Answer:
what is that?
Explanation:
Determine the magnitude of force F so that the resultant force of the three forces is as small as possible. What is the magnitude of the resultant force?
Answer:
hello your question is incomplete attached below is the missing diagram
answer : 7.87 KN
Explanation:
The magnitude of the resultant force = 7.87 KN
from fig 1
First step :
calculate the resultant force along the x- direction
Frx = ∑Fx = +8 kN - Fcos45° - 14cos 30°
= -4.1243 - 0.7071 F
next :
calculate the resultant force along the y- direction
Fry = ∑Fy = -Fsin45° + 14sin30°
= -0.7071 F + 7
attached below is the remaining part of the detailed solution
For heat transfer purposes, a standing man can be mod-eled as a 30-cm-diameter, 170-cm-long vertical cylinderwith both the top and bottom surfaces insulated and with theside surface at an average temperature of 34°C. For a con-vection heat transfer coefficient of 15 W/m2·K, determinethe rate of heat loss from this man by convection in still airat 20°C. What would your answer
Answer:
Rate of Heat Loss = 336 W
Explanation:
First, we will find the surface area of the cylinder that is modelled as the man:
[tex]Area = A = (2\pi r)(l)[/tex]
where,
r = radius of cylinder = 30 cm/2 = 15 cm = 0.15 m
l = length of cylinder = 170 cm = 1.7 m
Therefore,
[tex]A = 2\pi(0.15\ m)(1.7\ m)\\A = 1.6\ m^2[/tex]
Now, we will calculate the rate of heat loss:
[tex]Rate\ of\ Heat\ Loss = hA\Delta T[/tex]
where,
h = convective heat tranfer coefficient = 15 W/m²K
ΔT = Temperature difference = 34°C - 20°C = 14°C
Therefore,
[tex]Rate\ of\ Heat\ Loss = (15\ W/m^2K)(1.6\ m^2)(14\ K)\\[/tex]
Rate of Heat Loss = 336 W
A fill covering a wide area is to be placed at the surface of this profile. The fill has a total unit weight of 20 kN/m^3 and is 3 m thick. Assume that the data for the sample at 7.0 m are representative of the entire clay profile. Also assume that the clay is heavily over consolidated and that the danse sands at the surface of the profile are so stiff that they do not contribute to the settlement. Find the settlement of the surface due to compression of the clay layer
Answer:
hello your question lacks some information attached below is the complete question with the required information
answer : 81.63 mm
Explanation:
settlement of the surface due to compression of the clay ( new consolidated )
= 81.63 mm
attached below is a detailed solution to the given problem
Truckco manufactures two types of trucks: 1 and 2. Each truck must go through the painting shop and assembly shop. If the painting shop were completely devoted to painting Type 1 trucks, then 800 per day could be painted; if the painting shop were completely devoted to painting Type 2 trucks, then 700 per day could be painted. If the assembly shop were completely devoted to assembling truck 1 engines, then 1,500 per day could be assembled; if the assembly shop were completely devoted to assembling truck 2 engines, then 1,200 per day could be assembled. Each Type 1 truck contributes $300 to profit; each Type 2 truck contributes $500. How much capacity in percent does a single truck of each type uses at each shop?
Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is being charged, or a cold gas if it is being discharged. In a charging process, heat transfer from the hot gas increases thermal energy stored within the colder spheres; during discharge, the stored energy decreases as heat is transferred from the warmer spheres to the cooler gas. Consider a packed bed of 75-mm-diameter aluminum spheres (p = 2,700 kg/m^3; c = 950 J/kg*K; k = 240 W/m*K) and a charging process for which gas enters the storage unit at a temperature of 300 degrees C. The initial temperature of the spheres is Ti = 25 degrees C and the convection heat transfer coefficient is h = 75 W/m^2*K.
a. How long does it take a sphere near the inlet of the system to accumulate 90% of the maximum possible thermal energy? What is the corresponding temperature at the center of the sphere?
b. Is there any advantage to using copper (p = 8,900 kg/m^3; c = 380 J/kg*K; k = 390W/m*K) instead of aluminum?
c. Consider the same packed bed operating conditions, but with Pyrex (p = 2,200 kg/m^3; c = 840 J/kg*K; k = 1.4 W/m*K) used instead of aluminum. How long does it take a sphere near the inlet of the system to accumulate 90% of the maximum possible thermal energy? What is the corresponding temperature at the center of the sphere?
