Answer:
V = 20.6 m/s
Explanation:
Given that the temperature of the neon molecules = 356 Kelvin
Pressure = 0.9 bar
The mass number of Neon = 21.
Using the formula below
1/2 m v^2 = (3kT)/2
Where
T = temperature = 356 k
K = Bolzmann constant
= 1.38 × 10^-23jk^-1
NA = 6.02214076×10²³ mol⁻¹
Where NA = Avogadro number
Substitute all the parameters Into the formula
1/2 × 21/NA × v^2 = (3 × k × 356)/2
1.744×10^-23V^2 = 7.3692 × 10^-21
Make V^2 the subject of formula
V^2 = (7.37×10^-21)/1.744×10^-23
V^2 = 422.55
V = sqrt( 422.55)
V = 20.55
Therefore, the average (root mean square) velocity (m/s) of Neon molecules at 356 Kelvins and 0.9 bars is 20.55 m/s
Two technicians are discussing the a series-parallel brake lamp circuit that does not illuminate. Technician A states that an open in one of the parallel branches could be the cause. Technician B states that a shorted (or stuck closed) brake lamp switch wired in series could be the cause. Which technician is correct? Group of answer choices
Answer:
Technician B is correct.
Explanation:
Two technicians are discussing the a series-parallel brake lamp circuit that does not illuminate.
The statement made by Technician A saying that an open in one of the parallel branches could be the cause has no ground because an opening in any branches has not to do with the issue since there is no disconnection The statement made by Technician B saying that a shorted (or stuck closed) brake lamp switch wired in series could be the cause is correct.
The correct area in sq. Inches and sq. Feet is: Select one: a. 966.76 sq. Inches and 8.056 sq. Feet b. 96.676 sq. Inches and 8.056 sq. Feet c. 96.676 sq. Inches and 0.671 sq. Feet
Answer:
c. 96.676 sq. Inches and 0.671 sq. Feet
Explanation:
From the list of the given option, we are told to chose the correct area in sq. inches that correspond to sq. Feet.
If we recall from the knowledge of our conversion table that,
1 sq feet = 144 sq inches
Then, let's confirm if the option were true.
a. 966.76 sq. Inches and 8.056 sq. Feet
Assuming
if 1 sq feet = 8.056
in sq inches, we have ( 8.056 × 144 ) sq inches
= 1160.064 sq. inches
So, 1160.064 sq. inches is equal to 8.056 sq. Feet. Then option 1 is wrong
b. 96.676 sq. Inches and 8.056 sq. Feet
if 1 sq feet = 8.056
in sq inches, we have ( 8.056 × 144 ) sq inches
= 1160.064 sq. inches
So, 1160.064 sq. inches is equal to 8.056 sq. Feet. Then option 2 is wrong/
c. 96.676 sq. Inches and 0.671 sq. Feet
if 1 sq feet = 0.671
in sq inches, we have ( 0.671 × 144 ) sq inches
= 96.624 sq. Inches which is closely equal to 96.676 sq. Inches
Therefore, this is the correct answer as it proves that 96.676 sq. Inches = 0.671 sq. Feet
What are the Parts of a hydroelectric Power plant ?
Answer:
Trash rack
Open channel
Fore bay
Pen stock
Inlet valve
Turbine
Tailrace
Generator
Power house
When choosing a respirator for your job, you must conduct a _____ test.
A) Fit B) Practice C) Breathing D) Weight?
Answer:
Fit test
Explanation:
it's on the automotive technology sight : )
MATLAB can solve a variety of engineering problems including those requiring simulating differential equations and iterative numerical calculations.A. TrueB. False
Answer:
A. True
Explanation:
MATLAB may be defined as a programming platform that is designed specifically for the engineers as well as the scientists to carry out different analysis and researches.
MATLAB makes use of a desktop environment which is tuned for certain iterative analysis and the design processes with a programming language which expresses matrix as well as array mathematics directly.
Thus the answer is true.
What is best for a busy student to do for better results in school?
