/tex]
[tex]P'=0266btu/s[/tex]
Since
[tex]1btu/s=1.4148hp[/tex]
Therefore
[tex]P'=0.377hp[/tex]
A turbine of a fossil fuel burning installation delivers 1,500 hp of mechanical energy to a generator. The generator then converts 80.0% of the mechanical energy into electrical energy. If the terminal potential difference of the generator is 1790 V, what current does it deliver (in A)
Answer:
The generator delivers current of 500.11 A
Explanation:
Given the data in the question;
mechanical energy delivered to the generator = 1500 hp
efficiency η = 80.0 %
terminal potential difference of the generator = 1790 V
we know that;
1 hp = 746 W
so
the mechanical energy delivered to the generator will be
Generator Input = ( 1500 × 746 )W = 1119000 W
So the generator output will be;
Generator Output = Generator Input × η
we substitute
Generator Output = 1119000 W × 80.0 %
Generator Output = 1119000 W × 0.8
Generator Output = 895200 W
So the Current will be;
[tex]I[/tex] = Generator Output / terminal potential difference of the generator
we substitute
[tex]I[/tex] = 895200 W / 1790 V
[tex]I[/tex] = 500.11 A
Therefore, The generator delivers current of 500.11 A
A ceramic specimen with an elastic modulus of 300 GPa is under a tensile stress of 800 MPa. Will it fracture if its most severe flaw is an internal crack of 0.30 mm long with a tip radius of curvature in the amount of 0.0015 mm? Please justify your conclusion. (Hint: Compare the largest stress in the specimen around the crack to the theoretical strength which is roughly E/10).
Answer:
16Gpa < 30 Gpa
there would be no fracture
Explanation:
fracture can occur if the maximum strength at the top of the biggest flaw is more than the theoretical fracture
to get the theoretical strength =
e/10 = 300/10
= 30 Gpa
we get the magnitude at the buggest flaw
= 2σ√a/ρt
σ = 800
ρτ = 0.0015
a= 0.3/2
[tex]=2*800\sqrt{\frac{\frac{0.3}{2} }{0.0015} }[/tex]
= [tex]=2*800*\sqrt{100} \\=2*800*10\\=16000MPa[/tex]
= 16Gpa < 30 Gpa
the fracture is not going to happen given that the maximum strenght is smaller than the theoretical fracture strength.
Can you use isentropic efficiency for a non-adiabatic compressor?
Can you use isothermal efficiency for an adiabatic compressor?
An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 5.25 kW. Determine (a) the COP of this air conditioner and (b) the rate of heat transfer to the outside air.Answers:(a) 2.38, (b) 1065 kJ/min
Answer:
a) the COP of this air conditioner is 2.38
b) the rate of heat transfer to the outside air is 1065 kJ/min
Explanation:
Given the data in the question;
[ Outdoor ] ← Q[tex]_H[/tex] [ W[tex]_{net, in[/tex] ] Q[tex]_L[/tex] ← [ House ]
Rate of heat removed from the house; Q[tex]_L[/tex] = 750 kJ/min = ( 750 kJ/min × ( 1 kW / 60 kJ/min ) ) = 12.5 kW
Net-work input; W[tex]_{net, in[/tex] = 5.25 kW
a) The coefficient of performance of the air conditioner; COP.
COP = Q[tex]_L[/tex] / W[tex]_{net, in[/tex]
we substitute
COP = 12.5 kW / 5.25 kW
COP = 2.38
Therefore, the COP of this air conditioner is 2.38
b) the rate of heat transfer to the outside air.
Q[tex]_H[/tex] = Q[tex]_L[/tex] + W[tex]_{net[/tex]
we substitute
Q[tex]_H[/tex] = 12.5 kW + 5.25 kW
Q[tex]_H[/tex] = 17.75 kW
Q[tex]_H[/tex] = ( 17.75 × 60 ) kJ/min
Q[tex]_H[/tex] = 1065 kJ/min
Therefore, the rate of heat transfer to the outside air is 1065 kJ/min
4. An aluminum alloy fin of 12 mm thick, 10 mm width and 50 mm long protrudes from a wall, which is maintained at 120C. The ambient air temperature is 22C. The heat transfer coefficient and conductivity of the fin material are 140 W/m2K and 55 W/mk respectively. Determine a. Temperature at the end of the fin b. Temperature at the middle of the fin. c. Calculate the heat dissipation energy of the fin
Answer:
a) 84.034°C
b) 92.56°C
c) ≈ 88 watts
Explanation:
Thickness of aluminum alloy fin = 12 mm
width = 10 mm
length = 50 mm
Ambient air temperature = 22°C
Temperature of aluminum alloy is maintained at 120°C
a) Determine temperature at end of fin
m = √ hp/Ka
= √( 140*2 ) / ( 12 * 10^-3 * 55 )
= √ 280 / 0.66 = 20.60
Attached below is the remaining answers
A square plate of titanium is 12cm along the top, 12cm on the right side, and 5mm thick. A normal tensile force of 15kN is applied to the top side of the plate. A normal tensile force of 20kN is applied to the right side of the plate. The elastic modulus, E, is 115 GPa for titanium.If the left and bottom edges of the plate are fixed, calculate the normal strain and elongation of both the TOP and RIGHT side of the plate.