Answer:
A) i) 984.32 sec
ii) 272.497° C
B) It has an advantage
C) attached below
Explanation:
Given data :
P = 2700 Kg/m^3
c = 950 J/kg*k
k = 240 W/m*K
Temp at which gas enters the storage unit = 300° C
Ti ( initial temp of sphere ) = 25°C
convection heat transfer coefficient ( h ) = 75 W/m^2*k
A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere
First step determine the Biot Number
characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125
Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3
Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method
attached below is a detailed solution of the given problem
B) The physical properties are copper
Pcu = 8900kg/m^3)
Cp.cu = 380 J/kg.k
It has an advantage over Aluminum
C) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere
Given that:
P = 2200 Kg/m^3
c = 840 J/kg*k
k = 1.4 W/m*K
Water is pumped from a lake to a storage tank 18 m above at a rate of 70 L/s while consuming 20.4 kW of electric power. Disregard any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of the pump-motor unit (5-point), and (b) the pressure difference between the inlet and the exit of the pump (5-point).
Two fluids, A and B exchange heat in a counter – current heat exchanger. Fluid A enters at 4200C and has a mass flow rate of 1 kg/s. Fluid B enters at 200C and also has a mass flow rate of 1 kg/s, Effectiveness of heat exchanger is 75%. Determine the heat transfer rate and exit temperature of fluid B.
Answer:
Your question has some missing information below is the missing information
Given that ( specific heat of fluid A = 1 kJ/kg K and specific heat of fluid B = 4 kJ/kg k )
answer : 300 kW , 95°c
Explanation:
Given data:
Fluid A ;
Temperature of Fluid ( Th1 ) = 420° C
mass flow rate (mh) = 1 kg/s
Fluid B :
Temperature ( Tc1) = 20° C
mass flow rate ( mc ) = 1 kg/s
effectiveness of heat exchanger = 75% = 0.75
Determine the heat transfer rate and exit temperature of fluid B
Cph = 1000 J/kgk
Cpc = 4000 J/Kgk
Given that the exit temperatures of both fluids are not given we will apply the NTU will be used to determine the heat transfer rate and exit temperature of fluid B
exit temp of fluid B = 95°C
heat transfer = 300 kW
attached below is a the detailed solution
A steady state filtration process is used to separate silicon dioxide (sand) from water. The stream to be treated has a flow rate of 50 kg/min and contains 0.22% sand by mass. The filter has a cross-sectional area of 9 square meters and successfully filters out 90% of the input sand by mass. As the filter is used, a cake forms, which we will assume is pure sand (SG of sand = 2.25). The filter needs to be replaced once this cake has a thickness of 0.25 meters.
Required:
How long can a new filter be used before it needs to be replaced, in units of days?
Answer:
3.19 days
Explanation:
Given data :
stream flow rate = 50 kg/min
stream contains ; 0.22% sand by mass
Cross sectional area of filter = 9 m^2
Filter successfully filters out 90% of the input sand by mass
SG = 2.25
thickness of cake formed = 0.25 meters
Determine how long a new filter can be used before replacement
Given that for every 1 minute 5.43*10^-5 m thickness of sand layer(cake) forms on the filter and the replacement of filter is done once the cake thickness = 0.25 meters
To determine the number of days ( X ) before replacing filter we apply the relationship below
5.43 * 10^-5 = 1 min
0.25 m = X
hence ; X = 0.25 / (5.43 * 10^-5 ) = 4604.051 minutes ≈ 3.19 days
Attached below is the beginning part of the detailed solution
What is the importance of ethics in emerging technologies?
Explanation:
the Ethics of emerging Technology can only make use of speculative data about future products,uses and impacts.
