According to 14 CFR 91, helicopters may be flown at altitudes less than those prescribed for minimum safe altitude if the operation is conducted without hazard to persons or property on the surface.
This regulation acknowledges the unique capabilities of helicopters, allowing them to operate closer to the ground compared to fixed-wing aircraft.
This flexibility is due to the helicopter's ability to hover, maneuver, and land vertically, which can be advantageous in various situations such as emergency response, search and rescue, or utility work. However, it is crucial for pilots to ensure that their operations do not pose any risk to individuals or property on the ground.
When operating at lower altitudes, helicopter pilots must exercise increased vigilance and situational awareness to identify and avoid potential hazards. These can include obstacles like power lines, buildings, and trees, as well as the possibility of causing disturbance to individuals or animals on the ground.
In summary, 14 CFR 91 permits helicopters to fly at altitudes below the minimum safe altitude, as long as the operation is conducted without hazard to persons or property on the surface. This allowance provides flexibility for helicopter pilots in specific scenarios while emphasizing the importance of safety and risk mitigation.
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c) Three AC voltages are as follows:
e1 = 80 sin ωt volts;
e2 = 60 sin (ωt + π/2) volts;
e3 = 100 sin (ωt – π/3) volts.
Find the resultant e of these three voltages and express it in the form
Em sin (ωt ± φ). [5 MARKS]
When this resultant voltage is applied to a circuit consisting of a 10-Ω resistor and a capacitance of 17.3 Ω reactance connected in series, find an expression for the instantaneous value of the current flowing, expressed in the same form. [4 MARKS]
Answer:
E = 132.69 sin(ωt -11.56)
i(t) = 6.64 sin (ωt +48.44) A
Explanation:
given data
e1 = 80 sin ωt volts 80 < 0
e2 = 60 sin (ωt + π/2) volts 60 < 90
e3 = 100 sin (ωt – π/3) volts 100 < -60
solution
resultant will be = e2 + e2 + e3
E = 80 < 0 + 60 < 90 + 100 < -60
\(\bar E\) = 80 + j60 + 50 - j50\(\sqrt{3}\)
\(\bar E\) = 130 + (-j26.60)
\(\bar E\) = 132.69 that is less than -11.56
so
E = 132.69 sin(ωt -11.56)
and
as we have given the impedance
z = (10-j17.3)Ω
z = 19.982 < -60
and
i(t) = \(\frac{132.69}{19.982}\) sin(ωt -11.56 + 60)
i(t) = 6.64 sin (ωt +48.44) A
when must a competent person conduct an inspection of a sling?
Before each use and at least once every 12 months, a sling must be inspected by a qualified person.
What is a sling?An arm that has been hurt is supported with a sling, a bandage. Applying a sling Hold the arm still above and below the injury.
To create the sling, place the triangle bandage beneath the hurt arm and over the shoulder that is not hurt. Sling ends should be tied at the side of the neck.
A competent person should also examine a sling following any incident that might compromise its integrity, such as dropping it or exposing it to a potentially damaging environment.
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A competent person inspects a sling before each use and upon change in application. Periodic inspections are conducted based on frequency and conditions of use. If wear or damage is detected, the sling is removed from service.
Explanation:A
competent person
must conduct an inspection of a sling before each shift begins and whenever there is a change in the sling's application that could affect its operation. Additionally, a
periodic inspection
should be performed at intervals based on frequency of use, severity of service conditions, nature of lifts being made, and experience gained from the service life of slings used in similar circumstances. Factors to consider include visible damage, wear, deformation and increase in length. If any of these are detected, the sling should be removed from service.
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Which of the following sensors is used to provide suspension control module with feedback regarding vehicle cornering forces?
Question content area bottom
Part 1
A.
lateral accelerometer sensor
B.
vehicle speed sensor
C.
yaw rate sensor
D.
pressure sensor
Answer:
A
Explanation:
which of the following domains in bloom's taxonomy encompasses students' attitudes, values, and emotions?
The person's attitudes, feelings, and emotions are all part of the affective domain. Receiving phenomena is one of the categories of the affective realm.
What does Bloom's taxonomy's psychomotor domain entail?Physical movement, coordination, and the application of motor skills are all part of the psychomotor domain. These abilities must be developed via practice and are evaluated based on their executional speed, accuracy, distance, procedures, or techniques.
Which of the following domains in Bloom's taxonomy best represents the attitudes, values, and feelings of students?The emotional domain. The way we react with things emotionally, such as feelings, values, admiration, enthusiasms, motivations, and attitudes, is included in the affective domain (Krathwohl, Bloom, Masia, 1973). From the most basic behavior to the most complicated, the five main categories are listed.
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How are new slip rings attached to the rotor core during alternator rebuilding?