Answer:
Set high and clear expectations for quality work
Don't attempt to cram all your studying into one session
Explanation:
Participate in Class
eat a well balanced diet
Find perfect place to Study.
HavE a NIce dAY ;}
2.24 Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same state (a cycle). Process 1-2: Expansion from state 1 where p1 = 10 bar, V1 = 1 m3, to state 2 where V2 = 4 m3. During the process, pressure and pV1.5 volume are related by = constant. Process 2-3: Constant volume heating to state 3 where p3 = 10 bar. Process 3-1: Constant pressure compression to state 1. Sketch the processes on p –V coordinates and evaluate the work for each process, in kJ. What is the net work for the cycle, in kJ?
Answer:
Explanation:
Given that:
From process 1 → 2
[tex]P_1 = 10 bar \\ \\ V_1 = 1 m^3 \\ \\ V_2 = 4 m^3[/tex]
[tex]PV^{1.5} = \ constant[/tex]
[tex]\gamma = 1.5[/tex]
Process 2 → 3
The volume is constant i.e [tex]V_2 =V_3 = 4m^3[/tex]
[tex]P_3 = 10 \ bar[/tex]
Process 3 → 1
P = constant i.e the compression from state 1
Now, to start with 1 → 2
[tex]P_1V_1^{1.5} = P_2V_2^{1.5}[/tex]
[tex]P_2 = P_1 (\dfrac{V_1}{V_2})^{1.5}[/tex]
[tex]P_2 = 10 \times (\dfrac{1}{4})^{1.5}[/tex]
[tex]P_2 =1.25[/tex]
The work-done for the process 1 → 2 through adiabatic expansion is:
[tex]W = \dfrac{1}{1-\gamma}[P_2V_2-P_1V_1][/tex]
We know that 1 bar = [tex]10^5 \ N/m^2[/tex]
∴
[tex]W = \dfrac{1}{1-1.5}[1.25 \times 10^5 \times 4- 10 \times 10^5 \times 1][/tex]
[tex]W =1000000 \ J[/tex]
[tex]W_{1 \to 2} = 1000 kJ[/tex]
For process 2 → 3
Since V is constant
Thus:
W = PΔV = 0
[tex]W_{2 \to 3} = 0[/tex]
For process 3 → 1
W = PΔV
[tex]W _{3 \to 1} = P_3(V_1-V_3)[/tex]
[tex]W _{3 \to 1} = 10 \times 10^5 (1-4)[/tex]
[tex]W _{3 \to 1} = 10 \times 10^5 (-3)[/tex]
[tex]W _{3 \to 1} = -3 \times 10^6 \ J[/tex]
[tex]W _{3 \to 1} = -3000 \ kJ[/tex]
The net work-done now for the entire system is :
[tex]W_{net} = W_{1 \to 2} + W_{2 \to 3 } + W_{ 3 \to 1 }[/tex]
[tex]W_{net} = (1000 + 0 + (-3000)) \ kJ[/tex]
[tex]W_{net} =-2000 \ kJ[/tex]
The sketch of the processes on p -V coordinates can be found in the image attached below.