Past evidence shows that when a customer complains of an out-of-orderphone there is an 8% chance that the problem is with the inside wiring. During a 1-month period,100 complains are lodged. Assume that there have been no wide-scale problems that could beexpected to affect many phones at once, and that, for this reason, these failures are consideredto be independent.
Required:
a. Find the expected number of failures due to a problem with the inside wiring.
b. Find the probability that at least 10 failures are due to a problem with the inside wiring.
c. Would it be unusual if at most 5 were due to problems with the inside wiring? Explain, based on the probability of this occurring.
Answer:
a. The expected number of failures due to a problem with inside wiring is 8 failures
b. The probability that at least 10 failures are due to inside wiring is approximately 0.176
c. It will be not unusual
Explanation:
The probability that the problem of an out of order is the inside wiring, P(x) = 8%
The number of complaints in a month period, x = 100
a. The expected number of failures due to a problem with inside wiring, E(x) = x·P(x)
∴ E(x) = 100 × 8% = 8
The expected number of failures due to a problem with inside wiring, E(x) = 8 failures
b. The probability that at least 10, P₁₀, failures are due to inside wiring is given as follows;
The probability of success, P = 0.08, therefore, the probability of failure, q = 1 - 0.08 = 0.92
P = [tex]_nC_r[/tex]·[tex]P^r[/tex]·[tex]q^{n-r}[/tex]
P₀ = ₁₀₀C₀·(0.08)⁰·(0.92)¹⁰⁰ = 0.000239211874657
P₁ = ₁₀₀C₁·(0.08)¹·(0.92)⁹⁹ = 0.00208010325
P₂ = ₁₀₀C₂·(0.08)²·(0.92)⁹⁸ = 0.00895348793
P₃ = ₁₀₀C₃·(0.08)³·(0.92)⁹⁷ = 0.02543309616
P₄ = ₁₀₀C₄·(0.08)⁴·(0.92)⁹⁶ = 0.0536306593
P₅ = ₁₀₀C₅·(0.08)⁵·(0.92)⁹⁵ = 0.0895398833653
P₆ = ₁₀₀C₆·(0.08)⁶·(0.92)⁹⁴ = 0.123279549561
P₇ = ₁₀₀C₃·(0.08)⁷·(0.92)⁹³ = 0.143953759735
P₈ = ₁₀₀C₈·(0.08)⁸·(0.92)⁹² = 0.145518474516
P₉ = ₁₀₀C₉·(0.08)⁹·(0.92)⁹¹ = 0.129349755125
P₁₀ = ₁₀₀C₁₀·(0.08)¹⁰·(0.92)⁹⁰ = 0.10235502362
∴ P₀ + P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + P₈ + P₉ + P₁₀ = 0.82433004464
The probability of at least 10 failures are due problem with the inside wiring = 1 - (P₀ + P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + P₈ + P₉ + P₁₀) = 1 - 0.82433004464 = 0.175666995536
The probability of at least 10 failures are due problem with the inside wiring = 0.175666995536 ≈ 0.176
c. The probability of at most 5 failures are due problem with the inside wiring = P₀ + P₁ + P₂ + P₃ + P₄ + P₅ = 0.179876441908
Therefore, given that probability of at most 5 failures > The probability of 8 failures it will be not unusual since the cause of failure is more (92%) due to other causes which are more likely and therefore increase in the probability that there are fewer failures due inside wiring
Hi all, could you solve this please?
What is the value of the resistance R
Answer:
try to 36v power and take 1a and intersect to 3
Hydraulic fluid is primarily used to
in a hydraulic system.
