Answers
Answer:
C
Explanation:
Related Questions
What is discharge? Derive an expression for it.
Answers
Answer:
Discharge = area of the pipe or channel × velocity of the liquid
Q = Av
Explanation:
The flow rate of a liquid is a measure of the volume of liquid that moves in a certain amount of time. The flow rate depends on the area of the pipe or channel that the liquid is moving through, and the velocity of the liquid. If the liquid is flowing through a pipe, the area is A = πr2, where r is the radius of the pipe. For a rectangle, the area is A = wh where w is the width, and h is the height. The flow rate can be measured in meters cubed per second (m3/s), or in liters per second (L/s). Liters are more common for measures of liquid volume, and 1 m3/s = 1000 L/s.
1. At the end of the day, a bakery gives everything that is unsold to food banks for the needy. If it has 12 apple pies left at the end of a given day, in how many different ways can it distribute these pies among six food banks for the needy?
2. In how many different ways can the bakery distribute the 12 apple pies if each of the six food banks is to receive at least one pie?
Answers
Answer:
The answer ix below
Explanation:
There are 12 apple pies left.
Given that:
n = number of apple pies left = 12
x = number of food banks = 6
1) For the 12 apple pies to be distributed among 6 food banks. The number of ways in which this can be done is:
C(n + x - 1, x - 1) = C(12 + 6 -1, 6 - 1) = C(17, 5) = [tex]\frac{17!}{(17-5)!5!} =\frac{17!}{12!5!}=6188 \ ways[/tex]
12 apple pies can be distributed among 6 food banks in 6188 ways
2) For the 12 apple pies to be distributed among 6 food banks if each food bank must receive one pie, 6 pies would be remaining. The number of ways in which this can be done is:
C((n - x) + x - 1, n - x) = C(12 - 6 + 6 - 1, 12 - 6) = C(11, 6) = [tex]\frac{11!}{(11-6)!6!} =\frac{11!}{6!5!}=462 \ ways[/tex]
12 apple pies can be distributed among 6 food banks if each food bank must receive one pie in 462 ways
In each situation, write a recurrence relation, including base case(s), that describes the recursive structure of the problem. You do not need to solve the recurrence.
a) Let B(n) be the number of length n bit sequences that have no three consecutive 0s (i.e., do not contain the substring "000"). Write a recurrence for B(n).
b) Let S(n) be the number of subsets of {1, 2, ..., n} having the following property: no two elements in the subset are consecutive integers. The empty set with no elements should be included in your count. Write a recurrence for S(n).
c) Say you are tiling a 2 times n rectangle with L-shaped tiles of area 3 (trominoes). To tile the rectangle is to cover it with tiles so that no tiles overlap and every cell is covered by some tile. Let T(n) denote the number of ways to tile the rectangle. Write a recurrence for T(n).
d) A ternary string is like a binary string except it uses three symbols, 0, 1, and 2. For example, 12210021 is a ternary string of length 8. Let T(n) be the number of ternary strings of length n with the property that there is never a 2 appearing anywhere after a 0. For example, 12120110 has this property but 10120012 does not. Write a recurrence for T(n).
Answers
Explanation:
a) Given B(n) is the number of the length 'n' bit sequences which have no three consecutive 0s(i.e., they does not contain substring “000”)
Any bit string which has no 000 should have a 1 in at least one of the 1st three positions. Then, the n we will break all the bit strings by avoiding the 000 by when the 1st 1 occurs. i.e., each of the bit of string of the length of n will avoid 000 falls into the exactly any one of these cases:
1 is followed by the bit string of the length of (n-1) avoiding the 000.
01 is followed by some bit string of the length of (n-2) avoiding a 000.
001 which is followed by any bit string of the length(n-3) avoiding the 000.
Therefore, recurrence is
B(n)=B(n-1)+ B(n-2)+ B(n-3), with B(0)=1, B(1)=2, B(2)=4
Or
B(n)=B(n-1)+ B(n-2)+ B(n-3), with B(1)=2, B(2)=4, B(3)=7
b) Let the S(n)={1,2,3,…,n}. We will say that the subset A of S(n) is very good if A does not have any of the two integers that are consecutive.
For any k, let a(k) be a number of any good subsets of a S(k).
There are two types of good subsets of S(n).
Type 1 of good subsets of S(n) which contain the element n,
Type 2 of good subsets of S(n) which do not contain n.