A) The work done for each process are :
Process (1 - 2) = 1000 kJ Process (2 - 3) = 0 kJ process (3 - 1) = -3000 kJB) The net work for the cycle = -2000 kJ
Given Data :
For process (1 -2) For process ( 2 - 3 ) process ( 3 - 1 )
P₁ = 10 bar P₃ = 10 bar constant pressure compression
V₁ = 1 m³ constant volume heating
V₂ = 4 m³
PV[tex]^{1.5}[/tex] = constant
A) Determine work done for each process
Calculate work done for process (1 - 2)
W₁ ₋ ₂ = [tex]\frac{P_{1}V_{1} - P_{2}V_{2} }{n -1 }[/tex] * 100
= [ ( 10*1 ) - ( 1.25 * 4 ) ] / 1.5 - 1
= [ 10 - 5 ] / 0.5
= 10 * 100 = 1000 kJ
Calculate work done for process ( 2-3 )
given that there is constant volume heating
W₂₋₃ = 0 kJ
Calculate work done for process ( 3-1)
W₃₋₁ = P ( Δ V ) given that p = constant
= 10 * 100 ( -3 )
= - 3000 kJ
B) The net work for the cycle
W₁ ₋ ₂ + W₂₋₃ + W₃₋₁
= 1000 kJ + 0 kJ + - 3000 kJ
= - 2000 kJ
Hence we can conclude that the ) The work done for each process are :
Process (1 - 2) = 1000 kJ Process (2 - 3) = 0 kJ process (3 - 1) = -3000 kJand The net work for the cycle = -2000 kJ
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Attached below is the P-V sketch of the process
List 10 uses for surveying in areas other than land surveying. Select all that apply. a. topographic surveying b. satellite surveying c. aerial surveying d. optical tooling e. marketing surveying f. control surveying g. as-built surveying h. statistical surveying i. telephone surveying j. construction surveying k. alignment surveying l. mine surveying m. solar surveying
Answer:
b. satellite surveying
c. aerial surveying
d. optical tooling
e. marketing surveying
f. control surveying
h. statistical surveying
i. telephone surveying
k. alignment surveying
l. mine surveying
m. solar surveying
Explanation:
A Survey is an act of examination of the features of a subject or material under consideration. Land surveying refers to the examination of the natural and man-made features of a piece of land using scientific and mathematical methods.
Land surveying finds application in construction where a survey is made on all the structures found in a constructed property. Topographic surveying deals with examining the natural and man-made feature of a piece of land. As-built survey as the name implies examines the features and location of a building during or recently after construction. These three are examples of land surveys.
Some type of surveying other than land surveying includes satellite, aerial, optical tooling, marketing, control surveying, statistical, telephone, alignment, mine, and solar surveying
A Survey means an examination of a features, subject or material under consideration.
Land surveying refers to the examination of the natural and man-made features of a piece of land using scientific and mathematical methods.
However, some other type of surveying other than land surveying includes satellite, aerial, optical tooling, marketing, control surveying, statistical, telephone, alignment, mine, and solar surveying.
Read more about surveying
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When storing used oil, it need to be kept in________ container?
Answer: In a clean plastic or metal container with a tightly sealed lid
In mechanics of materials, the bending stress of a beam in bending can be determined by the equation σ = MyIwhere expressed in terms of SI base units M is the bending moment in Newton-meters (N-m), y is the distance from the neutral axis in meters (m), and I is the moment of inertia in meters to the fourth power (m4). The bending stress σ has the same units as those of:_____.a) pressure.b) density.c) force.d) spring constant.
Answer:
Explanation:
In mechanics of materials, the bending stress of a beam in bending can be determined by the equation
The Stokes-Oseen formula for drag force F on a sphere of diameter D in a fluid stream of low velocity V, density p and viscosity μ is
F=3πμDV+9π/16∗pV2d2
Is this formula dimensionally homogenous?
Answer:
[tex]\frac{ML}{T^2}=\frac{ML}{T^2}[/tex]
Hence it is proved that Stokes-Oseen formula is dimensionally homogenous.
Explanation:
For equation to be dimensionally homogeneous both side of the equation must have same dimensions.
For given Equation:
F= Force, μ= viscosity, D = Diameter, V = velocity, ρ= Density
Dimensions:
[tex]F=\frac{ML}{T^2}[/tex]
[tex]\mu=\frac{M}{LT}[/tex]
[tex]D=L\\\\V=\frac{L}{T}\\ \\\rho=\frac{M}{L^3}[/tex]
Constants= 1
Now According to equation:
[tex]\frac{ML}{T^2}=[\frac{M}{LT}][L] [\frac{L}{T}] + [\frac{M}{L^3}][\frac{L^2}{T^2}][L^2][/tex]
Simplifying above equation, we will get:
[tex]\frac{ML}{T^2}=2*\frac{ML}{T^2}[/tex]
Ignore "2" as it is constant with no dimensions. Now:
[tex]\frac{ML}{T^2}=\frac{ML}{T^2}[/tex]
Hence it is proved that Stokes-Oseen formula is dimensionally homogenous.
what type of comptuer was name for the cabinet where the brains
Answer: Central Processing Unit
By balancing information security and access, a completely secure information system can be created.A. TrueB. False
Answer: true
Explanation:
Which of the following is NOT a part of an "I" statement?
some connecting rods have ____ to help lubricate the cylinder wall or piston pin.
