Suppose you have two arrays: Arr1 and Arr2. Arr1 will be sorted values. For each element v in Arr2, you need to write a pseudo code that will print the number of elements in Arr1 that is less than or equal to v. For example: suppose you are given two arrays of size 5 and 3 respectively. 5 3 [size of the arrays] Arr1 = 1 3 5 7 9 Arr2 = 6 4 8 The output should be 3 2 4 Explanation: Firstly, you should search how many numbers are there in Arr1 which are less than 6. There are 1, 3, 5 which are less than 6 (total 3 numbers). Therefore, the answer for 6 will be 3. After that, you will do the same thing for 4 and 8 and output the corresponding answers which are 2 and 4. Your searching method should not take more than O (log n) time. Sample Input Sample Output 5 5 1 1 2 2 5 3 1 4 1 5 4 2 4 2 5
Answer:
The algorithm is as follows:
1. Declare Arr1 and Arr2
2. Get Input for Arr1 and Arr2
3. Initialize count to 0
4. For i in Arr2
4.1 For j in Arr1:
4.1.1 If i > j Then
4.1.1.1 count = count + 1
4.2 End j loop
4.3 Print count
4.4 count = 0
4.5 End i loop
5. End
Explanation:
This declares both arrays
1. Declare Arr1 and Arr2
This gets input for both arrays
2. Get Input for Arr1 and Arr2
This initializes count to 0
3. Initialize count to 0
This iterates through Arr2
4. For i in Arr2
This iterates through Arr1 (An inner loop)
4.1 For j in Arr1:
This checks if current element is greater than current element in Arr1
4.1.1 If i > j Then
If yes, count is incremented by 1
4.1.1.1 count = count + 1
This ends the inner loop
4.2 End j loop
Print count and set count to 0
4.3 Print count
4.4 count = 0
End the outer loop
4.5 End i loop
End the algorithm
5. End
A series of end-milling cuts is currently used to produce an aluminum part that is an aircraft component. The purpose of the machining operation is to remove 95% of the part weight to create a structural frame. A total of 4.0 min is lost during the milling cycle due to tool repositioning. The part has a length = 1.6 m, width=0 m, and mm. The operation uses a four-tooth indexable-insert end mill with mm at a cutting speed 600 m/min, chip load 0.15 mm/tooth, and average cross-sectional area of cut-240 mm. High-speed machining has to replace the conventional milling process. The same chip load and average area of cut will be used, but the cutting speed will be increased to 3600 m/min , and the time lost for tool repositioning will be reduced to 2.0 min.
Determine :
(a) The cycle time of the current milling operation and
(b) The cycle time of the proposed HSM operation.
(c) Is this part a good candidate for high-speed machining?
Answer:
what-
Explanation:
I dont understand
When you see a school bus stop with its stop sign extended or its lights flashing, you must stop, except for cases when:
Answer:
Whenever you approach a school bus from any direction, which has stopped to pick up or let off passengers while operating its flashing red lights, you must stop your vehicle at least 25 feet from the school bus. The only time you do not have to stop is when you are on the other side of a divided highway. You must stay stopped until the bus has started again or the bus driver stops operating the flashing red lights.
List down 10 items made from plastic in your home. Describe why each item is made from plastic.
Answer:
each item is made from plastic are:
water bottlebucketjuggallonplastic bags as poly thin bagto cover wirehandle of knife,pressure cooker,etc plastic roofdustbinplastic plates ,cupsExplanation:
Items made from plastic in our home are
water bottlesjugbucketfood wrappersdisposable plastic cupshand sanitizer bottlesplastic grocery bagsstraws fruit basket Dustbinitems and their uses
Water bottles are used to drink waterfood wrappers are used to wrap the different types of food itemsplastic grocery bags are used to carry grocery itemshand sanitizer bottles are used to store sanitizerfruit basket is used to store and carry fruits .hope it is helpful to you
The 5-lb cylinder is falling from A with a speed vA = 10 ft/sonto the platform. Determine the maximum displacementof the platform, caused by the collision. The spring has anunstretched length of 1.75 ft and is originally kept incompression by the 1-ft long cables attached to the platform. Neglect the mass of the platform and spring and any energy lost during the collision.
Answer: hello some pictorial details related to your question is missing attached below is the missing detail
answer : 0.0735 ft
Explanation:
weight of cylinder = 5 IB
speed ( v ) = 10 ft/s
Calculate the maximum displacement of platform
Express the initial energy of the system as
potential energy of cylinder + potential energy of spring + Kinetic energy of cylinder
= mgh + 1/2 ky^2 + 1/2 mv^2
y = initial spring compression = 1.75 - 1 = 0.75
hence the initial energy can be expressed as
= 5 ( 3 + x ) + 1/2 (400)(0.75)^2 + 1/2 ( 5/32.2) (10)^2 ------------ ( 1 )
Next : determine the Final energy of the system
Final energy of the system = potential energy of the system
= 1/2 k ( x + 0.75 )^2 ----- ( 2 )
Equating equations ( 1 ) and ( 2 ) to determine the value of x
5 ( 3 + x ) + 1/2 (400)(0.75)^2 + 1/2 ( 5/32.2) (10)^2 = 1/2 k ( x + 0.75 )^2
solve for x
x ( max displace of platform ) = 0.0735 ft
What type of sensor is a crankshaft position sensor?