We will first get an expression for a number of Type 1 of good subsets of the S(n) , where n≥2. This a subset does not contain n-1. So any Type 1 of the good subset of the S(n) is obtainable when adding n to good subset of the S(n-2) . Also, on adding n to a good subset of the S(n-2) , we always get a Type 1 good subset of the S(n). Thus there are exactly as many good Type 1 of subsets of S(n) as there is good subsets of S(n-2) . By the definition there are a(n-2) good subsets of S(n-2) .
A good subset of a S(n) is either Type 1 or Type 2. Thus the number of the good subsets of S(n) is a(n-2)+a(n-1)
We have therefore shown that a(n) = a(n-2)+a(n-1)
c). Here firstly see that if the n is not the multiple of a 3, then there will be no chance to tile the rectangle. And also if n is a multiple of a 3, then there may be two ways to tile the 1st three columns:
And the rest of tilling is the tilling of 2x(n-3) rectangle for which there are T(n-3).
So, the recurrence is
T(n )= {2T(n-3), with T(0)=1 if n=0(mod 3) or 0 else
We could use the base case T(3)=2
So, the following recurrence can be aslo equivalent to T(n)= 2T(n-3), with T(0)=1, T(1)=0,T(2)=0
Or
T(n)= 2T(n-3), with T(1)=0, T(2)=0,T(3)=2
d)A ternary string may be defined as a sequence of some digits, where each digit has either 0,1,or 2.
According to the problem given, we do not have a 2 anywhere after 0, and the dot which represents the binary string of length(n-1) with the property which we cannot use 2 anywhere after we use the 0.
Now for base case, we should note that any of the ternary string which has a length of 1 satisfies the given required property. Hence the recurrence is
T(1)=3
T(n)=2T(n-1)+2^(n-1)
Thermoplastics that exhibit a definite TM point have this kind of crystalline structure:
A. semi-crystalline
B. isotactic
C.amorphous
D. syndiotactic
Answers
Answer:
The correct option is A
Explanation:
Thermoplastics are polymers that can be recycled. They can be melted with heat and moulded into any shape which becomes hard when cooled. The heat applied to it in the process of recycling does not affect its chemical composition.
There are two main classes of thermoplastics; semi-crystalline and amorphous.
An amorphous thermoplastic has a disorderly arranged molecular structure which makes it difficult to have a definite melting temperature (Tm). However, a semi-crystalline thermoplastic has a well arranged molecular structure which makes it melt after a certain/definite amount of heat is absolved. Hence, the correct answer is A
Give the general layout of centrifugal pump.
Answers
Answer:
A centrifugal pump is a rotating machine that pumps liquid by forcing it through a paddle wheel or propeller called an impeller. It is a most common type of industrial pump. By the effect of the rotation of the impeller, the pumped fluid is drawn axially into the pump, then accelerated radially, and finally discharged tangentially.
Explanation:
Suppliers generally offer charts in the plan, which present the various things at the nominal operating point using two main arrangements: the inductors and the balancing pants.
Steam at 175 [C] and 300 kPa flows into a steam turbine at rate of 5.0 kg/sec. Saturated mixture of liquid and steam at 100 kPa flows out at the same rate. The heat loss from the turbine is 1,500 kW. Assuming that 60% of the steam is condensed into liquid at the outlet, how much shaft work can the turbine produce?
Answers
Answer:
The amount of shaft work the turbine cam do per second is 3660.29 kJ
Explanation:
The given parameters are;
Pressure at entry p₁ = 300 kPa
The mass flow rate, [tex]\dot {m}[/tex] = 5.0 kg/sec
The initial temperature, T₁ = 175°C
Therefore;
The enthalpy at 300 kPa and 175°C, h₁ = 2,804 kJ/kg
At the turbine exit, we have;
The pressure at exit, p₂ = 100 kPa
The quality of the steam at exit, in percentage, x₂ = 60%
Therefore, for the enthalpy, h₂ for saturated steam at 100 kPa and quality 60%, we have;
h₂ = 417.436 + 0.6 × 2257.51 = 1771.942 kJ/kg
Heat loss from the turbine, [tex]h_l[/tex]= 1,500 kW
By energy conservation principle we have;
dE/dt = [tex]\dot Q[/tex] - [tex]\dot W[/tex] + ∑[tex]m_i \cdot (h_i[/tex]+ [tex]ke_i[/tex] +[tex]pe_i[/tex]) - ∑[tex]m_e \cdot (h_e[/tex] + [tex]ke_e[/tex] +[tex]pe_e[/tex] )
0 = -[tex]h_l[/tex] - [tex]\dot W[/tex] + [tex]m_i \cdot h_i[/tex] - [tex]m_e \cdot h_e[/tex]
[tex]\dot W[/tex] = [tex]\dot {m}[/tex] × (h₁ - h₂) - [tex]h_l[/tex] = 5.0×(2,804 - 1771.942) - 1500 = 3660.29 kJ/s
The rate of work of the shaft = 3660.29 kJ/s
The amount of shaft work the turbine cam do per second = 3660.29 kJ.