The new slip rings is been attached to the rotor core during alternator rebuilding by b. Soldered on.
What is the process of Soldering ?Soldering is the use of Solder which is a metal alloy consisting tin and lead and is the way of joining different types of metals together by melting solder.
This is been used on new slip rings to get it attached to the rotor core during alternator rebuilding .
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CHECK THE COMPLETE QUESTION BELOW:
How are new slip rings attached to the rotor core during alternator rebuilding?
a. Machine clamped
b. Soldered on
c. Crimped on
d. epoxy glue
vital role of maritime english among seaferers
Answer:
uehgeg7djw7heidiisosowiuisiejei2k
Give one example of each of the following structures. Your answer should include relevant
sketches, loads (no numerical values required), equations, and calculations. (a) A statically determinate beam. (b) A statically determinate truss or frame. (c) A truss statically indeterminate to degree 1. (d) A beam statically indeterminate to degree 2. (e) An unstable beam with two spans
Answer:
The above three conditions are commonly referred to as the equations of equilibrium for planar structures. ∑ Fx and ∑ Fy are the summation of the x and y components of all the forces acting on the structure, and ∑ Mz is the summation of the couple moments and the moments of all the forces about an axis z, perpendicular to the plane xy of the action of the forces.
Explanation:
how do you fix this code? python
from random import randint
class Character:
def __init__(self):
self.name = ""
self.health = 1
self.health_max = 1
def do_damage(self, enemy):
damage = min(
max(randint(0, self.health) - randint(0, enemy.health), 0),
enemy.health)
enemy.health = enemy.health - damage
if damage == 0: print "%s evades %s's attack." % (enemy.name, self.name)
else: print "%s hurts %s!" % (self.name, enemy.name)
return enemy.health <= 0
The correct code that fixes this bug-filled python code is:
from random import randint
class Character:
def __init__(self):
self.name = ""
self.health = 1
self.health_max = 1
def do_damage(self, enemy):
damage = min(
max(randint(0, self.health) - randint(0, enemy.health), 0),
enemy.health)
enemy.health = enemy.health - damage
if damage == 0:
print("%s evades %s's attack." % (enemy.name, self.name))
else:
print("%s hurts %s!" % (self.name, enemy.name))
return enemy.health <= 0
class Enemy(Character):
def __init__(self, player):
Character.__init__(self)
self.name = 'a goblin'
self.health = randint(1, player.health)
class Player(Character):
def __init__(self):
Character.__init__(self)
self.state = 'normal'
self.health = 10
self.health_max = 10
def quit(self):
print(
"%s can't find the way back home, and dies of starvation.\nR.I.P." % self.name)
self.health = 0
def help(self): print(Commands.keys())
def status(self): print("%s's health: %d/%d" %
(self.name, self.health, self.health_max))
def tired(self):
print("%s feels tired." % self.name)
self.health = max(1, self.health - 1)
def rest(self):
if self.state != 'normal':
print("%s can't rest now!" % self.name)
self.enemy_attacks()
else:
print("%s rests." % self.name)
if randint(0, 1):
self.enemy = Enemy(self)
print("%s is rudely awakened by %s!" %
(self.name, self.enemy.name))
self.state = 'fight'
self.enemy_attacks()
else:
if self.health < self.health_max:
self.health = self.health + 1
else:
print("%s slept too much." % self.name)
self.health = self.health - 1
def explore(self):
if self.state != 'normal':
print("%s is too busy right now!" % self.name)
self.enemy_attacks()
else:
print("%s explores a twisty passage." % self.name)
if randint(0, 1):
self.enemy = Enemy(self)
print("%s encounters %s!" % (self.name, self.enemy.name))
self.state = 'fight'
else:
if randint(0, 1):
self.tired()
def flee(self):
if self.state != 'fight':
print("%s runs in circles for a while." % self.name)
self.tired()
else:
if randint(1, self.health + 5) > randint(1, self.enemy.health):
print("%s flees from %s." % (self.name, self.enemy.name))
self.enemy = None
self.state = 'normal'
else:
print("%s couldn't escape from %s!" %
(self.name, self.enemy.name))
self.enemy_attacks()
def attack(self):
if self.state != 'fight':
print("%s swats the air, without notable results." % self.name)
self.tired()
else:
if self.do_damage(self.enemy):
print("%s executes %s!" % (self.name, self.enemy.name))
self.enemy = None
self.state = 'normal'
if randint(0, self.health) < 10:
self.health = self.health + 1
self.health_max = self.health_max + 1
print("%s feels stronger!" % self.name)
else:
self.enemy_attacks()
def enemy_attacks(self):
if self.enemy.do_damage(self):
print("%s was slaughtered by %s!!!\nR.I.P." %
(self.name, self.enemy.name))
Commands = {
'quit': Player.quit,
'help': Player.help,
'status': Player.status,
'rest': Player.rest,
'explore': Player.explore,
'flee': Player.flee,
'attack': Player.attack,
}
p = Player()
p.name = input("What is your character's name? ")
