Set up the differential equation that describes the motion under the assumptions of this section. Solve the differential equation. State whether the motion of the spring-mass system is harmonic, damped oscillation, critically damped, oroverdamped. If the motion is a damped oscillation, rewrite in the form (22).33. The spring-mass system has an attached mass of 10 g. The spring constant is30 g/s 2 . A dashpot mechanism is attached, which has a damping coefficient of 40 g/s. The mass is pulled down and released. At time t = 0, the mass is 3 cmbelow the rest position and moving upward at 5 cm/s.34. A long spring has a mass of 1 slug attached to it. The spring stretches 16/13 ftand comes to rest. The damping coefficient is 2 slug/s. The mass is subjected toan impulsive force at time t = 0, which imparts a velocity of 5 ft/s downward.
Answers
Answer:
Explanation:
The principle and application of differential equation is applied and the steps with detailed and appropriate substitution is as carefully shown in the attached files.
m = mass attached
k = spring constant
c = damping coefficient
Related Questions
The flow of a liquid in a 2 inch nominal diameter steel pipe produces a pressure drop due to friction of 78.86 kPa. The length of pipe is 40 m and the mean velocity is 3 m/s. if the density of the liquid is 1000 kg/m^3, then a) determine the reynolds number b) determind if the flow is laminar or tubulent c) compute viscosity of the liquid d) compute the mass flow rate (assume ?, equivalent roughness factor, for Steel pipe to be 45.7 x 10-6 m)
Answers
Answer:
Explanation:
The detailed steps and careful analysis is as shown in the attached file.
Based on the calculations, the Reynolds number is equal to [tex]2.9 \times 10^3[/tex]
Given the following data:
Length of pipe = 40 meters.Diameter of pipe = 2 inches to m = 0.0508 m.Pressure drop = 78.86 kPa.Mean velocity = 3 m/s.Density of liquid = 1000 [tex]kg/m^3[/tex].Roughness factor = [tex]45.7 \times 10^{-6}[/tex] m.How to calculate the Reynolds number.Reynolds number has a direct relationship with friction factor. Thus, we would determine the friction factor by using this formula:
[tex]f=\frac{2 \Delta P D}{ \rho Lu^2}[/tex]
Where:
D is the diameter.L is the length.[tex]\Delta P[/tex][tex]\DeltaP[/tex][tex]\DeltaP[/tex] is the pressure drop.u is the mean velocity.[tex]\rho[/tex] is the density.Substituting the given parameters into the formula, we have;
[tex]f=\frac{2 \times 78.86 \times 10^3 \times 0.0508}{ 1000 \times 40 \times 3^2}\\\\f=\frac{8012.176}{360000}[/tex]
f = 0.0223.
For the Reynolds number:
[tex]N_{Re}=\frac{64}{f} \\\\N_{Re}=\frac{64}{0.0223}[/tex]
Reynolds number = [tex]2.9 \times 10^3[/tex]
Note: Fluid flow is turbulent when Reynolds number is greater than 2000 ([tex]N_{Re} > 2000[/tex]) and it is laminar when it is lesser than 2000 ([tex]N_{Re} < 2000[/tex]).
b. The flow of this liquid is turbulent.
c. To determine the viscosity:
[tex]V=\frac{\rho uD}{N_{Re}} \\\\V=\frac{1000 \times 3 \times 0.0508}{2.9 \times 10^3} \\\\V=\frac{152.4}{2.9 \times 10^3}[/tex]
V = 0.0526 Kgm/s.
d. To determine the mass flow rate:
[tex]m=\rho A u=\rho u\frac{\pi}{4} D^2\\\\m=1000 \times 3 \times 0.7854 \times 0.0508^2[/tex]
m = 6.081 Kg/s.
Read more on Reynolds number here: https://brainly.com/question/14306776
Let m be an integer in the set {0,1,2,3,4,5,6,7,8, 9}, and consider the following problem: determine m by asking 3-way questions, i.e. questions with at most 3 possible answers. For instance, one could ask which of 3 specific subsets m belongs to.
Give a decision tree argument showing that at least 3 such questions are necessary in worst case. In other words, prove that no correct algorithm can solve this problem by asking only 2 questions in worst case.
Answers
Answer:
Take any algorithm if that algorithm solves this problem it can be represented as a ternary decision tree. Therefore each question has at most three answers.
There are ten possible verdicts, the height of such kind of tree should satisfy
ℎ >= ⌈log3(10)⌉ = 3
Hence no such algorithm can ask less than three questions in the worst case.
---
b)
Each and every internal node represents a question asking whether m belongs to one of three possible subset of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} or not
For example 0123|456|789 represented the questionDoes “m: belongs to {0, 1, 2, 3}, to {4, 5, 6}, or to {7, 8, 9}?"