Which phase of DevSecOps emphasizes reliability, performance, and scaling
Answers
Answer:
Test Phase
Explanation:
DevSecOps is an organizational software engineering culture and practice that unifies software development (Dev), security (Sec) and operations (Ops). The main characteristic of DevSecOps is to improve customer outcomes and mission value by automating, monitoring, and applying security at all phases of the software lifecycle.
There are nine phases of the software lifecycle which are: plan, develop, build, test, release, deliver, deploy, operate, and monitor.
The Performance test in the test phase will ensure that applications will perform well under the expected workload. The test focus is on application response time, reliability, resource usage and scalability.
In development phases, database design, development, and testing activities generate database artifacts, which are data models, database schema files, trigger definitions, view definition, test data, test data generation scripts, test scripts, etc. These database artifacts must be under configuration management control. During test phase, database functional test is like application code unit test and functional test to validate the schema, triggers, and data compliance. The non-functional test includes load testing, stress test, and performance test. The security test focuses on vulnerability scan, user authentication and authorization, unauthorized access to data, data encryption, privilege elevation, SQL injection, and denial of service.
In a bid eliminate the vulnerability experienced during the traditional development process, DevSecOps emphasizes reliability, performance and Scaling with the integration of Security phase.
The integration of Security infrastructure into the Development operation(DevOps) process ensures that security challenges experienced by softwares are tackled immediately hence ensuring reliability and reduced vulnerability.
DevSecOps ensures that performance isn't sacrificed for security, hence, softwares are continously checked for security at every phase of the development process during testing.
Therefore, the security phase of the DevSecOps pipeline ensures that satisfactory security and Performance levels are met.
Learn more : https://brainly.com/question/17205994
In which of the following branches of engineering is the practice not restricted?
a) civil engineering.
b) mechanical engineering.
c) nuclear engineering.
Answers
Answer:
a) civil engineering.
Explanation:
Civil engineering is a professional engineering program that deals with the construction, design, and maintenance of all the natural and man-made environments including dams, buildings, railways, and roads.
Civil engineering is the branch of engineering that is the practice not restricted because civil engineer is not restricted to academic profession but practice in designing and construction can make someone a professional civil engineer.
Hence, the correct answer is "a)."
The branch of engineering in which the practice is not restricted is; Civil Engineering
Civil Engineering is a branch of engineering that deals with the design, construction and maintenance of the physical and naturally built environment.
These physically and naturally built environment includes; houses, roads, bridges, airports, railways, canals, dams, sewage systems, pipelines e.t.c
Now, mechanical engineering involves design, production, operation and maintenance of mechanical systems or machineries.
In conclusion, we see that the mechanical branch of engineering is restricted to machinery unlike civil engineering that is not restricted to only buildings but also includes pipelines, bridges, roads, railways, dams, sewage systems e.t.c
Finally, the nuclear engineering branch is also restricted to only nuclear fission and fusion applications.
Read more about civil engineering at; https://brainly.com/question/14235678
What is the minimum amplitude at time zero sine wave??
Answers
Answer:
One cycle for a sine wave is 360o or 2 radians. a) A sine wave with maximum amplitude at time t=0. The amplitude of a sine wave is maximum at the peak of the wave. Case 1: assuming that the wave is starting its cycle at t=0 then there is no phase shift for the wave at time t=0 without considering the amplitude...
Define water hammer. Give four effects of water hammer.