print("(type help to get a list of actions)\n")
print("%s enters a dark cave, searching for adventure." % p.name)
while(p.health > 0):
line = input("> ")
args = line.split()
if len(args) > 0:
commandFound = False
for c in Commands.keys():
if args[0] == c[:len(args[0])]:
Commands[c](p)
commandFound = True
break
if not commandFound:
print("%s doesn't understand the suggestion." % p.name)
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Estimate properties and pipe diameter Determine the diameter of a steel pipe that is to carry 2000 gal/min of gasoline with a pressure drop of 5 psi per 100 ft of horizontal pipe. Pressure drop is a function of flow rate, length, diameter, and roughness. Either iterative methods OR equation solvers are necessary to solve implicit problems. Total head is the sum of the pressure, velocity, and elevation. What is the density of gasoline
Answer:
Diameter of pipe is 0.535 ft
Explanation:
see attachment, its works out 1st half
Which of the following reduces friction in an engine A)wear B)drag C)motor oil D)defractionation
It is motor oil, as oil is used to reduce friction
6 A series of vane shear tests has been performed in a stratum of inorganic clay that has a plasticity index of 50. The vane had a diameter of 50 mm and a height of 100 mm. The test results were as follows: Depth (m) Torque at Failure (N-m) 3.4 12.7 4.1 18.1 5.0 15.8 6.6 20.1 Compute the undrained shear strength, su, for each test, then combine this data to determine a single su value for this stratum.
Your job is to perform the steps of MapReduce to calculate a count of the number of squares, stars, circles, hearts and triangles in a dataset.Step 0: Store the dataset across 4 partitions in HDFS.
Note: we have already done one partition for you. Hint: Balance the load, but there is more than on possible "correct" partitioning.
Step 1: Map the data.
Hint: Mapping involves clustering like keys together. Show this in the visual placement of keys within a partition.
Step 2: Sort and Shuffle.
Note: as mentioned in lecture, you don't have to use the same number of nodes in this step as you did before. Let's use three instead. Hint: Balance the load.
Step 3: Reduce to calculate the final counts.
Hint: Fill in the blank lines to finalize the key-value pairs
Answer:
Explanation:
Step 0: Store the dataset across 4 partitions in HDFS.
Partition 1 (already done):
- Partition 1 contains a subset of the dataset.
Partition 2:
- Partition 2 contains another subset of the dataset, balancing the load across the partitions.
Partition 3:
- Partition 3 holds a portion of the dataset to distribute the data evenly.
Partition 4:
- Partition 4 stores the remaining part of the dataset, ensuring load balance across all partitions.
Step 1: Map the data.
In this step, we cluster similar keys together within each partition.
Partition 1:
- Map function clusters keys: squares, stars, circles, hearts, triangles.
Partition 2:
- Map function clusters keys: squares, stars, circles, hearts, triangles.
Partition 3:
- Map function clusters keys: squares, stars, circles, hearts, triangles.
Partition 4:
- Map function clusters keys: squares, stars, circles, hearts, triangles.
Step 2: Sort and Shuffle.
In this step, we sort and shuffle the data across three nodes while maintaining load balance.
Node 1:
- Receives and processes data from Partition 1, Partition 2, and Partition 3.
Node 2:
- Receives and processes data from Partition 1, Partition 2, and Partition 4.
Node 3:
- Receives and processes data from Partition 3 and Partition 4.
Step 3: Reduce to calculate the final counts.
In this step, we perform the final reduction to calculate the counts for each key.
Node 1:
- Reduce function calculates the count of squares: count_squares = _____.
- Reduce function calculates the count of stars: count_stars = _____.
Node 2:
- Reduce function calculates the count of circles: count_circles = _____.
- Reduce function calculates the count of hearts: count_hearts = _____.
Node 3:
- Reduce function calculates the count of triangles: count_triangles = _____.
By following these steps of MapReduce, we can calculate the final counts for squares, stars, circles, hearts, and triangles in the dataset.
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Effective operation of roller chain and gear couplings
requires proper
A. lubrication and cooling.
B. lubrication and balance.
C. alignment and balance.
D. alignment and lubrication.
Effective operation of roller chain and gear couplings requires proper alignment and lubrication.
Roller chain coupling are compact and flexible couplings. They are used for low to moderate speed/torque applications. For effective operation they need to be maintained by applying lubrication and aligning.
Gear couplings are used for high speed and torque applications. For effective performance they need to be lubricated frequently and proper alignment is needed.