of
Select one:
a. drive the pump
b. heat up the pump
c. reduce torque
d. transfer energy
Hll
Answer:
hydraulic flute is energy transfer medium in all hydraulic system this simple function is only achieved by flute that does not easily trap gases trapped gas and forming problems should bring a higher level of compressibility to a fluid than its usual at a point to support a very steep pass trekking system
A thin silicon chip and an 8-mm-thick aluminum substrate are separated by a 0.02-mm-thick epoxy joint. The chip and substrate are each 10 mm on a side, and their exposed surfaces are cooled by air, which is at a temperature of 25 C and provides a convection coefficient of 100 W/m2 K. If the chip dissipates 104 W/m2 under normal conditions, will it operate below a maximum allowable temperature of 85 C
Answer:
The chip will operate below a maximum allowable temperature of 85°C
Explanation:
Given data:
8-mm-thick aluminum
0.02 mm-thick epoxy joint
chip and substrate = 10 mm on a side
temperature = 25°C
attached below is a detailed solution
Tc = 75.3 ° c which is less than 85°c . hence the chip will operate below a maximum allowable temperature of 85°C
The dry weather average flow rate for a river is 8.7 m3/s. During dry weather flow, the average COD concentration in the river is 32 mg/L. An industrial source continuously discharges 18,000 m3/d of wastewater contains an average 342 mg/L COD concentration into the river. What is the COD mass loading in the river upstream of the industrial source discharge
Answer:
6156 kg /day
Explanation:
Determine the COD mass loading in the river upstream of the industrial source discharge
Given data:
Flow rate of river = 8.7 m^3/s
Average COD concentration in river = 32 mg/L
Industrial source continuous discharge ( Qw )= 18,000 m^3/d
Yw = 342 mg/l
since :
1 m^3 = 1000 liters
Qw = 18 * 10^6 liters = ( 18 million per day )
Hence the COD mass loading
= Yw * Qw
= 342 * 18 liters
= 6156 kg /day
A demand factor of _____ percent applies to a multifamily dwelling with ten units if the optional calculation method is used.
A 3-phase , 1MVA, 13.8kV/4160V, 60 Hz, transformer with Y-Delta winding connection is supplying a3-phase, 0.75 p.u. load on the 4160V side of the transformer. The load has leading power factor of 0.9. It issupplied by 1 p.u. voltage on the 13.8kV side. The transformer per unit impedance is j0.12 referred to thesecondary side.
a. Find the load impedance.
b. Find the input current on the primary side in real units.
c. Find the input power factor
Answer:
a) 23.89 < -25.84 Ω
b) 31.38 < 25.84 A
c) 0.9323 leading
Explanation:
A) Calculate the load Impedance
current on load side = 0.75 p.u
power factor angle = 25.84
[tex]I_{load}[/tex] = 0.75 < 25.84°
attached below is the remaining part of the solution
B) Find the input current on the primary side in real units
load current in primary = 31.38 < 25.84 A
C) find the input power factor
power factor = 0.9323 leading
attached below is the detailed solution
A signal is assumed to be bandlimited to kHz. It is desired to filter this signal with an ideal bandpass filter that will pass the frequencies between kHz and kHz by a system for processing analog signals composed of a digital filter with frequency response sandwiched between an ideal A/D and an ideal D/A, both operating at sampling interval . 1. Determine the Nyquist sampling frequency, (in kHz), for the input signal. 2. Find the largest sampling period (in s) for which the overall system comprising A/D, digital filter and D/A realize the desired band pass filter.
Answer:
Hello your question is poorly written attached below is the complete question
answer :
1) 60 kHz
2) Tmax = ( 1 / 34000 ) secs
Explanation:
1) Determine the Nyquist sampling frequency, (in kHz), for the input signal.
F(s) = 2 * Fmax
Fmax = 30 kHz ( since Xa(t) is band limited to 30 kHz )
∴ Nyquist sampling frequency ( F(s) ) = 2 * 30 = 60 kHz
2) Determine the largest sampling period (in s) .
Nyquist sampling period = 1 / Fs = ( 1 / 60000 ) s
but there is some aliasing of the input signal ( minimum aliasing frequency > cutoff frequency of filter ) hence we will use the relationship below
= 2π - 2π * T * 30kHz ≥ 2π * T * 4kHz
∴ T ≤ [tex]\frac{1}{34kHz}[/tex]
largest sampling period ( Tmax ) = ( 1 / 34000 ) secs
A company purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%. 1) The inspector randomly picks 20 items from a shipment. What is the probability that there will be at least one defective item among these 20? 2) Suppose that the company receives 10 shipments in a month and the inspector randomly tests 20 devices per shipment. What is the probability that there will be exactly 3 Shipments each containing at least one defective device among the 20 that are selected and tested from the shipment?
