i need some help with this
Java programing
def digit_sum(number):
sumOfDigits = 0
while number > 0:
sumOfDigits += number % 10
number = number // 10
return sumOfDigits
or you can use,
In [2]: digit_sum(10)
Out[2]: 1
In [3]: digit_sum(434)
Out[3]: 11
or you can use,
digits = "12345678901234567890"
digit_sum = sum(map(int, digits))
print("The equation is: ", '+'.join(digits))
print("The sum is:", digit_sum)
Have a great day <3
For a steel alloy it has been determined that a carburizing heat treatment of 9-h duration will raise the carbon concentration to 0.38 wt% at a point 1.2 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 6.4 mm position for an identical steel and at the same carburizing temperature.
Answer:
t2 = 256 hours
Explanation:
Given data:
Carbon concentration ( C ) at 1.2mm from surface, C = 0.38 wt%
Duration( t ) of heat treatment for 0.38wt% at 1.2mm = 9-hr
Estimate the time (in h) necessary to achieve the same concentration at a 6.4 mm
Assuming same concentration of 0.38wt% we will apply Fick's second law for constant surface concentration
attached below is the remaining part of the solution
x1 = 1.2 mm
x2 = 6.4 mm
t1 = 9-hr
t2 = ?
t2 = [tex](\frac{6.4}{1.2} )^{2} * 9[/tex] = 256 hours
The efficiency of a steam power plant can beincreased by bleeding off some of the steam thatwould normally enter the turbine and usingit topreheat the water entering the boiler. In this process,liquid water at 50oC and 1000 kPa is mixed withsuperheated steam at 200oC and 1000 kPa. If the plantoperators want to produce a saturated liquid at 1000kPa, what ratio of mass flow rates of water andsuperheatedsteam are required
Answer:
Explanation:
This is Answer....
Car pool vehicles are those with
Answer:
The pavement of the carpool lanes is marked with the diamond symbol. These carpool lanes are reserved for buses and vehicles with a minimum of two or three people and that includes the driver
Explanation:
hope this helped
A shaft is made from a tube, the ratio of the inside diameter to the outside diameter is 0.6. The material must not experience a shear stress greater than 500KPa. The shaft must transmit 1.5MW of mechanical power at 1500 revolution per minute. Calculate the shaft diameter
Answer:
shaft diameter = [tex]\sqrt[3]{0.3512}[/tex] mm = 0.7055 mm
Explanation:
Ratio of inside diameter to outside diameter ( i.e. d/D )= 0.6
Shear stress of material ( Z ) ≤ 500 KPa
power transmitted by shaft ( P ) = 1.5MW of mechanical power
Revolution ( N ) = 1500 rev/min
Calculate shaft Diameter
Given that: P = [tex]\frac{2\pi NT}{60}[/tex] ---- 1
therefore; T = ( 1.5 *10^3 * 60 ) / ( 2[tex]\pi[/tex] * 1500 ) = 9.554 KN-M
next
[tex]\frac{T}{I_{p} } = \frac{Z}{R}[/tex]
hence ; T = Z[tex]_{p} *Z[/tex]
attached below is the remaining part of the solution
A switch that can open or close an electric circuit can be used to?
Answer:
When a switch is in the "off" position the circuit is open. Electric charges cannot flow when a switch is in the off position.
Explanation:
A switch that can open or close an electric circuit can be used to stop the flow of current.
What is a switch?A switch can be defined as an electrical component (device) that is typically designed and developed for interrupting the flow of current or electrons in an electric circuit.
This ultimately implies that, a switch that can open or close an electric circuit can be used to stop the flow of current.
Read more on switch here: https://brainly.com/question/16160629
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In a metal-oxide-semiconductor (MOS) device, a thin layer of SiO2 (density = 2.20 Mg/m3) is grown on a single crystal chip of silicon. How many Si atoms and how many O atoms are present per square millimeter of the oxide layer? Assume that the layer thickness is 160 nm.
Answer:
3.52×10⁶ atoms of Si and 7.05×10⁶ atoms of O
Explanation:
It is all about unit conversions.
The area of our MOS is 1 mm². So, we know that the thickness is 160 nm. This data can give us the volume. We convert nm to mm.