Hence effective operation of roller chain and gear couplings requires proper lubrication and alignment.
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A boy riding a bicycle can be modeled as a spring-mass-damper system with an equivalent weight, stiffness, and damping constant of 800 N, 50,000 N/m, and 1,000 N-s/m, respectively. The differential setting of the concrete blocks on the road caused the level surface to decrease suddenly, as indicated in Figure shown. If the speed of the bicycie is 5 m/s (18 km/hr), determine the displacement of the boy in the vertical direction Assume that the bicycle is free of vertical vibration before encountering the step change in the vertical displacement. Hint: Assume that the bicycie and the boy fall as a rigid body by 5 cm at point A. Thus the mass will be subjected to an initial downward displacement of 5 cm (t-0 assumed at point A). ed keq cm
The youngster has moved about 0.065 meters in the vertical direction (6.5 cm).
What was the purpose of the query?The front wheel will drop down as the bicycle hits the abrupt level change, giving the entire system a startling jolt. The initial downward displacement of the mass is 5 cm since it is expected that the bicycle and the youngster will fall as a rigid body at point A by a distance of 5 cm. With the provided parameters of weight, stiffness, damping constant, and beginning velocity, we can calculate the displacement of the boy in the vertical direction as roughly 0.065 m (6.5 cm) using the equation for the displacement of a spring-mass-damper system.
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A river has an average rate of water flow of 59.6 M3/s. This river has three tributaries, tributary A, B and C, which account for 36%, 47% and 17% of water flow respectively. How much water is discharged in 30 minutes from tributary B?
Answer:
50421.6 m³
Explanation:
The river has an average rate of water flow of 59.6 m³/s.
Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:
Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s
The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken
time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds
The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³
what feature on some network switches can be used to detect faked arp messages?
Some network switches have a feature called Dynamic ARP Inspection (DAI) that can be used to detect faked ARP messages. DAI helps protect the network by validating ARP packets and ensuring the content loaded in them is accurate and legitimate.
On some network switches, a function known as "content loaded" can be used to recognise false ARP messages. With the use of this functionality, the switch can check for any anomalies or discrepancies in the contents of packets that are going through it, including ARP messages, that might point to a phoney or malicious message. The switch can protect the network's security and integrity by employing this capability to assist stop ARP spoofing attacks.In a network, Address Resolution Protocol (ARP) packets are verified by Dynamic ARP Inspection (DAI), a security feature. A network administrator can intercept, log, and delete ARP messages that have faulty MAC address to IP address bindings using DAI. With this feature, the network is shielded against some "man-in-the-middle" assaults.
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What are the disadvantages of genetic engineering towards traditional farming practices
Genetic engineering involves the manipulation of an organism’s genetic material, which alters its traits and properties. Its use in agriculture has led to the development of genetically modified crops, animals, and microorganisms, which have been used to increase food production and boost economic growth.
Environmental Risks-The release of genetically engineered organisms into the environment poses a significant risk to biodiversity. These organisms may interbreed with their natural counterparts, leading to the extinction of certain species.Increased Cost-Genetically modified seeds are more expensive than traditional seeds. Human Health Risks-There is a lack of knowledge about the long-term effects of genetically modified organisms on human health.
Ethical Concerns-The modification of an organism’s genetic material has raised ethical concerns, particularly regarding animal welfare. In conclusion, genetic engineering presents some negative effects on traditional farming practices. The use of genetically modified crops has increased production and food security, but it also poses risks to human health, animal welfare, and the environment. As such, careful consideration of the risks and benefits of genetic engineering is required to ensure that it does not pose harm to traditional farming practices and the environment.
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Think of five people with high self-esteem and write their names and how they show high self - esteem.
The functionality of a building can be affected by three types of non-structural damage: (1) damage to partitions, (2) ceiling damage and (3) damage to MEP equipment (e.g., HVAC, water pipes). If any two of these three types of non-structural damage exceeds their specified limits under earthquake shaking, then the building will not be functional. P= partition damage exceeds limit given that an earthquake occurs C= ceiling damage exceeds limit given that an earthquake occurs M= MEP equipment damage exceeds limit given that an earthquake occurs Based on historical data, the following probabilities are known P(P)=0.20
P(C)=0.05
P(M)=0.10
P(P∣C)=0.40
P(PM)=0.05
P(C∪M)=0.145
P(PCM)=0.001
Answer the following questions: a) [5 pts] Are the events P and M independent. To get full credit, you must provide calculations to support your answer. b) [5 pts] Are the events C and M independent. To get full credit, you must provide calculations to support your answer. c) [15 pts] What is the probability that the building will not be functional ( F
ˉ
) if an earthquake occurs. Use the following relationship to solve this problem: P( F
ˉ
)=P(PC)+P(PM)+P(CM)−2P(PCM)
a) To determine if events P and M are independent, we need to check if the probability of P occurring is affected by the occurrence of M, and vice versa. The probability that the building will not be functional if an earthquake occurs is 0.448, or 44.8%.