Answer:
some connecting rods have spit holes
Some connecting rods have crankshaft to help lubricate the cylinder wall or piston pin.
What are crankshaft?A crankshaft is a mechanical component that converts reciprocating motion in a piston engine into rotational motion.
The crankshaft is a rotating shaft with one or more crankpins that drive the pistons via the connecting rods.
Reduced fuel economy is one issue you'll face if your car's crankshaft isn't working properly. The reason for this is that your vehicle's fuel injectors will not efficiently direct gas to the engine.
The connecting rods' big ends are lubricated with oil that travels up an oilway drilled through the crankshaft.
From the main bearings to the big end bearings, oil is pushed. It will lubricate the cylinder walls in some engines by passing through a passage drilled in the connecting rod.
Thus, the answer is crankshaft.
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A rocket is shot straight up from the earth with a net acceleration (= acceleration by the rocket engine - gravitational pullback) of 7m/sec during the initial stage of flight until the engine cut out at t = 10 sec. How high will it go, air resistance neglected?
Answer:
599.7 m
approximately 600 m
Explanation:
initial speed of the rocket = 0
net acceleration upwards = 7 m/s^2
the engine cuts out 10 sec after take off
maximum height reached = ?
we neglect air resistance
To get the velocity of the rocket at the point where the engine cuts off, we use the equation
v = u + at
where
v is the velocity at this point where the engine stops = ?
u is the initial velocity of the rocket from rest = 0 m/s
a is the net acceleration upwards = 7 m/s^2
t is the time the engine runs = 10 s
substituting, we have
v = 0 + (7 x 10)
v = 70 m/s
to get the distance from the ground to this point, we use the equation
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as
where
v is the final velocity at the the height where the engine is cut out = 70 m/s
u is the initial speed at the ground = 0 m/s
a is the net acceleration on the rocket = 7 m/s^2
s is the distance from the ground to this point
substituting, we have
[tex]70^{2}[/tex] = [tex]0^{2}[/tex] + 2(7 x s)
4900 = 14s
s = 4900/14 = 350 m
After this point when the engine cuts out, the rocket experiences an acceleration proportional to the acceleration due to gravity 9.81 m/s^2 downwards, and slows down gradually before coming to a stop at the maximum height.
To get this height, we use the equation
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] - 2gs (the negative sign is due to the downward direction of the acceleration g)
where
v is the final velocity at the maximum height = 0 m/s (it comes to a stop)
u is the speed at the instance that the engine is cut out = 70 m/s
g is the acceleration due to gravity = 9.81 m/s^2
s is the distance from this point to the maximum height
substituting values, we have
[tex]0^{2}[/tex] = [tex]70^{2}[/tex] - 2(9.81 x s)
0 = 4900 - 19.62s
4900 = 19.62s
s = 4900/19.62 = 249.7 m
The maximum height that will be reached = 350 m + 249.7 m = 599.7 m
approximately 600 m
The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is 225∘C and the change in plate temperature with time (dT/dt) is -0.022 K/s. The ambient air temperature is 25∘C and the plate measures 0.3×0.3with a mass of 3.75 kg and a specific heat of 2770J/kg⋅K.
Answer:
h = 6.35 W/m².k
Explanation:
In order to solve this problem, we will use energy balance, taking the thin hot plate as a system. According to energy balance, the rate of heat transfer to surrounding through convection must be equal to the energy stored in the plate:
Rate of Heat Transfer Through Convection = Energy Stored in Plate
- h A (Ts - T₀) = m C dT/dt
where,
h = convection heat transfer coefficient = ?