160 nm . 1×10⁻⁶ mm /1nm = 1.6×10⁻⁴ mm
By the way, now we can determine the volume of MOS, in order to work with density.
1.6×10⁻⁴ mm . 1 mm² = 1.6×10⁻⁴ mm³
But density is mg/m³, so we convert mm³ to m³
1.6×10⁻⁴ mm³ . 1×10⁻⁹ m³/mm³ = 1.6×10⁻¹³ m³
Now, we apply density to determine the mass of MOS
Density = mass /volume → Density . volume = mass
1.6×10⁻¹³ m³ . 2.20mg/m³ = 3.52×10⁻¹³ mg
To make more easier the calculate, we convert mg to g.
3.52×10⁻¹³ mg . 1g /1000mg = 3.52×10⁻¹⁶ g
To count the atoms, we determine molar mass of SiO₂ → 60.08 g/mol
We need to know moles of Si and O₂ in the MOS
Firstly, we determine amount of MOS: 3.52×10⁻¹⁶ g / 60.08 g/mol = 5.86×10⁻¹⁸ moles
1 mol of SiO₂ has 1 mol of Si and 2 mol of O so:
5.86×10⁻¹⁸ mol of SiO₂ may have:
(5.86×10⁻¹⁸ . 1) /1 = 5.86×10⁻¹⁸ moles of Si
(5.86×10⁻¹⁸ .2) /1 = 1.17×10⁻¹⁷ moles of O₂
Let's count the atoms (1 mol of anything contain NA particles)
5.86×10⁻¹⁸ mol of Si . 6.02×10²³ atoms/ mol = 3.52×10⁶ atoms of Si
1.17×10⁻¹⁷ mol of O₂ . 6.02×10²³ atoms/ mol = 7.05×10⁶ atoms of O
The number of Si and O atoms present per mm² of the oxide layer are respectively; 3.52 × 10⁶ atoms of Si and 7.05×10⁶ atoms of O
What is the number of atoms present?We are given;
Area of MOS device; A = 1 mm²
Thickness of layer; t = 160 nm = 1.6 × 10⁻⁴ mm
Formula for volume is;
V = Area * thickness
V = 1.6 × 10⁻⁴ mm × 1 mm² = 1.6 × 10⁻⁴ mm³
Converting volume to m³ gives;
V = 1.6 × 10⁻¹³ m³
Now, to get the mass of MOS, we will use the formula;
Mass = Density * volume
we are given density = 2.20mg/m³
Thus;
Mass = 1.6 × 10⁻¹³ m³ × 2.20mg/m³
Mass = 3.52 × 10⁻¹³ mg = 3.52×10⁻¹⁶ g
From periodic table, the molar mass of SiO₂ = 60.08 g/mol
Then;
Number of moles of MOS = (3.52 × 10⁻¹⁶ g)/60.08 g/mol
Number of moles of MOS = 5.86 × 10⁻¹⁸ moles
Now, 1 mol of SiO₂ is formed from 1 mol of Si and 2 mol of O. Thus;
5.86×10⁻¹⁸ mol of SiO₂ will have;
5.86×10⁻¹⁸ moles of Si
and (5.86×10⁻¹⁸ .2) = 11.72 × 10⁻¹⁸ moles of O₂
From Avogadro's number that 1 mol equals 6.02×10²³ atoms , we can say that;
5.86 × 10⁻¹⁸ mol of Si × 6.02 × 10²³ atoms/mol = 3.52 × 10⁶ atoms of Si
11.72 × 10⁻¹⁸ mol of O₂ × 6.02 × 10²³ atoms/mol = 7.05×10⁶ atoms of O
Read more about number of atoms at; https://brainly.com/question/3157958
The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod is d = 25 mm and the outside diameter of the steel tube is D= 45 mm. The length of the composite column is L = 761 mm. A force P = 88 kN is applied at the top surface, distributed across both the rod and tube.
Required:
Determine the normal stress σ in the steel tube.
Answer:
Explanation:
From the information given:
[tex]E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN[/tex]
The total load is distributed across both the rod and tube:
[tex]P = P_1+P_2 --- (1)[/tex]
Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.