If P and M are independent, then P(P∩M) = P(P) * P(M).
Given:
P(P) = 0.20 (probability of partition damage exceeding limit)
P(M) = 0.10 (probability of MEP equipment damage exceeding limit)
P(PM) = 0.05 (probability of both partition and MEP equipment damage exceeding limits)
P(P∩M) = P(PM) = 0.05
P(P) * P(M) = 0.20 * 0.10 = 0.02
Since P(P∩M) = P(P) * P(M), we can conclude that events P and M are independent.
b)Similarly, to determine if events C and M are independent, we need to check if the probability of C occurring is affected by the occurrence of M, and vice versa.
If C and M are independent, then P(C∩M) = P(C) * P(M).
Given:
P(C) = 0.05 (probability of ceiling damage exceeding limit)
P(M) = 0.10 (probability of MEP equipment damage exceeding limit)
P(CM) = 0.001 (probability of both ceiling and MEP equipment damage exceeding limits)
P(C∩M) = P(CM) = 0.001
P(C) * P(M) = 0.05 * 0.10 = 0.005
Since P(C∩M) = P(C) * P(M), we can conclude that events C and M are independent.
c) To calculate the probability that the building will not be functional (Fˉ) if an earthquake occurs, we can use the provided relationship:
P(Fˉ) = P(PC) + P(PM) + P(CM) - 2P(PCM)
Given:
P(PC) = 0.40 (probability of partition and ceiling damage both exceeding limits)
P(PM) = 0.05 (probability of partition and MEP equipment damage both exceeding limits)
P(CM) = 0.001 (probability of ceiling and MEP equipment damage both exceeding limits)
P(PCM) = 0.001 (probability of all three types of damage exceeding limits)
P(Fˉ) = 0.40 + 0.05 + 0.001 - 2 * 0.001
P(Fˉ) = 0.448
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A construction company distributes its products by trucks loaded at its loading station. A backacter in conjunction with trucks are used for this purpose. If it was found out that on an average of 12 trucks per hour arrived and the average loading time was 3 minutes for each truck. A truck must queue until it is loaded. The backacter’s daily all-in rate is GH¢ 1000 and that of the truck is GH¢ 400.
a) Compute the operating characteristics: L, Lq, W, Wq, and P.
b) The company is considering replacing the backacter with a bigger one which will have an average service rate of 1.5 minutes to serve trucks waiting to have their schedules improved. As a manager, would you recommend the new backacter if the daily all-in rate is GH¢ 1300.
c) The site management is considering whether to deploy an extra backwater to assist the existing one. The daily all-in-rate and efficiency of the new backwater is assumed to be the same as that of the existing backwater. Should the additional backwater be deployed?
Answer:
a) \(L = 1.5\)
\(L_q = 0.9\)
\(W = \dfrac{1 }{8 } \, hour\)
\(W_q = \dfrac{3}{40 } \, hour\)
\(P = \dfrac{3}{5 }\)
b) The new backacter should be recommended
c) The additional backacter should not be deployed
Explanation:
a) The required parameters are;
L = The number of customers available
\(L = \dfrac{\lambda }{\mu -\lambda }\)
μ = Service rate
\(L_q\) = The number of customers waiting in line
\(L_q = p\times L\)
W = The time spent waiting including being served
\(W = \dfrac{1 }{\mu -\lambda }\)
\(W_q\) = The time spent waiting in line
\(W_q = P \times W\)
P = The system utilization
\(P = \dfrac{\lambda }{\mu }\)
From the information given;
λ = 12 trucks/hour
μ = 3 min/truck = 60/3 truck/hour = 20 truck/hour
Plugging in the above values, we have;
\(L = \dfrac{12 }{20 -12 } = \dfrac{12 }{8 } = 1.5\)
\(P = \dfrac{12 }{20 } = \dfrac{3}{5 }\)
\(L_q = \dfrac{3}{5 } \times \dfrac{3}{2 } = \dfrac{9}{10 } = 0.9\)
\(W = \dfrac{1 }{20 -12 } = \dfrac{1 }{8 } \ hour\)
\(W_q = \dfrac{3}{5 } \times \dfrac{1}{8 } = \dfrac{3}{40 } \, hour\)
(b) The service rate with the new backacter = 1.5 minutes/truck which is thus;
μ = 60/1.5 trucks/hour = 40 trucks/hour
\(P = \dfrac{12 }{40 } = \dfrac{3}{10}\)
\(W = \dfrac{1 }{40 -12 } = \dfrac{1 }{38 } \, hour\)
\(W_q = \dfrac{3}{10 } \times \dfrac{1}{38 } = \dfrac{3}{380 } \, hour\)
λ = 12 trucks/hour
Total cost = \(mC_s + \lambda WC_w\)
m = 1
\(C_s\) = GH¢ = 1300
\(C_w\) = 400
Total cost with the old backacter is given as follows;
\(1 \times 1000 + 12 \times \dfrac{1}{8} \times 400 = \$ 1,600.00\)
Total cost with the new backacter is given as follows;
\(1 \times 1300 + 12 \times \dfrac{1}{38} \times 400 = \$ 1,426.32\)
The new backacter will reduce the total costs, therefore, the new backacter is recommended.