Answers
Answer:
Water Hammer is a knocking sound in an water pipe which occurs when the tap is turned off briskly
Explanation:
A three-phase, 60-Hz, completely transposed 345-kV, 200-km line has two 795,000-cmil (403-mm2) 26/2 ACSR conductors per bundle and the following positive-sequence line constants: z 0.032 + 10.35 /km y j4.2 x 10-6 S/km Full load at the receiving end of the line is 700 MW at 0.99 p.f. leading and at 95% of rated voltage. Assuming a medium-length line, determine the following:
a. ABCD parameters of the nominal π circuit
b. Sending-end voltage Vs, current Is, and real power Ps
c. Percent voltage regulation
d. Transmission-line efficiency at full load
Answers
Answer:
B) sending end voltage : Vs-l-l = 345.8 ∠ 26.14⁰ kv
sending end current : Is = 1.241 ∠ 15.5⁰ KA
real power = 730.5 Mw
C) percent voltage regulation = 8.7%
D) Transmission line efficiency = 95.8%
Explanation:
attached is the detailed solution to the problem
Given data:
l = 200 km
z = 0.032 + j0.35 Ω/km
y = j4.2 * 10^-6 S/km
A) find the total series impedance and shunt admittance
B) sending end voltage : Vs-l-l = 345.8 ∠ 26.14⁰ kv
sending end current : Is = 1.241 ∠ 15.5⁰ KA
real power = 730.5 Mw
C) percent voltage regulation = 8.7%
D) Transmission line efficiency = 95.8%
Students would like to sell cold drinks to raise money for a field trip. They need to keep the drinks cold for 3 hours at the ball game. Which step of the engineering design process are the students at?
Answers
Answer:
Problem definition
Explanation:
The engineering design is a series of step that the engineer gradually take in other to create a functioning product or process. There are many different models of the engineering design process , but they can be shrunk into four basic steps which are
problem definition conceptual designpreliminary designdetailed designdesign communication.The students are within the first step, and the problem here is 'an appropriate means for the students to keep their drinks cold, for 3 hours for the ball game.'
A turbine in a simple ideal Rankine cycle with water as the working fluid operates with an inlet temperature of 500°C and pressure of 600 kPa and an exit temperature of 45°С. Find the turbine work in MW if the mass flow rate of the steam is at 11 kg/s.
Answers
Answer:
the turbine work in MW is [tex]\mathbf{W_t = 10.47299 \ kW}[/tex]
Explanation:
Given that:
the inlet temperature = 500°C
pressure = 600 kPa
exit temperature = 45°С
mass flow rate = 11 kg/s
The objective is to find the turbine work in MW given that the mass flow rate of the steam is at 11 kg/s.
From the steam tables, the data obtained for the enthalpies at the inlet temperature of 500°C and a pressure of 600 kPa is:
[tex]\mathtt{h_1 : 3482.7 \ kJ/kg}[/tex]
The turbine expansion process is also the isentropic process, as such:
Inlet entropy [tex]\mathbf{s_1}[/tex] = Exit entropy [tex]\mathbf{s_2}[/tex]
The data obtained from the steam table for [tex]\mathbf{s_1}[/tex] = 8.003 kJ/kg.K
[tex]s_2 = s_{f2}+ x_2 *s_{fg2}[/tex]
The data obtained from the steam tables for this entities are as follows:
[tex]\mathsf{s_{f2} = 0.638 \ kJ/Kg/K }[/tex]
[tex]\mathtt{s_{fg2}=7.528 \ kJ/kg/K}[/tex]
since [tex]\mathbf{s_1}[/tex] = [tex]\mathbf{s_2}[/tex] = 8.003 kJ/kg.K
Therefore;
[tex]8.003 = 0.638 + x_2 * 7.528[/tex]
[tex]8.003 -0.638 = 7.528x_2[/tex]
[tex]7.365= 7.528x_2[/tex]
[tex]x_ 2= \dfrac{ 7.365}{7.528}[/tex]
[tex]x_2 = 0.978[/tex]
From the steam tables, the data obtained for the enthalpies at the exit temperature of 45°C and a pressure of 600 kPa is:
[tex]h_{f2}[/tex] = 188.4
[tex]h_{fg2[/tex] =2394.9
Thus;
[tex]h_ 2= 188.4 +0.978*2394.9[/tex]
[tex]h_ 2=2530.61[/tex] kJ/kg
The workdone for the turbine process can be computed as:
[tex]\mathsf{W_t = m(h1-h_2)}[/tex]
[tex]\mathsf{W_t = 11(3482.7-2530.61)}[/tex]
[tex]\mathsf{W_t = 11(952.09)}[/tex]
[tex]\mathsf{W_t = 10472.99}[/tex] kW
To MW, we have
[tex]\mathbf{W_t = 10.47299 \ kW}[/tex]
what is advantage and disadvantage of flat top sampling in digital signal trasmission?