[tex]\delta_1=\delta_2[/tex]
[tex]\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}[/tex]
[tex]\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}[/tex]
[tex]P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})[/tex]
[tex]P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}[/tex]
[tex]P_2 = 6.6212 \ P_1[/tex]
Replace [tex]P_2[/tex] into equation (1)
[tex]P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\ -88 = 7.6212 \ P_1 \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\ P_1 = -11.547 \ kN[/tex]
Finally, to determine the normal stress in aluminum rod:
[tex]\sigma _1 = \dfrac{P_1}{A_1} \\ \\ \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}[/tex]
[tex]\sigma_1 = - 23.523 \ MPa}[/tex]
Thus, the normal stress = 23.523 MPa in compression.
the removed soil at an excavation site is also called spoil?
Answer:
True, That is correct. Soil removed from an excavation site is indeed called spoil.
Spoil definition: The waste material (such as soil) brought up during the course of an excavation.
Hope I helped! Sorry if not. Have a wonderful week and follow me for more help! Remember your worth and love yourself! Adíos! ;D
Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.
Thus, The debris that is dug up during an excavation, such as soil. Depending on the location and the type of excavation, the composition of the spoil material can change.
It could consist of dirt, gravel, boulders, debris, or other substances that were removed or moved during the excavation process.
Typically, the spoil is put aside and used for a variety of tasks, including backfilling, grading, or disposal, as necessary for the project and permitted by local laws.
Thus, Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.
Learn more about Soil, refer to the link:
https://brainly.com/question/31227835
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Air at 25 m/s and 15°C is used to cool a square hot molded plastic plate 0.5 m to a side having a surface temperature of 140°C. To increase the throughput of the production process, it is proposed to cool the plate using an array of slotted nozzles with width and pitch of 4 mm and 56 mm, respectively, and a nozzle-to-plate separation of 40 mm. The air exits the nozzle at a temperature of 15°C and a velocity of 10 m/s.
Required:
a. Determine the improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate.
b. Would the heat rates for both arrangements change significantly if the air velocities were increased by a factor of 2?
c. What is the air mass rate requirement for the slotted nozzle arrangement?
Answer:
a. 2.30
b. decreases with increasing velocity.
c. 0.179 kg/s.
Explanation:
Without mincing let's dive straight into the solution to the question above.
[a].
The improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate can be determined by calculating turbulent flow:
The turbulent flow over the plate= 10 × 0.5/ 20.92 × 10⁻6 = 2.39 × 10⁵.
While the turbulent flow correlation = 0.037( 2.39 × 10⁵)^[tex]\frac{4}{5}[/tex] (0.7)^[tex]\frac{1}{3}[/tex] = 659.6.
Array of slot noozle = [10 × (2 × 0.004)]/ 20.92 × 10^-6] = 3824.
where A = 4/56 =0.714.
And Ar = [ 60 + 4 (40/2 × 4) - 2 ]^2 ]-1/2 = 0.1021.
N = 2/3 (0.1021)^3/4 [ 2 × 3824/ ( 0.0714 / 0.1021) + (.1021/0.0714)] (0.700)^0.42 =24.3.
h = 24.3 × 0.030/0.004 = 91.1 W/m^2k.
Therefore; 659.6 × 0.030/0.5 = 39.0 W/m²k.
The turbulent flow = 0.5 × 39.6 × 0.5( 140 -15) = 1237.5 W.
The slot noozle = 91.1 × 0.5 × 0.5 [ 140 -15] = 2846.87W.
The improvement in cooling rate = 2846.87/ 1237.5 = 2.30.
[b].
2.3 [ (2^2/3)/ 2^4/5] = 2.1
Thus, it decreases with increasing velocity
[c].
The air mass rate requirement for the slotted nozzle arrangement = 9 × 0.995 (0.5 × 0.004)10 = 0.179 kg/s.
After earning a bachelor's degree, one must do which of the following before taking the PE examination to receive a Professional Engineering license?
electrical engineering
Answer:
Electrical engineering is an engineering discipline concerned with the study, design and application of equipment, devices and systems which use electricity, electronics, and electromagnetism.
Explanation:
The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator. The elevation difference between the free surfaces upstream and downstream of the dam is 165 ft. Water is to be supplied at a rate of 7000 lbm/s. The electrical power generated is measured to be 1546 hp and the generator efficiency of 92%. (1 hp = 550 lbf.ft/s).