c)
Here μ = 3 min/ 2 trucks = 2×60/3 truck/hour = 40 truck/hour
\(\therefore W = \dfrac{1 }{40 -12 } = \dfrac{1 }{38 } \, hour\)
Total cost with the one backacter is given as follows;
\(1 \times 1000 + 12 \times \dfrac{1}{8} \times 400 = \$ 1,600.00\)
Total cost with two backacters is given as follows;
\(2 \times 1000 + 12 \times \dfrac{1}{38} \times 400 = \$ 2,126.32\)
The additional backacter will increase the total costs, therefore, it should not be deployed.
2. A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit
of surface area is twice as great for the for the hemisphere as it is for the cylindrical side wall. Determine the dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction.
An extremum is a point where the function has its highest or lowest value, and at which the slope is zero.
The dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum is as follows;
Height of cylinder = 2 × Radius of cylinderReason:
The given parameters are;
Form of silo = Cylinder surmounted by a hemisphere
Cost of construction of hemisphere per square unit = 2 × The cost of of construction of the cylindrical side wall
Required:
The dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum
Solution:
The fixed volume of the silo, V, can be expressed as follows;
\(V = \pi \cdot r^2 \cdot h + \dfrac{2}{3} \cdot \pi \cdot r^3\)
\(h = \dfrac{V - \dfrac{2}{3} \cdot \pi \cdot r^3 }{\pi \cdot r^2 }\)Where;
h = The height of the cylinder
r = The radius of the cylinder
The surface area is, Surface Area = 2·π·r·h + 2·π·r²
The cost, C = 2·π·r·h + 2×2·π·r² = 2·π·r·h + 4·π·r²
Therefore;
\(C = 2 \times \pi \times r\times \dfrac{V - \dfrac{2}{3} \cdot \pi \cdot r^3 }{\pi \cdot r^2 } + 4 \cdot \pi \cdot r^2 = \dfrac{8 \cdot r^3 \cdot \pi + 6 \cdot V}{3 \cdot r}\)
\(C = \dfrac{8 \cdot r^3 \cdot \pi + 6 \cdot V}{3 \cdot r}\)At the minimum value, we have;
\(\dfrac{dC}{dr} =0 = \dfrac{d}{dr} \left(\dfrac{8 \cdot r^3 \cdot \pi + 6 \cdot V}{3 \cdot r} \right) = \dfrac{16 \cdot r^3 \cdot \pi - 6 \cdot V}{3 \cdot r^2}\)Which gives;
16·π·r³ = 6·V
\(V = \dfrac{16 \cdot \pi \cdot r^3}{6} = \dfrac{8 \cdot \pi \cdot r^3}{3}\)Which gives;
\(h = \dfrac{\dfrac{8 \cdot \pi \cdot r^3}{3} - \dfrac{2}{3} \cdot \pi \cdot r^3 }{\pi \cdot r^2 } = 2 \cdot r\)
h = 2·r
The height of the silo, h = 2 × The radius, 2
Therefore, the dimensions to be used if the volume is fixed and the cost is to be kept to a minimum is, height, h = 2 times the radius
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How do designers use the elements of design?
Shape, colour, space, form, line, value, and texture are some of the basic components of any visual design, and they are known as design elements. To generate a picture, graphic designers employ design components.
That can provoke a variety of emotions, express a certain atmosphere, or lead the attention in a particular way.
Any visual piece of art, whether it be for interior design, a logo, an advertising, or online design, should take into account a number of graphic design components. Design's fundamental components are:
1. Use of colour to set the tone for your composition. Color is the feeling that people experience when light waves strike an item and reflect back to the optic nerve in their eyes. In order to illustrate and characterise a subject, artists and designers employ colour. Designers utilise colour to convey feel, light, depth, and viewpoint. Designers apply colour theory and the colour wheel.