Answer:
[tex]X_t=2.17391304*10^{-4}[/tex]
[tex]X_r=2.89855072*10^{-4}[/tex]
[tex]e_t=0.0026[/tex]
[tex]e_r=0.0035[/tex]
Explanation:
From the question we are told that:
Dimension [tex]12*12[/tex]
Thickness [tex]l_t=5mm=5*10^-3[/tex]
Normal tensile force on top side [tex]F_t= 15kN[/tex]
Normal tensile force on right side [tex]F_r= 20kN[/tex]
Elastic modulus, [tex]E=115Gpap=>115*10^9[/tex]
Generally the equation for Normal Strain X is mathematically given by
[tex]X=\frac{Force}{Area*E}[/tex]
Therefore
For Top
[tex]X_t=\frac{Force_t}{Area*E}[/tex]
Where
[tex]Area=L*B*T[/tex]
[tex]Area=12*10^{-2}*5*10^{-3}[/tex]
[tex]Area=6*10^{-4}[/tex]
[tex]X_t=\frac{15*10^3}{6*10^{-4}*115*10^9}[/tex]
[tex]X_t=2.17391304*10^{-4}[/tex]
For Right side[tex]X_r=\frac{Force_r}{Area*E}[/tex]
Where
Area=L*B*T
[tex]Area=12*10^{-2}*5*10^{-3}[/tex]
[tex]Area=6*10^{-4}[/tex]
[tex]X_r=2.89855072*10^{-4}[/tex]
[tex]X_r=2.89855072*10^{-4}[/tex]
Generally the equation for elongation is mathematically given by
[tex]e=strain *12[/tex]
For top
[tex]e_t=2.17391304*10^{-4}*12[/tex]
[tex]e_t=0.0026[/tex]
For Right
[tex]e_r=2.89855072*10^{-4} *12[/tex]
[tex]e_r=0.0035[/tex]
How to Cancel prescription
are there engineering students here?
Technician A says white smoke in the exhaust of a diesel engine can be the result of a cylinder misfire in a warm engine. Technician B says blue smoke in the exhaust of a diesel engine can be caused by scored cylinder walls. Who is correct?
Answer:
Both
Explanation:
Because of water, fuel does not burn completely. This brings about water fumes that are white in color and looks like white smoke. If engine is cold and water is heating, it leads to steam formation like water vapor. The white times are because of not firing properly in the heated engine. Technician A is right.
Blue fine is caused by this scoring. It is also caused by dirty oil. Technician b is right too
A cylindrical block of wood 1 m in diameter and 1 m long has a specific weight of 7500 N/m^3. Will it float in water with its axis vertical?
Answer:
The block will float with its axis vertical.
Explanation:
For it to float, the upward force on the cylindrical block must be equal to the weight of an equal volume of water. Also, this upward force must be greater than or equal to the weight of the cylindrical block for it to float.
So, weight of cylindrical block, W = specific weight × volume
specific weight = 7500 N/m³
volume = πd²h/4 where d = diameter of block and h = height of block
volume = π(1 m)² × 1 m/4 = π/4 m³ = 0.7854 m³
W = 7500 N/m³ × 0.7854 m³ = 5890.5 N
Since the density of water = 1000 kg/m³, its specific weight W' = 1000 kg/m³ × 9.8 m/s² = 9800 N/m³
Since the volume of the cylinder = volume of water displaced, the weight of water displaced W' = upward force = specific weight of water × volume of water displaced = 9800 N/m³ × 0.7854 m³ = 7696.92 N
Since W' = 7696.92 N > W = 5890.5 N, the block will float with its axis vertical since the upward force is greater than the weight of the cylindrical block.
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Explanation:
mitski
Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm and 5.0 cm, find the volume flow rate and the average velocity in each pipe section.
Answer:
volumetric flow rate = [tex]0.0251 m^3/s[/tex]
Velocity in pipe section 1 = [tex]6.513m/s[/tex]
velocity in pipe section 2 = 12.79 m/s
Explanation:
We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.