Answer:
1) The probability of at least 1 defective is approximately 45.621%
2) The probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is approximately 16.0212%
Explanation:
The given parameters are;
The defective rate of the device = 3%
Therefore, the probability that a selected device will be defective, p = 3/100
The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;
The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97
The probability of 0 defective in 20 = ₂₀C₀(0.03)⁰·(0.97)²⁰ ≈ 0.543794342927
The probability of at least 1 = 1 - The probability of 0 defective in 20
∴ The probability of at least 1 = 1 - 0.543794342927 = 0.45621
The probability of at least 1 defective ≈ 0.45621 = 45.621%
2) The probability of at least 1 defective in a shipment, p ≈ 0.45621
Therefore, the probability of not exactly 1 defective = q = 1 - p
∴ q ≈ 1 - 0.45621 = 0.54379
The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;
P(Exactly 3 with at least 1) = ₁₀C₃(0.45621)³(0.54379)⁷ ≈ 0.160212
Therefore, the probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is 16.0212%
f. Where is the electrical panel located for the west apartment
A parallel circuit has a resistance of 280 and an inductive reactance of 360 02. What's this circuit's impedance?
Answer:
540 W
Explanation:
Answer the question on the image and a brianiest will be given to the person that provided the right answer to it.
Answer:
(a) The distance up the slope the wagon moves before coming to rest is approximately 21.74 m
(b) The distance the wagon comes to rest from the starting point is approximately 12.06 m
(c) The value of 'U' at which the wagon should be propelled if it is to come finally to rest at its starting point is approximately 3.214 m/s (the difference in value can come from calculating processes)
Explanation:
The wagon motion parameters are;
The mass of the wagon, m = 7,200 kg
The initial velocity with which the wagon is projected along the horizontal rail, v = U
The length of the horizontal portion of the rail = 100 m
The angle of inclination of the inclined portion of the rail, θ = sin⁻¹(0.01)
The exerted frictional resistance to motion of the rail, [tex]F_f[/tex] = 140 N
∴ θ = sin⁻¹(0.01)
The work done by the frictional force on the horizontal portion of the rail = 140 N × 100 m = 14,000 J
(a) If U = 3 m/s, we have;
Kinetic energy = 1/2·m·v²
The initial kinetic energy of the wagon, K.E. is given with the known parameters as follows;
K.E. = 1/2 × 7,200 kg × (3 m/s)² = 32,400 J
The energy, E, required to move a distance, 'd', up the slope is given as follows;
E = [tex]F_f[/tex] × d + m·g·h
Where;
[tex]F_f[/tex] = The friction force = 140 N
m = The mass of the wagon = 7,200 kg
g = The acceleration due to gravity ≈ 9.81 m/s²
h = The height reached = d × sin(θ) = d × 0.01
Therefore;
E = 140 N × d₁ + 7,200 kg × 9.81 m/s² × d₁ × 0.01 = 846.32 N × d
The energy, [tex]E_{NET \ horizontal}[/tex], remaining from the horizontal portion of the rail is given as follows;
[tex]E_{NET \ horizontal}[/tex] = Initial kinetic energy of the wagon - Work done on frictional resistance on the horizontal portion of the rail
∴ [tex]E_{NET \ horizontal}[/tex] = 32,400 J - 14,000 J = 18,400 J
[tex]E_{NET \ horizontal}[/tex] = 18,400 J
Therefore, for the wagon with energy, [tex]E_{NET \ horizontal}[/tex] to move up the train, we get;
[tex]E_{NET \ horizontal}[/tex] = E
∴ 18,400 J = 846.32N × d
d₁ = 18,400 J/(846.36 N) ≈ 21.7401579 m
d₁ ≈ 21.74 m
The distance up the slope the wagon moves before coming to rest, d₁ ≈ 21.74 m