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Which is a squeezing force? *
O Compression
Tension
Shear
O Torsion
O Bending
Answer:
compression
Explanation:
Compression is when you push two things together, which squeezes them
Using the table below, draw the network diagram and answer the questions. 1. How many paths are in the network, and what are they? (1 mark) 2. What is the critical path and its duration? (1 mark) 3. What is the float on activity U? (1 mark) 4. What is the impact to the project if activity B takes three weeks longer than planned? (1 mark) Activity Predecessor Estimate in Weeks Start - 0 C Start 6 B Start 4 P Start 3 A C,B,P 7 U P 4 T A 2 R A 3 N U 6 End T,R,N 9 Question 2 Your group was tasked to produce a research manual and listed the activities that must be completed, and the time estimates for each activity as shown in the table below: Note: You are required to use the last non-zero digit of your group leader’s student ID number for A,B,C,D,E. (where relevant based on your group size. Crash value cannot be ‘0’. Thus use the next bigger non zero digit from your ID number. Activity Immediate Predecessor Time (weeks) Cost (RM) Normal Crash Normal Crash A - a *a-1 700 800 B - 2 or b *b-1 1200 1900 C A 8 or c *c-2 1600 2200 D B 5 or d *d-2 1500 2300 E C 6 or e *e-2 2000 2800 F C,D 6 5 800 1200 G E,F 2 1 500 1000 (a) Draw a project network diagram. (1 mark) (b) Find the critical path.(1 mark) (c) What is the normal expected project completion time? (1 mark) (d) What is the total project cost using the normal time? (1 mark) (e) If the project is to be completed 2 weeks faster, determine which activities to crash at a minimum cost. What additional cost is incurred? What is the total cost? (3 marks) Question 3 Diddy Corp would like to develop an order quantity and reorder policy that would minimize their total costs of their inventory of printers. The probability distribution for demand for ace drill on a daily basis is shown below. Demand 0 1 2 3 4 5
The number of paths in the network is six.
The paths are:
C - A - T - N - EndC - A - T - R - N - EndC - B - P - U - N - EndC - B - P - U - T - R - N - EndC - B - P - A - T - N - EndC - B - P - A - T - R - N - End
2. The critical path is C - A - T - N - End with a duration of 24 weeks.
3. The float on activity U is 1 week.
4. If activity B takes three weeks longer than planned, it will affect the duration of path 3 and the entire project. It will elongate the project duration to 29 weeks.
2. b) The critical path is A-C-E-G with a duration of 18 weeks.
(c) The normal expected project completion time is 18 weeks.
(d) The total project cost using normal time is RM 7600.
(e) To complete the project two weeks faster, activities C and E can be crashed for the minimum cost.
The additional cost incurred will be RM 600.
The total cost is RM 8200.
3 The inventory cost that will minimize the total cost of inventory is obtained using the formula,
TAC = IC + OC + HCS
Where TAC
= Total Annual CostIC = Inventory CostOC = Ordering CostHCS = Holding and Carrying Cost.
The probability distribution is shown below. Demand 0 1 2 3 4 5Probability 0.20 0.15 0.25 0.20 0.10 0.10
To determine the optimal order quantity and reorder policy that would minimize total cost,
the EOQ (Economic Order Quantity) model will be used.
The EOQ can be determined using the formula,
Q = √((2DCO)/CH')
Where Q = Economic Order Quantity
D = Demand'
C0 = Cost per order
H = Holding cost
C = IC per unit
Let’s assume the cost per unit, C = RM 500,
the holding cost, H = 0.2,
and demand, D = 3 units per day.
Hence,CO = RM 800 (C0 = Cost per order, which is the same as ordering cost)
Then,Q = √((2 x 3 x RM 800)/RM 100)Q = 24 units per order
Then, the reorder level can be determined using the formula,
ROL = d * LWhere L = Lead time
ROL = 3 x 5 = 15 units
The total annual cost can be determined by substituting the values into the formula,
TAC = IC + OC + HCS
where IC = 0.5 x RM 500
= RM 250OC
= (365/EOQ) x C0OC
= (365/24) x RM 800OC
= RM 12133.33HCS
= (Q/2) x H x dHCS
= (24/2) x 0.2 x 3HCS
= RM 7.20
TAC = RM 250 + RM 12133.33 + RM 7.20TAC = RM 12390.53
Hence, the order quantity that will minimize the total cost of inventory is 24 units.