2 Select the correct answer from each drop-down menu. Identify the jobs with the type of risks they include. Michael is a/n engineer with the risk of inhaling very small crystalline particles that can cause silicosis, pulmonary diseases, or lung cancer. Tommy is a/n engineer with the risk of exposure to poisonous pollutants in production units when a certain reaction does not follow h. textile Reset Next chemical automobile
Answer:
2nd option is correct
Explanation:
as Tommy is an engineer only exposed to poisonous pollutants while Michael is an engineer exposed to small crystalline particles that can cause lung cancer and other diseases
I Hope You Got Your Answer
Blood pressure is conventionally measured in the dimensions of millimeters in a column of mercury, and the readings are expressed as two numbers, for example, 120 and 80. The first number is called the systolic value, and it is the maximum pressure developed as the heart contracts. The second number (called the diastolic reading) is the pressure when the heart is at rest. In the units of kPa and psi, what is the difference in pressure between the given systolic and diastolic readings? The density of mercury is 13.54 Mg/m3.
Answer:
- the difference in pressure between the given systolic and diastolic readings in KPa is 5.313 KPa
- the difference in pressure between the given systolic and diastolic readings in psi is 0.77 psi
Explanation:
Given the data in the question;
blood pressure reading = 120 and 80 { systolic and diastolic }
To determine the difference in pressure between the two readings, we use the equation as follows;
change in pressure ΔP = p × g × h
where p is mercury density, g is acceleration due to gravity and h is difference of height in mercury column.
Frist,
difference of height in mercury column h = 120 - 80 = 40 mm = 0.04 m
given that; The density of mercury is 13.54 Mg/m³ = 13.54 × 10³ kg/m³
Not that Mg is Megagrams not Milligrams }
we know that g = 9.81 m/s²
so we substitute into our equation;
change in pressure ΔP = (13.54 × 10³) × 9.81 × 0.04
ΔP = 5313.096 kg/m-s² ≈ 5313.096 N/m²
ΔP = 5.313 KPa
Therefore, the difference in pressure between the given systolic and diastolic readings in KPa is 5.313 KPa.
In psi,
ΔP = 5.313 KPa
ΔP = 5313 Pa
ΔP = 5313 pa × ( 1.45 × 10⁻⁴ psi / 1 Pa )
ΔP = 0.770385 psi ≈ 0.77 psi
Therefore, the difference in pressure between the given systolic and diastolic readings in psi is 0.77 psi
Three single-phase, 10 kVA, 460/120 V, 60 Hz transformers are connected to form a three-phase 460/208 V transformer bank. The equivalent impedance of each transformer referred to HV side is 1.0 +j2.0 Ω. The transformer delivers 20 kW at 0.8 pf leading. Answer the following questions:
(a) Draw a schematic diagram showing the transformer connection.
(b) Determine the magnitude of transformer primary and secondary winding currents.
(c) Determine the primary voltage magnitude for this operating condition. Determine the voltage regulation
Answer:
A) attached below
B) I₁ = 18.1 A , I₂ = 69.39 A
C) V( magnitude) = 454.5 ∠ 5.04° V , Voltage regulation = ≈ -1.2%
Explanation:
A) Schematic diagram attached below
attached below
B) magnitude of primary and secondary winding currents
I₂ ( secondary current ) = P / √3 * VL * cos∅ ---------- ( 1 )
VL = Line voltage = 208
cos∅ ( power factor ) = 0.8
P = 20 * 10^3 watts
insert values into equation 1
I₂ = 69.39 A
I₁ ( primary current ) = I₂V2 / V1
I₁ = ( 69.39 * 120 ) / 460 = 18.1 A
C ) Calculate the Primary voltage magnitude and the Voltage regulation
V(magnitude ) = Vp + ( I₁ ∠∅ ) Req ( 1 + j2 = 2.24 ∠63.43° )
= 460 + ( 18.1 * cos^-1 (0.8) ) ( 1 + j2 )
= 460 + 40.544 ∠ 100.3°
∴ V( magnitude) = 454.5 ∠ 5.04° V
Voltage regulation
= ((Vmag - V1) / V1 )) * 100
= (( 454.5 - 460 / 460 )) * 100
= -1.195 % ≈ -1.2%