A = Surface area of plate through which heat transfer takes place = 2 x 0.3 m x 0.3 m (2 is multiplied for two sides of thin plate) = 0.18 m²
Ts = Surface Temperature of hot thin plate = 225⁰C
T₀ = Ambient Temperature = 25°C
m = mass of plate = 3.75 kg
C = Specific Heat = 2770 J/kg. k
dT/dt = rate of change in plate temperature = - 0.022 K/s
Therefore,
- h (0.18 m²)(225 - 25) k = (3.75 kg)(2770 J/kg.k)(- 0.022 k/s)
h = (- 228.525 W)/(- 36 m².k)
h = 6.35 W/m².k
Calculate the value of ni for gallium arsenide (GaAs) at T = 300 K. The constant B = 3. 56 times 1014 9cm -3 K-3/2) and the bandgap voltage Eg = 1. 42 eV.
This question is incomplete, the complete question is;
Calculate the value of ni for gallium arsenide (GaAs) at T = 300 K.
The constant B = 3.56×10¹⁴ (cm⁻³ K^-3/2) and the bandgap voltage E = 1.42eV.
Answer: the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³
Explanation:
Given that;
T = 300k
B = 3.56×10¹⁴ (cm⁻³ K^-3/2)
Eg = 1.42 eV
we know that, the value of Boltzmann constant k = 8.617×10⁻⁵ eV/K
so to find the ni for gallium arsenide;
ni = B×T^(3/2) e^ ( -Eg/2kT)
we substitute
ni = (3.56×10¹⁴)(300^3/2) e^ ( -1.42 / (2× 8.617×10⁻⁵ 300))
ni = (3.56×10¹⁴)(5196.1524)e^-27.4651
ni = (3.56×10¹⁴)(5196.1524)(1.1805×10⁻¹²)
ni = 2.1837 × 10⁶ cm⁻³
Therefore the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³
The steering column connects the steering wheel to the steering gear.
a. True
b. False
A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at 2308C by rejecting its waste heat to cooling water that enters the condenser at 188C at a rate of 0.25 kg/s and leaves at 268C. The refrigerant enters the condenser at 1.2 MPa and 658C and leaves at 428C. The inlet state of the compressor is 60 kPa and 2348C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator, and (d) the theoretical maximum refrigeration load for the same power input to the compressor.
Answer:
hello your question is incomplete attached below is the missing part and also attached is the solution
answer: a) 0.4801
b) 5.398 kw
c) 2.14
d) 12.72
Explanation:
The quality of the refrigerant at the evaporator inlet
h4 = hf4 + x4(hfx4)
Refrigeration load
Ql = m(h1-h4)
COP of the refrigerator
Ql / m(h2-h1) - Qm
Theoretical maximum refrigeration load
( Ql )max = COPr.rev * [m(h2-h1) - Qin]
The quality that will exist at the inlet of the refrigerant's evaporator would be:
a). [tex]0.4801[/tex]
The load of the refrigeration would be
b). [tex]5.398 kW[/tex]
The refrigerator's COP would be:
c). [tex]2.14[/tex]
The maximum refrigeration load would be as follows:
d) [tex]12.72[/tex]
a). The determination of the quality of the refrigerant at the inlet of the evaporator would be:
[tex]h4 = hf_{4} + x_{4}(hfx_{4})[/tex]
As given,
[tex]hf_{4}[/tex] [tex]= 3.841[/tex]
[tex]hf_{2,4}[/tex][tex]= 223.95[/tex]
[tex]h_{4} = 111.37[/tex]
Now,
solving for [tex]x_{4}[/tex] [tex]= (111.37 - 3.841)/223.95[/tex]
[tex]= 0.4801[/tex]
b). Refrigeration load
[tex]Q_{l}[/tex] [tex]= m(h_{1} - h_{4})[/tex]
[tex]= 0.0455(230.01-111.37)[/tex]
[tex]= 5.398 kW[/tex]
c). COP of the refrigerator
[tex]Q_{l}[/tex]/[tex]m(h_{2} - h_{1}) - Q_{m}[/tex]
by putting the values, we get
∵ COP [tex]= 2.14[/tex]
d). Theoretical maximum refrigeration load
[tex](Q_{l})[/tex]max [tex]= COPr.rev[/tex] × [tex]{m(h_{2} - h_{1}) - Q_{in}[/tex]
by putting the values, we get
∵ [tex]Q_{L}max = 12.72[/tex]
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what are PAT&E tests on production sysems used for.