Determine:
a. the overall efficiency of the turbine-generator.
b. the mechanical efficiency of the turbine.
Answer:
a) 75%
b) 82%
Explanation:
Assumptions:
[tex]\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}[/tex]
Properties: The density of water [tex]\delta = 1000 kg/m^3[/tex]
Conversions:
[tex]165 \ ft \ to \ meters = 50 m \\ \\7000 \ lbm/s \ to \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \ to \ kilowatt = 1166 kw \\ \\[/tex]
Analysis:
Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:
[tex]e_{mech_{in}} - e_{mech_{out}} = gh - 0[/tex]
Then;
[tex]gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}[/tex]
gh = 0.491 kJ/kg
[tex]\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg[/tex]
= 1559 kW
Therefore; the overall efficiency is:
[tex]\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}[/tex]
[tex]= \dfrac{1166 \ kW}{1559 \ kW}[/tex]
= 0.75
= 75%
b) mechanical efficiency of the turbine:
[tex]\eta_{turbine- generator} = \eta_{turbine}\times \eta_{generator}[/tex]
thus;
[tex]\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%[/tex]
1. Two aluminium strips and a steel strip are to be bonded together to form a composite bar. The modulus of elasticity of steel is 200 GPa and 75 GPa for aluminium. The allowable normal stress in steel is 220 MPa and 100 MPa in aluminium. Determine the largest permissible bending moment when the composite bar is bent about horizontal axis. a
Answer:
1.933 KN-M
Explanation:
Determine the largest permissible bending moment when the composite bar is bent horizontally
Given data :
modulus of elasticity of steel = 200 GPa
modulus of elasticity of aluminum = 75 GPa
Allowable stress for steel = 220 MPa
Allowable stress for Aluminum = 100 MPa
a = 10 mm
First step
determine moment of resistance when steel reaches its max permissible stress
next : determine moment of resistance when Aluminum reaches its max permissible stress
Finally Largest permissible bending moment of the composite Bar = 1.933 KN-M
attached below is a detailed solution
what's nested piezometer?
Explanation:
Nested piezometers indicate an upward flow if the elevation of the top of the water in the piezometer tube that penetrates the aquifer to the deeper point is greater than the elevation of the water in the shallower tube
Allura Red Moles
Question of the Day {QOD}:
If you decided to drink massive amounts of Cherry Kool-AidTM on a dare, would you die from Allura Red toxicity or water intoxication first?
5. Answer the Question of the Day. Support your answer with specific evidence from your experimentation including LD50 calculations. In your discussion, calculate both the amount of grams and moles of Allura red in Kool-Aid as well as the volume of Kool-Aid you would need to ingest to reach the median lethal dose of Allura red
The concentration of Allura red used was 1.89x10^-4 M
Kool aid absorbance (after dilution: 1mL kool aid, 9mL water): 0.453
Kool aid concentration: 1.97x10^-4 M
LD50 water: 90g/kg
LD50 (rats and mice): 6,000 - 10,000 mg/kg
Answer:
Die of intoxication by water first
Explanation:
We assume that the weight of the man is 154.35 pounds which is 70 kg
LD50 water = 90g per kg
Maximum concentration = 90x70
= 6300grams
Convert grams to liters
6300/100
= 6.3 litres
From here we get amount of kool aid
6.3 x 1.97x10^-4
= 1.24x10^-3
= 1.24grams
1.24 grams is below 420 kool aid is lower than LD50 with about 6 grams for 1 kg (6x70kg = 420). So 420 is lethal dose. But 1.24 is less than this so the man has to die of water intoxication first.
The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel. Determine the magnitude of force P so that the rigid beam tilts 0.015∘.
Answer:
Magnitude of force P = 25715.1517 N
Explanation:
Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.
To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.