Answers
A well-mixed sewage lagoon is receiving 500 m3/d of sewage. The lagoon has a surface area of 10 hectares and a depth of 1 m. The pollutant concentration in the raw sewage is 200 mg/L. The organic matter in the sewage degrades biologically (decays) in the lagoon according to first-order kinetics. The reaction rate constant (decay coefficient) is 0.75 d-1. Assuming no other water losses or gains (evaporation, seepage, or rainfall) and that the lagoon is completely mixed, find the steady-state concentration of the pollutant in the effluent.
Answers
Answer:
Steady-state concentration of the pollutant in the effluent= 1.3245033 mg/L
Explanation:
Given Data:
Amount of sewage received=500 m^3/d
Surface Area= 10 hectares=10*10^4 m^2
Depth=1 m
Pollutant concentration=200 mg/L
Decay coefficient=0.75 d-1
Required:
Steady-state concentration of the pollutant in the effluent= ?
Solution:
Volume=Surface Area * Depth
[tex]Volume=10*10^4 *1\\Volume=10*10^4\ m^3[/tex]
Time to fill the lagoon=[tex]\frac{Volume}{Amount\ received\ per\ day}[/tex]
[tex]Time\ to\ fill\ the\ lagoon=\frac{10*10^4\ m^3}{500\ m^3}\\ Time\ to\ fill\ the\ lagoon= 200\ days[/tex]
Formula for steady State:
[tex]A_t=\frac{A_0}{1+kt}[/tex]
where:
A_t is the steady state concentration
A_0 is the initial concentration
k is the decay constant
t is the time
[tex]A_t=\frac{200\ mg/L}{1+0.75*200}\\ A_t=1.3245033\ mg/L[/tex]
Steady-state concentration of the pollutant in the effluent= 1.3245033 mg/L
what is the total inductance of a circuit that contains two 10 uh inductors connected in a parallel?
Answers
Answer:
5 microhenries
Explanation:
The effective value of inductors in parallel "add" in the same way that resistors in parallel do. The value is the reciprocal of the sum of the reciprocals of the inductances that are in parallel.
10 uH ║ 10 uH = 5 uH
The effective inductance is 5 uH.
Complete the grading of fine aggregate table given below. Plot grading curve and calculate
fineness modulus. Also comment on the type of the grading curve.
Answers
Answer:
Attachment...?
Explanation:
Which 3-input gate outputs a false value only when all inputs are true?
a) AND
b) NAND
c) OR
d) NOR
e) XOR
f) XNOR
Answers
Answer:
(b) NAND
(d) NOR
(f) XNOR
Explanation:
This is best explained using a truth table.
A truth table containing 3 hypothetical inputs and their corresponding outputs using the AND, NAND, OR, NOR, XOR, and XNOR gates, has been attached to this response.
From the table, it can be seen that the NAND, NOR and XNOR gates produce a false value (0) when all inputs are true.
Note:
On the table,
For the AND column, a 1 is being output only when all three inputs are 1. Otherwise, a 0 results.
For the NAND column (which is the negation of the AND), a 0 is being output only when all three inputs are 0. Otherwise, a 1 results.
For the OR column, a 1 is being output when at least one of the three inputs is 1. Otherwise, a 0 results.
For the NOR column (which is the negation of the OR), a 0 is being output when at least one of the three inputs is 0. Otherwise, a 1 results.
For the XOR column, a 1 is being output when the number of inputs that are true (1) is odd. i.e when the number of 1s is 1 or 3. Otherwise, a 0 results.
For the XNOR column (which is the negation of the XOR), a 0 is being output when the number of inputs that are true (1) is odd. i.e when the number of 1s is 1 or 3. Otherwise, a 1 results.
Air flow at 15 m/s past a thin flat plate. Estimate the distance x from the leading edge at which the boundary layer thickness will be (a) 10 cm and (b) 1 mm. Use a transition Reynolds number of 5 x 105.