Answer:
Portable appliance testing (PAT) is the term used to describe the examination of electrical appliances and equipment to ensure they are safe to use. Most electrical safety defects can be found by visual examination but some types of defect can only be found by testing.
please give brainlist
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( ꈍᴗꈍ)
The wall is 15 ft high and consists of 2 x 4 in. studs, plastered on one side. On the other side there is 4-in. clay brick. Determine the average load in lb>ft of length of wall that the wall exerts on the floor.
Answer:
645 lb/ft
Explanation:
The height of the wall = 15 ft
For the 2 x 4 in. studs:
The minimum dead load for a 2 x 4 in. studs (from table 1-3) = 4 psf = 4 lb/ft²
Therefore for a 15 ft tall wall, the dead load for a 2 x 4 in. studs = 15 ft × 4 lb/ft² = 60 lb/ft
For the 4 in. clay brick:
The minimum dead load for a 4 in. clay brick (from table 1-3) = 39 psf = 39 lb/ft²
Therefore for a 15 ft tall wall, the dead load for a 4 in. clay brick = 15 ft × 39 lb/ft² = 585 lb/ft
Therefore the average load = load for the 4 in. clay brick + load for the 2 x 4 in. studs = 585 lb/ft + 60 lb/ft = 645 lb/ft
A set of experiments is run on an op amp that is ideal except for having a finite gain A. The results are tabulated below. Are the results consistent? If not, are they reason-able, in view of the possibility of experimental error? What do they show the gain to be? Using this value, predict values I of the measurements that were accidentally omitted (the blank entries).Experiment # v1 v2 vo 1 0.00 0.00 0.00 2 1.00 1.00 1.00 3 1.00 1.00 4 1.00 1.10 10.1 5 2.01 2.00 -0.99 6 1.99 2.00 1.00 7 5.10 -5.10
Answer:
i)The results are consistent at ; Rows ( 1,2,4,6 ), which simply means that it is fairly consistent and makes sense
ii) Row 5 shows a gain of -99 which is ≈ -100 and there is a possibility of experimental error of (1%)
ii ) The omitted values are : row 3 = 0.99, Row 7 = 5.049
Explanation:
To calculate/determine the missing/omitted values we have to apply this formula:
Vo = ( V1 - V2 ) G
Vo = output = 1
G = gain = - 100
V1 =
we find V1 along line 3 ( row 3 )
(V1 - V2) G = 1
( V1 - 1 ) * -100 = 1
hence V1 = 0.99
we find V2 along row 7
( V1 - V2 ) G = 1
( 5.10 - V2 ) * -100 = 1
hence V2 = 5.049
A civil engineer designs a wheelchair accessible ramp next to a set of steps leading up to a building. The height from the ground to the top of the stairs is 3ft. Based on ADA codes, the slope must be 1:12 or less. (Slope is equal to the rise of the ramp divided by the run of the ramp.) What is the IMA of this ramp if the engineer uses a slope of 1:12?
Answer: IMA = 12.042
Explanation:
Given that;
Height from the ground h is 3 ft
Slope of ramp s is 1:12
Horizontal length of the ramp x will be 3 × 12 = 36 ft
Now to get the IMA ( ideal mechanical advantage,)
IMA = length of the ramp over / height of the ramp
IMA = (√( 36² + 3²)) / 3
IMA = (√ 1305 ) / 3
IMA = 36.124 / 3
IMA = 12.042
What should you do before you start welding?
Explanation:
1. Weld only in authorized areas. Make sure the area is dry, chemical free, and well ventilated.
2. Inspect the equipment before starting to use it.
3. Keep other people away, unless they are authorized to be there and are wearing the appropriate personal protective equipment.