Proof -
Given that,
Diameter = 12 mm = 0.012 m
Length = 0.6 m
[tex]\theta[/tex] = 0.015°
Youngs modulus of elasticity of 34 stainless steel is 193 GPa
Now,
By applying the conditions of equilibrium, we have
∑fₓ = 0, ∑[tex]f_{y}[/tex] = 0, ∑M = 0
If ∑[tex]M_{A}[/tex] = 0
⇒[tex]F_{BC}[/tex]×0.9 - P × 0.6 = 0
⇒[tex]F_{BC}[/tex]×3 - P × 2 = 0
⇒[tex]F_{BC}[/tex] = [tex]\frac{2P}{3}[/tex]
If ∑[tex]M_{B}[/tex] = 0
⇒[tex]F_{AD}[/tex]×0.9 = P × 0.3
⇒[tex]F_{AD}[/tex] ×3 = P
⇒[tex]F_{AD}[/tex] = [tex]\frac{P}{3}[/tex]
Now,
Area, A = [tex]\frac{\pi }{4} X (0.012)^{2}[/tex] = 1.3097 × 10⁻⁴ m²
We know that,
Change in Length , [tex]\delta[/tex] = [tex]\frac{P l}{A E}[/tex]
Now,
[tex]\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }[/tex] = 9.1626 × 10⁻⁹ P
[tex]\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }[/tex] = 1.83253 × 10⁻⁸ P
Given that,
[tex]\theta[/tex] = 0.015°
⇒[tex]\theta[/tex] = 2.618 × 10⁻⁴ rad
So,
[tex]\theta = \frac{\delta_{BC} - \delta_{AD}}{0.9}[/tex]
⇒2.618 × 10⁻⁴ = ( 1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9
⇒P = 25715.1517 N
∴ we get
Magnitude of force P = 25715.1517 N
The magnitude of Force P is; P = 25715.15 N
What is the magnitude of the force?
If we draw a free body diagram of the rigid beam system, then for beam AB we can take moments in the following manner;
Taking moments about point A, we have;
(F_bc * 0.9) - P(0.6) = 0
F_bc = ²/₃P
Taking moments about B gives;
P(0.3) - F_ad * 0.9 = 0
F_ad = ¹/₃P
Normal stress for BC is;
σ_bc = F_bc/A_bc
σ_bc = (²/₃P)/(π * 0.006²)
σ_bc = (²/₃P)/(1.131 × 10⁻⁴) N/m²
σ_ad = (¹/₃P)/(π * 0.006²)
σ_ad = (¹/₃P)/(1.131 × 10⁻⁴) N/m²
We know that;
Elongation is; ΔL = PL/AE = (P/A) * (L/E)
Where E for 304 stainless steel is 193 GPa = 193 × 10⁹ Pa
Thus;
ΔL_bc = (²/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))
ΔL_bc = 1.83253P × 10⁻⁸
Likewise;
ΔL_ad = (¹/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))
ΔL_ad = 9.1626P × 10⁻⁹ m
Converting the beam tilt angle from degrees to radians gives;
θ = 0.015° = 0.00026179939 rads
Using small angle analysis, we can say that;
θ = (ΔL_bc - ΔL_ad)/36
θ = P((1.83253P × 10⁻⁸) - (9.1626P × 10⁻⁹))/36
Solving gives P = 25715.15 N
Read more about Magnitude of Force at; https://brainly.com/question/13370981
An array of eight aluminum alloy long fins, each 3 mm wide, 0.4 mm thick, and 40 mm long, is used to cool a transistor. When the base is at 340 K and the ambient air is at 300 K, how much power do they dissipate if the combined convection and radiation heat transient coefficient is estimated to be 8 W/m2K? The alloy has a conductivity of 175 W/mKand the heat transfer from the tip is negligible.
Answer:
0.08704 W
Explanation:
converting the mm to m (1000mm = 1m)
cross-sectional area of the fins, Ac = (0.003) (0.0004) = 0.0000012m^2
The wetted perimeter of the cross-section, P = 2 (0.003 + 0.0004) = 0.0068m
Thickness of solid in direction of heat flow, B^2 = (heat transient coefficient, h) (The wetted perimeter of the cross-section, P) ÷ (Thermal conductivity, k) (cross-sectional area of the fins, Ac)
B^2 = (8 W/m2K)(0.0068m) ÷ (175 W/mK)(0.0000012m^2)
=259.0476m^-2
B= square root of the result
B = 16.09m^-1
we now look for:
The Coordinate, x = B, multiplied by Length, L
x = (16.09m^-1) (0.04m) = 0.6436
finding the side area of a fin = P multiplied by Length, L
= 0.0068m X 0.04m = 0.000272m^2
Neglecting inefficiency, assuming the fins are all 100% efficient, the power they would dissipate =
h, Heat-transfer coefficient (PL) (temperature of at the base - temperature at the ambient air)
= (8) (0.000272m^2)(340 K- 300k)
= 0.08704 W
A 5-m-long, 4-m-high tank contains 2.5-m-deep water when not in motion and is open to the atmosphere through a vent in the middle. The tank is now accelerated to the right on a level surface at 2 m/s2. Determine the maximum gage pressure in the tank. Mark that point at the interior bottom of the tank. Draw the free surface at this acceleration.