Answers
Answer:
a) x = 5.7791 m ( this a turbulent flow )
b) x = 0.04 m ( this is a Laminar flow )
Explanation:
For Air, take
p = 1.2 kg/m³ and u = 1.8×10⁻⁵ kg/m³
So for A , x = 10 cm
Guess turbulent flow;
(Sturb / x) = (0.16 / (p∪x / u)^1/7)
we substitute
(0.1/x = 0.16) / ( 1.2 × 15 × x / 1.8×10⁻⁵)^1/7
x = 5.7791 m
CHECK
Re = (1.2 × 15 × 5.7791) / 1.8×10⁻⁵ = 5.779×10⁶
Therefore this a turbulent flow
for B, x = 1mm
Guess Laminar flow
(S-laminar / x) = ( 5 / (Reₓ)^1/2)
x = (Stutb)²p∪ / 23u
x = ((0.001)² × 1.2 × 15) / (25 × 1.8×10⁻⁵)
x = 0.04 m
CHECK
Re = (1.2 × 15 × 0.04) / (1.8×10⁻⁵)
= 40000
Therefore this is a Laminar flow
A hollow shaft of diameter ratio 3/8 (internal dia to outer dia) is to transmit 375 kW power at 100 rpm. The maximum torque being 20% greater than the mean torque. The shear stress is not to exceed 60 N/mm2 and twist in a length of 4m not to exceed 2o. Calculate its external and internal diameters which would satisfy both the above conditions. (G= 0.85 X 105 N/mm2)
Answers
Answer:
External diameter = 158.15 mm mm
Internal diameter = 59.31 mm
Explanation:
We are given;
Diameter ratio; d_i = ⅜d_o
Where d_i is internal diameter and d_o is external diameter
Power;P = 375 KW = 375000 W
Rotational speed;N = 100 rpm
Max torque is 20% greater than mean torque; T_max = 1.2T_avg
Shear stress;τ = 60 N/mm²
Length; L = 4m = 4000 mm
Angle of twist; θ = 2° = 2π/180 radians
Modulus of rigidity;G = 0.85 X 10^(5) N/mm²
Formula for the power transmitted by the shaft is;
P = 2πNT_avg/60
Plugging in the relevant values, we have ;
375000 = 2π × 100T_avg/60
T_avg = (375000 × 60)/(2π × 100) = 35809.862 N.m = 35809862 N.mm
Since T_max = 1.20T_avg
Thus, T_max = 1.20(35809862) = 42971834.4 N.mm
Checking for strength, we'll use;
τ = Tr/J_p
Or since r = d/2
It can be written as;
τ = T(d_o)/2J_p - - - (1)
Where T is T_max
But Polar moment of inertia of hollow shaft is;
J_p = [π(d_o)⁴ - π(d_i)⁴]/32
Now, we are told that d_i = ⅜d_o
Thus;
J_p = [π(d_o)⁴ - π(⅜d_o)⁴]/32
J_p = (π/32) × d_o⁴(1 - 3⁴/8⁴)
J_p = 0.0926 d_o⁴
Plugging this for J_p in eq 1,we have;
τ = T(d_o)/2(0.0926d_o⁴)
Making d_o the subject gives;
d_o³ = T/(2 × 0.0926τ)
Plugging in the relevant values to give;
d_o³ = 42971834.4/(2 × 0.0926 × 60)
d_o³ = 3867155.7235421166
d_o = ∛3867155.7235421166
d_o = 156.96 mm
Thus, d_i = ⅜ × 156.96 = 58.86 mm
Checking for stiffness, we'll use;
T/J_p = Gθ/L
Again T is T_max
Plugging in the relevant values, we have;
42971834.4/0.0926 d_o⁴ = (0.85 × 10^(5) × 2π/180)/4000
464058686.825054/d_o⁴ = 0.7417649321
d_o⁴ = 464058686.825054/0.7417649321
d_o⁴ = 625614216.5028806
d_o = ∜625614216.5028806
d_o = 158.15 mm
d_i = ⅜ × 158.15 = 59.31 mm
So we will pick the highest valies.
Thus;
d_o = 158.15 mm
d_i = 59.31 mm
Class hello Sarah{
Pubic static void(“hello Sarah”)
Name=input(“hello Sarah”)
Print(“hello Sarah”).