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a 227-ft length of canal is to be lined with concrete for erosion control. with 12% allowance for waste and overexcavation, the volume (yd3 ) of concrete that must be delivered is most nearly:
Answer:
First, we need to calculate the volume of concrete needed to line the canal. The volume of a cylinder is given by the formula:V = πr^2hWhere V is the volume, r is the radius of the cylinder, and h is the height (length) of the cylinder.Since we are lining a canal, the cross-sectional shape of the canal is a rectangle, not a circle. However, we can still use the formula for the volume of a cylinder by treating the width of the canal as the radius and the height of the canal as the length.In this case, the width of the canal is w and the height is h, so the volume of the concrete needed to line the canal is:V = πw^2hWe are given that the length of the canal is 227 ft, so h = 227 ft. We are not given the width of the canal, so we cannot use the formula to calculate the volume of concrete needed.Next, we need to calculate the volume of concrete needed with the 12% allowance for waste and overexcavation. The volume of concrete needed with the allowance is the original volume plus 12% of the original volume.The volume of concrete needed with the allowance can be calculated using the following formula:V' = V + (V * 12%)Where V' is the volume of concrete needed with the allowance and V is the original volume.Since we do not know the original volume, we cannot use this formula to calculate the volume of concrete needed.To solve this problem, we need to know the width of the canal in order to calculate the volume of concrete needed. Without this information, it is not possible to determine the volume of concrete that must be delivered.
Explanation:
It is desired to obtain 500 VAR reactive power from 230 Vrms 50 Hz 1.5 KVAR reactor. What should be the angle of the AC to AC converter to be used? Calculate the THD of the current drawn from the mains (consider up to the 12th harmonic)?
Answer:
14.5° ; THD % = 3.873 × 100 = 387.3%.
Explanation:
Okay, in this question we are given the following parameters or data or information which is going to assist us in solving the question efficiently and they are;
(1). "500 VAR reactive power from 230 Vrms 50 Hz 1.5 KVAR reactor".
(2). Consideration of up to 12th harmonic.
So, let us delve right into the solution to the question above;
Step one: Calculate the Irms and Irms(12th) by using the formula for the equation below;
Irms = reactive power /Vrms = 500/230 = 2.174 A.
Irms(12th) = 1.5 × 10^3/ 12 × 230 = 0.543 A.
Step two: Calculate the THD.
Before the Calculation of the THD, there is the need to determine the value for the dissociation factor, h.
h = Irms(12th)/Irms = 0.543/ 2.174 = 0.25.
Thus, THD = [1/ (h)^2 - 1 ] ^1/2. = 3.873.
THD % = 3.873 × 100 = 387.3%.
Step four: angle AC - Ac converter
theta = sin^-1 (1.5 × 10^3/ 12 × 500) = 14.5°.
the maximum length of a double section ladder is
The maximum length of a double section ladder is 48 feet. It is important to adhere to this limit to ensure safety and stability when using ladders for any task.
Double section ladders are popular for their versatility and ability to extend to greater heights. They are commonly used in various industries, such as construction and maintenance. However, it is crucial to note that exceeding the maximum length limit can pose serious safety risks. Ladders that are too long may become unstable and cause accidents, which can lead to injuries or even fatalities. Therefore, it is important to carefully select the appropriate ladder length based on the task at hand and ensure that safety guidelines are followed at all times.
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What quantity measures the effect of change?
control variable
independent variable
relative variable
dependent variable
Answer:
The Dependent Variable is the Effect, Its value depends on changes from the Independent Variable
Hope this Helps
A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation of 0.009 in. The modulus of elasticity of the material is most nearly:
Answer:
E = 8.83 kips
Explanation:
First, we determine the stress on the rod:
\(\sigma = \frac{F}{A}\\\\\)
where,
σ = stress = ?
F = Force Applied = 1300 lb
A = Cross-sectional Area of rod = 0.5\(\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2\)
Therefore,
\(\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips\)
Now, we determine the strain:
\(strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon = 7.5\ x\ 10^{-4}\)
Now, the modulus of elasticity (E) is given as:
\(E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}\)
E = 8.83 kips
A mass of 39 lbm of helium undergoes a process from an initial state of 50 ft3/ lbm and 60°F to a final state of 20 ft3/lbm and 240°F. Determine the entropy change of helium during this process assuming the process is reversible. The gas constant of helium is R = 0.4961 Btu/lbm·R. The constant volume specific heat of helium at room temperature is cv = 0.753 Btu/lbm·R. The entropy change of helium during this process is Btu/R.
The entropy change of helium during this process is -9 Btu/R
How to solve for the enthropy changeThe formula is given as
\(Change in S = m[Cvln\frac{T_{2} }{T_{1} } + Rln\frac{V_{2} }{V_{1} }\)
Where the variables are M = mass = 39
Cv = specific heat of helium = 0.753
t2 = 240 + 460
T1 = 60 + 460
R = gas constant = 0.4961
V1 = initial state = 50
V2 = final state = 20
we would have to put these values in the formula that we have above
\(39[0.753\frac{240+460}{60+460} +0.4961ln\frac{20}{50}]\)
= -9 Btu/R.
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