A standard 20° pressure angle, 20 tooth pinion with a diametral pitch of 12 rotates at 1776 rpm driving a mating gear at 740 rpm.a) How many teeth are there on the gear?b) What are the pitch diameters of the pinion and gear?c) What are the standard addendum and dedendum diameters of the pinion and gear?d) What is the theoretically correct center distance?e) What are the radial, tangential and normal tooth loads when contact between the pinion and the gear occurs at the pitch point and the gearset is transmitting ½ hp?
Answer:
A) 48
B) Pitch diameters : pinion = 42.164 mm, Gear = 101.19 mm
C) standard addendum : pinion = 46.3804, Gear = 105.406
standard dedendum : pinion = 37.265 mm, Gear = 96.312 mm
D) 71.672 mm
E) 94.989 N , 101.0858 N, 34.573 N
Explanation:
Given Data :
∅ = 20⁰ , Tp = 20 ( tooth pinion ),
diameter pitch = 12, Np = 1776 rpm ,
Ng = 740 rpm,
attached below is the detailed solution of the given problems
what type of address do computer use to find something on network?
ip address
url address
mac address
isp address
Answer:
The answer would be ISP address
Explanation:
An Internet Protocol address is a numerical label assigned to each device connected to a computer network that uses the Internet Protocol for communication. An IP address serves two main functions host or network interface identification and location addressing which can help you a lot.
The type of address the computer use to find something on network is ISP address.
What is ISP?ISP is simply Internet service provider. They are companies which provides internet services to people. They charge based on the quantity of their service used at a particular time. ISP companies also provide softwares for its users.
What is IP address?An Internet Protocol address refers to a numerical label assigned to each device connected to a computer network that uses the Internet Protocol for communication.
Therefore, the correct answer is option D
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A proposed piping and pumping system has 20-psig static pressure, and the piping discharges to atmosphere 160 ft above the pump. If the piping friction loss is 20 ft head, the minimum pressure rating (psi) of the piping system is most nearly:
(A) 50
(B) 100
(C) 150
(D) 250
Answer: (B) 100
Explanation:
Given that;
Pstatic = 20 psig , hz = 160ft, hf = 20ft
Now total head will be;
T.h = hz + hf
T.h= 160 + 20
T.h = 180ft
Minimum pressure = Psatic + egh
we know that specific weight of water is 62.4 (lb/ft3)
so
P.min = (20 bf/in² ) + (62.4 b/ft³ × 180 fr
P.min = (20 bf/in² ) + ( 62.4 × 180 × 1 ft²/144 in²)
P.min = 20 + 78
P.min = 98 lbf/in²
Therefore the minimum pressure rating (psi) of the piping system is most nearly B) 100
A 20-foot-long W10 x 60 is suspended and hanging from one end. If the modulus of elasticity is 29,000 ksi, determine the following.A. What is the maximum tensile stress?
B. What is the maximum normal strain?
Answer:
(a) the maximum tensile stress is 68.2 psi
(b) the maximum normal strain is 2.35 x 10⁻⁶
Explanation:
Given;
modulus of elasticity, E = 29,000 ksi = 29 x 10⁶ psi
(a) the maximum tensile stress
[tex]\tau = \frac{f}{A}[/tex]
f is the maximum force suspended = 20 x 60 = 1200 lb
A is the area of W10 x 60 = 17.6 in²
[tex]\tau = \frac{1200}{17.6} \\\\\tau = 68.2 \ psi[/tex]
(b) the maximum normal strain.
According to Hook's law stress is directional to strain
τ = Eε
[tex]\epsilon = \frac{\tau}{E}\\\\\epsilon = \frac{68.2}{29*10^{6}}\\\\\epsilon = 2.35*10^{-6}[/tex]
When entering a freeway you should always:
A. Slow down and proceed when it is safe.
B. Stop and make sure there is no traffic approaching.
C. Accelerate to the same speed as the freeway traffic and merge smoothly.
D. Go as fast as you can and swing abruptly into traffic.