The density of water is = 997 kg/m³
density = mass/ volume
since we are given the mass, therefore, the volume will be mass/density
25/997 = [tex]0.0251 m^3/s[/tex]
volumetric flow rate = [tex]0.0251 m^3/s[/tex]
Average velocity calculations:
Pipe section A:
cross-sectional area =
[tex]\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2[/tex]
mass flow rate = density X cross-sectional area X velocity
velocity = mass flow rate /(density X cross-sectional area)
[tex]velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s[/tex]
Pipe section B:
cross-sectional area =
[tex]\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2[/tex]
mass flow rate = density X cross-sectional area X velocity
velocity = mass flow rate /(density X cross-sectional area)
[tex]velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s[/tex]
Safety Issues for Operators of Oil and Gas Exploiting Equipment when working off rigs
Answer:
Safety First, Safety Always. Safety stands out as a core value for the oil and natural gas industry, embedded in every process and decision for operations. The oil and natural gas industry and the federal government are working together to continuously improve the safety of offshore operations. ...
Other than applying a strain to the gauge, what is the primary external/environmental factor that will influence the readings of a strain gauge?
Answer:
Temperature of the environment
Explanation:
Aside the strain applied to the strain gauge there are several other factors that might affect the reading of the strain gauge, and they are either external or internal factors like ; resistivity, length, and material of the strain gauge.
But the primary external factor is the Temperature of the environment when reading is taken
Can some people answer these questions so i can get to know the age group i an making my target market for DT-GSCE thankyou if you do my deadline is tomorrow :D
Answer:
I think you might have forgotten to post the problems
A cylindrical space capsule lands in the ocean. This capsule is 2.44 m long, 1.10 m in diameter, and weighted at one end so that it floats with its long central axis vertical and 0.820 m of its length above the water surface. The mass density of sea water is 1025 kg/m3.What is the magnitude of the buoyant force exerted on the capsule?
Answer:
The correct answer is "15456.8 N".
Explanation:
According to the question,
The inside volume will be:
= [tex]3.14\times (\frac{1.1}{2} )^2\times (2.44-0.82)[/tex]
= [tex]3.14\times \frac{1.21}{4}\times 1.62[/tex]
= [tex]3.14\times 03025\times 1.62[/tex]
= [tex]1.538757 \ m^3[/tex]
hence,
The buoyant force will be:
= [tex]V\times Pw\times g[/tex]
= [tex]1.538757\times 1025\times 9.8[/tex]
= [tex]15456.8 \ N[/tex]
Why does the ceramic made from Thorium and Oxygen have the chemical ratio of 2 oxygen atoms to every thorium atom (ThO2)
The purpose of the international residential code is to
Answer:
The International Building Code (IBC) is a model code that provides minimum requirements to safeguard the public health, safety and general welfare of the occupants of new and existing buildings and structures.
Explanation:
The allowable tensile stress for a 6.25 mm diameter bolt with a thread length of 5.5 mm is 207 MPa. The allowable shear stress of the material is 103 MPa. Where and how will such a bolt be most likely to fail if placed in tension
Answer:
At the threads due to shear.
Explanation:
Given :
The allowable tensile stress = 207 MPa
The allowable shear stress = 103 MPa
If a tensile force is applied, the maximum shear stress occurs at the threads of the bolt. The bolt is most likely to fail at the critical section. The critical cross section is the section having the minimum cross sectional area.
The portion of the bolt having threads has the minimum cross sectional area.
So when the bolt is applied with a tensile force, failure is most likely to take place at the threads due to the shearing force.
The number of pulses per second from IGBTs is referred to as
An engineer is applying dimensional analysis to study the flow of air through this horizontal sudden contraction for the purpose of characterizing the pressure drop. The flow is being modeled as constant density and steady. What is the functional relationship of the variables that characterize this situation
Answer:
The answer is " [tex]\Delta p = f(V1, p, V2, d, D, L)[/tex]"
Explanation:
Please find the complete question in the attached file.
Its change in temperature in pipes depends on rate heads and loss in pipes owing to pipe flow, contractual loss, etc.
The temperature change thus relies on V1 v2 p d D L.
A brittle material is subjected to a tensile stress of 1.65 MPa. If the specific surface energy and modulus of elasticity for this material are 0.60 J/m2 and 2.0 GPa, respectively. What is the maximum length of a surface flaw that is possible without fracture
Answer:
The maximum length of a surface flaw that is possible without fracture is
[tex]2.806 \times 10^{-4} m[/tex]
Explanation:
The given values are,
σ=1.65 MPa
γs=0.60 J/m2
E= 2.0 GPa
The maximum possible length is calculated as:
[tex]\begin{gathered}a=\frac{2 E \gamma_{s}}{\pi \sigma^{2}}=\frac{(2)\left(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\right)(0.60 \mathrm{~N} / \mathrm{m})}{\pi\left(1.65\times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\right)^{2}} \\=2.806 \times 10^{-4} \mathrm{~m}\end{gathered}[/tex]
The maximum length of a surface flaw that is possible without fracture is
[tex]2.806 \times 10^{-4} m[/tex]