(b) Given that the initial velocity of the wagon, U = 3 m/s, the distance up the slope the wagon moves before coming to rest is given above as d₁ ≈ 21.74 m
The initial potential energy, PE, of the wagon while at the maximum height up the slope is given as follows;
P.E. = m·g·h = 7,200 kg × 9.81 m/s² × 21.74 × 0.01 m = 15,355.3968 J
The work done, 'W', on the frictional force on the return of the wagon is given as follows;
W = [tex]F_f[/tex] × d₂
Where d₂ = the distance moved by the wagon
By conservation of energy, we have;
P.E. = W
∴ 15,355.3968 = 140 × d₂
d₂ = 15,355.4/140 = 109.681405714
Therefore;
The distance the wagon moves from the maximum height, d₂ ≈ 109.68 m
The distance the wagon comes to rest from the starting point, d₃, is given as follows;
d₃ = Horizontal distance + d₁ - d₂
d₃ = 100 m + 21.74 m - 109.68 m ≈ 12.06 m
The distance the wagon comes to rest from the starting point, d₃ ≈ 12.06 m
(c) For the wagon to come finally to rest at it starting point, we have;
The initial kinetic energy = The total work done
1/2·m·v² = 2 × [tex]F_f[/tex] × d
∴ 1/2 × 7,200 × U² = 2 × 140 × d₄
d₄ = 100 + (1/2·m·U² - 140×100)
(1/2·m·U² - 140×100)/(m·g) = h = d₁ × 0.01
∴ d₁ = (1/2·m·U² - 140×100)/(m·g×0.01)
d₄ = 100 + d₁
∴ d₄ = 100 + (1/2·m·U² - 140×100)/(m·g×0.01)
∴ 1/2 × 7,200 × U² = 2 × 140 × (100 + (1/2 × 7,200 × U² - 140×100)/(7,200 × 9.81 ×0.01))
3,600·U² = 280·(100 + (3,600·U² - 14,000)/706.32)
= 28000 + 280×3,600·U²/706.32 - 280 × 14,000/706.32
= 28000 - 280 × 14,000/706.32 + 1427.11518858·U²
3,600·U² - 1427.11518858·U² = 28000 - 280 × 14,000/706.32
U²·(3,600 - 1427.11518858) = (28000 - 280 × 14,000/706.32)
U² = (28000 - 280 × 14,000/706.32)/(3,600 - 1427.11518858) = 10.3319363649
U = √(10.3319363649) = 3.21433295801
The value of 'U' at which the wagon should be propelled if it is to come finally to rest at its starting point is U ≈ 3.214 m/s
Percentage error = (3.214-3.115)/3.214 × 100 ≈ 3.1% < 5% (Acceptable)
The difference in value can come from difference in calculating methods
Air enters a compressor operating at steady state at 1 bar, 290 K, with a mass flow rate of 0.1 kg/s and exits at 980 K, 10 bar. The velocities at the inlet and exit are 10 m/s and 30 m/s, respectively. The air can be modeled as an ideal gas, and potential energy effects can be neglected. If the magnitude of the power input to the compressor is 77 kW, determine the rate of heat transfer, in kW.
Answer:
7.615 kW
Explanation:
Solution in pen paper form in the attachment section
Write a program that read two integers and display their MOD,VID and their floating-point division in both settings x/y and y/x
e.g 5/3 and 3/5
Answer:
#!/usr/bin/env python
def calculate(x, y):
return {
"MOD": x % y,
"DIV": x/y, # you mean div instead of “VID”, right?
"floating-point division": float(x)/y,
}
def calculateInBothSettings(x, y):
return {
"x/y": calculate(x, y),
"y/x": calculate(y, x),
}
if __name__ == "__main__":
x = int(input("x: "))
y = int(input("y: "))
print(calculateInBothSettings(x, y))
Explanation:
I wrote a python script. Example output:
x: 2
y: 3
{'x/y': {'MOD': 2, 'DIV': 0.6666666666666666, 'floating-point division': 0.6666666666666666}, 'y/x': {'MOD': 1, 'DIV': 1.5, 'floating-point division': 1.5}}
Which of the following components of a dwellings plumbing system blocks sewer gases from entering the building and sickening the occupants
Answer:
first of all I need the choices you have not given any choices to choose
how do we succeed in mechanical engineering?