Answer: hello your question lacks the required diagram attached below is the diagram
answer : 29528.1 N/m^2
Explanation:
Given data :
dimensions of tank :
Length = 5-m
Width = 4-m
Depth = 2.5-m
acceleration of tank = 2m/s^2
Determine the maximum gage pressure in the tank
Pa ( pressure at point A ) = s*g*h1
= 10^3 * 9.81 * 3.01
= 29528.1 N/m^2
attached below is the remaining part of the solution
Leland wants to work in a Production career operating heavy machinery. Which type of education or training should Leland seek?
a bachelor’s degree then a master’s degree
vocational school certificate or master’s degree
on-the-job training or vocational school certificate
associate’s degree then a bachelor’s degree
Answer:
it is indeed C
Explanation:
Answer:
c
Explanation:
A power winch is designed to raise a 4,961 N load 10 meter in 2 minutes. The winch is designed with a 283 mm diameter drum taking up the wire lifting the load. The drum will be connected to a 9 rpm motor through a gearbox. What is the minimum torque (Nm) that the motor shaft coupling should be designed to transmit
Answer:
T = 438.87 N.m
Explanation:
The power required to raise the 4961 N load in 10 meters for 2 minutes is:
[tex]P = \dfrac{4961*10}{2*60}\\ \\ P = 413.42 Nm/sec[/tex]
P = Torque × W
[tex]413.42 = T \times \dfrac{2 * \pi*9}{60}[/tex]
[tex]413.42 = T \times0.942 \\ \\ T = \dfrac{413.42}{0.942}[/tex]
[tex]T = \dfrac{413.42}{0.942}[/tex]
T = 438.87 N.m
Derive the next state equations for each type (D, T, SR, and JK) of basic memory element. The next state equation is a symbolic equation describing the next state (Q ) as a function of the inputs (D,T,SR, or JK) and state (Q). In order to determine the next state equations for a a JK memory element, build a 3-variable Kmap with Q, J, and K as the inputs. The entries in the Kmap should be Q . Solving this Kmap will yield the next state equation. Show all work for full credit.
Answer:
Attached below is the derived next state equations
Explanation:
Attached below is the derived next state equations
used for the solution of the given problem.
what’s your favorite color of the alphabet? if so, what fruit is it?
Hi sorry I need points I'm New
Answer: My favorite color is Red favorite letter C favorite fruit watermelon.
When choosing a respirator for your job, you must conduct a
test.
Automotive gas turbines have been under development for decades but have not been commonly used in automobiles. Yet helicopters routinely use gas turbines. Explore why different types of engines are used in these respective applications. Compare selection factors such as performance, power-to-weight ratio, space requirements, fuel availability, and environmental impact.
Required:
Summarize your findings in a report with at least three references.
Answer:
Gas turbines in Helicopters require lesser space.
Explanation:
[1] In terms of Space Requirements:
The gas used in helicopters requires lesser space as compared to Automotive gas turbines. The gas in automobile have higher thermal efficiency.
[2]. In terms of Environmental impact:
The occurrence of environmental solution is very slim when used in helicopters' engines.
[3]. In terms of power-to-weight ratio:
The vibrations in engines of helicopters make it to have lesser efficiency as compared to automobile.
[4]. In terms of Fuel availability:
Fuel is available. Automobile can make use of gas as fuel.
Which one of the following answer options are your employers responsibility
Where are your answer options?
Answer:
Implement a hazard communication program
Explanation: i took the quiz
What two units of measurement are used to classify engine sizes?