Answers
Answer:
class helloSarah {
public static void main(String [] args) {
Name = "Hello Sarah";
System.out.print(Name);
}
}
Explanation:
The question seem incomplete; However, since the program is incorrect, I'll assume the question is to correct the given code;
The program segment was written in Java and the correct program is in the answer section
See bold texts for line by line explanation
This line declares the class name
class helloSarah {
This line declares the main method of the program
public static void main(String [] args) {
This line initializes string variable Name
String Name = "Hello Sarah";
This line prints the initialized variable
System.out.print(Name);
}
}
Answer:
class helloSarah {
public static void main(String [] args) {
Name = "Hello Sarah";
System.out.print(Name);
]
Explanation:
a circuit contains four capacitors connected in parallel. The values of the capacitors are 0.1 microfarads, 0.6 microfarads 1.0 microfarads and 0.05 microfarads. what is the total capacitance of this circuit. A.0.46 Microfarads B. 1.75Microfarads C.1.25Microfarads D0.42Microfarads
Answers
Answer:
B. 1.75 microfarads
Explanation:
When capacitors are connected in parallel, the total capacitance is equivalent to the sum of each individual capacitor's capacitance.
With this in mind, we have the values, 0.1, 0.6, 1.0, and 0.05, all microfarads. So simply, we need to sum these values.
0.1 + 0.6 + 1.0 + 0.05
= 0.7 + 1.0 + 0.05
= 1.7 + 0.05
= 1.75
So the total capacitance for this circuit would be 1.75 microfarads.
Cheers.
In a 1-phase UPS, Vd = 350 V, vo(t) = 170 sin(2π * 60t) V, and io(t) = 10 sin(2π * 60t - 30ᴼ) A.Calculate and plot da(t), db(t), vaN(t), vbN(t), Id, id2(t) and id(t). Switching frequency fs = 20 kHz.
Answers
Answer: provided in the explanation section
Explanation:
The question says;
In a 1-phase UPS, Vd = 350 V, vo(t) = 170 sin(2π * 60t) V, and io(t) = 10 sin(2π * 60t - 30ᴼ) A.Calculate and plot da(t), db(t), vaN(t), vbN(t), Id, id2(t) and id(t). Switching frequency fs = 20 kHz.
Answer
From this we have come to this;
da = Ṽan/Vd = 0.5 + (0.5 * 0.4857) sin w,t
db = Ṽbn/Vd = 0.5 - (0.5 * 0.4857) sin w,t
Ṽan(t) = da * 350 V
Ṽbn(t) = db * 350 V
Id = 0.5 * Ṽo/ Vd * Îo cosΦ1 = 0.5 * (170/350) * 10 * cos 30ᴼ = 2.429 A
Id2 = -0.5 * 170/350 * 10 * sin(2 w,t - 30ᴼ) = -2.429 sin(2 w,t - 30ᴼ)
īd = Id + id2 = 2.429 A + -2.429 sin(2 w,t - 30ᴼ).
Note: Attached is a copy of an image showing and explaining thr plots.
cheers i hope this has been helpful !!!!
In a voltage-divider biased pnp transistor, there is no base current, but the base voltage is approximately correct. The most likely problem(s) is
Answers
Answer:
the base-emitter junction is open and the emitter resistor is open
Explanation:
Because here there will be no base current only when the the base emitter junction is kept open and. Also when emitter resistor is kept open or with thus there will be no voltage drop across the Resistor meaning the base voltage Will be equal to that in the voltage divider circuitry
Why some types of aggregate are susceptible to damage from repeated freezing and thawing? Explain.
Answers
Answer:
Some types of aggregate are susceptible to damage from repeated freezing and thawing due to their porosity. An aggregate being porous allows water molecules to enter in between the rocks.
When freezing occurs, water is known to expand. The expansion of this in the rocks creates a type of pressure which results in the fracture of the rocks. Subsequent freezing and thawing will allow for more fracture between the rock particles which will lead to its disintegration.
The spring-held follower AB has weight W and moves back and forth as its end rolls on the contoured surface of the cam, where the radius is r and z = asin(2θ). If the cam is rotating at a constant rate θ', determine the force at the end A of the follower when θ = θ 1. In this position the spring is compressed δ1. Neglect friction at the bearing C.