2 A SQUARE GIVEN
LENGTH OF DIAGONAL = 70 mm
Answer:
Area of square = 2,450 mm²
Explanation:
Given:
Length of diagonal = 70 mm
Find:
Area of square
Computation:
Area of square = diagonal² / 2
Area of square = 70² / 2
Area of square = 4900 / 2
Area of square = 2,450 mm²
A workpiece in the form of a bar 100 mm in diameter is to be turned down to 70 mm diameter for 50 mm of its length. A roughing cut using maximum power and a depth of cut of 12 mm is to be followed by a finishing cut using a feed of 0.1 mm and a cutting speed of 1.5 m/s. It takes 20 s to load and unload the workpiece and 30 s to set the cutting conditions, set the tool at the beginning of the cut and engage the feed. The specific cutting energy for the material is 2.3 GJ/m3 and the lathe has a 3-kW motor and a 70 percent efficiency. Estimate:
Answer:
Hello your question has some missing part below is the missing part
Estimate The matching time for the Finish cut
answer: 79.588 seconds
Explanation:
Calculate the matching time for the Finish cut
Diameter of workpiece before cutting = 100 mm
hence
Diameter of workpiece before Finishing cut ( same as after rough cut )
D2 = D1 - 2 * d = 100 - (2 * 12) = 76mm
step 1 : determine spindle speed
V = [tex]\frac{\pi D_{2}N_{s} }{60}[/tex] -------- ( 1 ). where : D2 = 76 mm , Ns = ? , V = 1.5 ( input values into equation 1 )
therefore Ns = 376.94 rpm
Finally : Determine the matching time
T = [tex]\frac{L + A}{Ft * Ns}[/tex] ---- ( 2 ) . where : Ft = 0.1 mm , Ns = 376.94 rpm, L = 50 mm , A = 0 ( input values into equation 2 )
note : A = allowance length ( not given ) , ft = feed rate
T = 50 / ( 0.1 * 376.94 )
= 1.3265 minutes ≈ 79.588 seconds
A customer complains that the car pulls to the left during hard braking. An inspection shows that the front pads and rear shoes are worn. After resurfacing the rotors and drums and replacing the pads and shoes, a road test reveals that the car still pulls to the left during hard braking. Technician A says the wheel alignment should be checked. Technician B says the tires could be the problem and should be swapped left to right to check whether the direction of the pull changes. Who is correct
Multiple Choice
A client requests Je tong battery life for a tablet. Which step in the design process is this request associated with?
O communicate the solution
O solve the problem
O evaluate the problem
O define the problem
Answer:
evaluate the problem
Explanation:
An engineering design process is a series of steps that engineers use to create products.
Therefore, if a client requests for the battery life for a tablet, the step in the design process the request is associated with is "evaluate the problem".
This is because, by finding out the battery life, you're already evaluating the problem.
1. A drilling operation is to be performed with a 10 mm diameter twist drill in a steel workpart. The hole is a blind hole at a depth of 60 mm and the point angle is 118. The cutting speed is 30 m/min and the feed is 0.25 mm/rev. Determine (a) the cutting time to complete the drilling operation, and (b) metal removal rate during the operation, after the drill bit reaches full diameter.
Answer:
The answer is below
Explanation:
v = velocity = 30 m/min = 30 * 10³ mm/min, D = diameter = 10 mm, f = feed = 0.25 mm/rev, point angle = 118, cutting time = Tm, d = depth = 60 mm
[tex]a)\\N=\frac{v}{\pi D}=\frac{30*10^3}{\pi * 10}=954.9\ rev/min\\\\f_r=Nf =954.9(0.25)=238\ mm/min\\\\A=0.5Dtan(90-\frac{point\ angle}{2} )=0.5*10*tan(90-\frac{118}{2} )=3\ mm\\\\T_m=\frac{(d+A)}{f_r} =\frac{60+3}{238}=0.265 \ s\\\\b)\\metal\ removal\ rate(R_{MR})=0.25\pi D^2f_r\\\\R_{MR}=0.25\pi (10)^2(238)=18692\ mm^3/min[/tex]