Answers
Answer:
some parts of your question is missing attached below is the missing part
Answer : Fa = 4.46 Ib
Explanation:
use the equation
Z = 0.1 sin2∅
next we will differentiate the equation to get the locus of the velocity
z = 0.2 cos2∅∅
differentiate the equation furthermore to get the locus of acceleration in the horizontal axis
z = -0.4sin2∅(∅)^2 + 0.2cos2∅∅
note : express ∅ as 6 rad.[tex]s^{-1}[/tex] for angular velocity and ∅ = 0 for angular acceleration
equation above becomes :
Z = - 0.4 sin 2∅ ( 6)^2 + 0.2 cos 2∅(0)
= - 14.4 sin 2∅ ( acceleration of the follower in horizontal direction )
next calculate The force at the end of A of the follower
Fa - Kx = mz
note: m = w / g hence : Fa - Kx = w/g z ------- (2)
w = weight of the spring-held follower = 0.75 Ib
x = compression of the spring = 0.4
k = spring stiffness = 12 Ib/ft
∅ = 45⁰
g = 32.2 ft/s^2
input these values into equation 2
hence : Fa = 4.46 Ib ( force at the end A of the follower )
help out with that question
Answers
Answer:
(523.74 lb)∠-48.59°
Explanation:
a) We can find the sum of the force vectors several ways, but first we need to know the direction of vector BT. The angle PBT is identical to the angle shown as α, which we can find from the lengths BD and AD.
α = arctan(BD/AD) = arctan((3√3)/6)) ≈ 40.893°
Using the Law of Cosines, we can find the magnitude of PT from ...
PT^2 = PB^2 +BT^2 -2·PB·BT·cos(α)
PT^2 = 800^2 +600^2 -2·800·600·2/√7 = 1000000 -1920000/√7
PT ≈ 523.74 . . . . lb
The direction of PT can be found from the Law of Sines.
∠BPT = arcsin(BT/PT·sin(α)) ≈ 1.145597sin(40.893°) ≈ 48.59°
Relative to the +x axis, the resultant force (R) is ...
R = (523.74 lb)∠-48.59°
__
b) The graphical solution is shown in the attachment. The graphing tool measures the segment lengths and angles. Those measures confirm the above result. (The pound values shown are scaled up by a factor of 100 from the segment lengths on the diagram. They are "captions" for the respective vectors.)
Consider a 2.4-kW hooded electric open burner in an area where the unit costs of electricity and natural gas are $0.10/kWh and $1.20/therm (1 therm = 105,500 kJ), respectively. The efficiency of open burners can be taken to be 73 percent for electric burners and 38 percent for gas burners. Determine the rate of energy consumption and the unit cost of utilized energy for both electric and gas burners.
Answers
Answer:
1.75 kW
$0.137 kWh
4.61 kW
$3.16 therm
Explanation:
Utilized power input of the burner is
P(ui) = total power input * efficiency
P(ui) = 2400 W * 0.73
P(ui) = 1752 W or 1.75 kW
Unit cost of utilized energy is
C(ui) = Unit cost of electricity/efficiency
C(ui) = $0.1 / 0.73 kWh
C(ui) = $0.137 kWh
Power input to the gas burner is
P(gi) = Utilized power input of the burner / efficiency of the burner
P(gi) = 1.75 / 0.38
P(gi) = 4.61 kW
Unit cost of utilized energy is
C(gi) = Unit cost of gas /efficiency
C(gi) = $1.2 / 0.38 kWh
C(gi) = $3.16 therm
how many crankshaft is in V8 engine?
Answers
Answer:
Explanation:
hola friend!!!!!
there is only one crankshaft in V8 engine
Hope this helps
plz mark as brainliest!!!!!!!!
Why do electricians require critical thinking skills? In order to logically identify alternative solutions to problems in order to understand the implications of new information in order to attend to what others are saying in order to install equipment and wiring to meet specifications
Answers
Answer:
In order to logically identify alternative solutions to problems
Explanation:
Electricians are specialized in electrical wiring of buildings, transmission lines, stationary machines, and related equipment. They are either employed in the installations of new electrical components, or to maintain an already installed component. The job of an electrician can be mentally tasking, especially in troubleshooting for fault, and methods of fixing of faults. Some problems might require an out-of-norm approach to solve, and the electrician has to be able to logically identify alternative solutions to problems.
Answer:
in order to logically identify alternative solutions to problems
Explanation:
it makes the most sense