Verdicts are placed in brackets "[ ]"
Explanation:
decision tree is attached below
Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime X (in weeks) has a gamma distribution with mean 28 weeks and standard deviation 14 weeks. (a) What is the probability that a transistor will last between 14 and 28 weeks? (Round your answer to three decimal places.) (b) What is the probability that a transistor will last at most 28 weeks? (Round your answer to three decimal places.) Is the median of the lifetime distribution less than 28? Why or why not? The median is 28, since P(X ≤ ) = .5. (c) What is the 99th percentile of the lifetime distribution? (Round your answer to the nearest whole number.) (d) Suppose the test will actually be terminated after t weeks. What value of t is such that only 0.5% of all transistors would still be operating at termination? (Round your answer to the nearest whole number.) t = weeks
Answers
Answer:
a The probability Needed is 0.424
b1 The probability Needed is 0.567
b2 The median of the life time distribution is less than 28 cause its probability is less than the probability of 28
c The Needed life time is 70 weeks
d The needed lifetime is t = 77 weeks
Explanation:
Let A represent the life time of the transistor in weeks and follows a gamma distribution with parameters ([tex]\alpha,\beta[/tex]).
Looking at what we are give we can say that
E(A) = 28 weeks and [tex]\sigma_A[/tex] = 14 weeks
Hence the mean of the gamma distribution is E(A) = [tex]\alpha\beta ------(1)[/tex]
and the variance of the gamma distribution is V(A) = [tex]\alpha\beta^2 ------(2)[/tex]
Looking at Equation 2
[tex]\alpha\beta^2 =V(A)[/tex]
[tex](\alpha\beta)\beta = V(A)[/tex]
[tex]E(A)\beta = V(A)[/tex]
[tex]\beta = \frac{V(A)}{E(A)}[/tex]
[tex]Note:\alpha\beta = E(A)\\\ and \ \sigma_A^2 = V(A)[/tex]
[tex]\beta = \frac{\sigma_A^2}{E(A)}[/tex]
[tex]\beta = \frac{14^2}{28} = \frac{196}{28} = 7[/tex]
Looking at Equation 1
E(A) = [tex]\alpha \beta[/tex]
28 = [tex]\alpha (7)[/tex]
=> [tex]\alpha = \frac{28}{7}[/tex]
=> [tex]\alpha = 4[/tex]
Now considering the first question
i.e to find the probability that a transistor will last between 14 and 28 weeks
P(14 < A < 28) = P(A ≤28) - P(A ≤ 14)
The general formula for probability of gamma distribution
= [tex]F[\frac{a}{\beta},\alpha ] - F[\frac{a}{\beta},\alpha ][/tex]
[tex]= F[\frac{28}{7},4 ] - F[\frac{14}{7},4 ][/tex]
[tex]= F(4,4) - F(2,4)[/tex]
[tex]= 0.567 - 0.143[/tex] (This is gotten from the gamma table)
[tex]= 0.424[/tex]
Now considering the second question
i.e to find the probability that a transistor will last at most 28 weeks
[tex]P(A \le 28)[/tex] = [tex]F[\frac{28}{7},4][/tex]
[tex]=F(4,4)[/tex]
[tex]= 0.567[/tex] (This is obtained from the gamma Table)
we need to determine the median of the life time distribution less than 28
from what we are given [tex]P(X \le \mu ) = 0.5[/tex] we can see that [tex]\mu[/tex]<28 since [tex]P(A \le 28)[/tex] = 0.567 and this is greater than 0.5
Now considering the Third question
i.e to find the [tex]99^{th}[/tex] percentile of the life time distribution
Generally
F(a : [tex]\alpha[/tex]) =99%
F(a : 4) = 0.99
Looking at the gamma table the tabulated values at [tex]\alpha =4[/tex] and the corresponding 0.99 percent value to a is 10
Now , find the product of [tex]\beta =7[/tex] with a = 10 to obtain the [tex]99^{th}[/tex] percentile
[tex]99^{th} \ percentile = a\beta[/tex]
[tex]= 10 * 7[/tex]
[tex]=70[/tex]
Now considering the Fourth question
i.e to determine the number of weeks such that 0.5% of all transistor will still be operating
The first thing to do is to find the value in the incomplete gamma function with [tex]\alpha = 4[/tex] in such a way that
F(a:4) = 1 - 0.5%
F(a: 4) = 1 - 0.005
F(x: 4) = 0.995
So we will now obtain the 99.5 percentile
Looking at the gamma tabulated values at [tex]\alpha = 4[/tex] and the corresponding 0.995 percent value for a (x depending on what you want to denoted it with) is 11
So we would the obtain the product of [tex]\beta =7[/tex] with a = 11 which is the [tex]99.5^{th}[/tex] percentile
[tex]t = a\beta[/tex]
[tex]= 11 *7[/tex]
[tex]= 77[/tex]
The probability will be:
(a) 0.4240
(b) 0.5670
(c) 70
(d) 77
According to the question,
→ [tex]\mu = \alpha \beta[/tex]
[tex]28= \alpha \beta[/tex] ...(equation 1)
→ [tex]\sigma^2 = \alpha \beta^2[/tex]
[tex]14^2 = \alpha \beta^2[/tex] ...(equation 2)
By putting "equation 1" in "equation 2", we get
→ [tex]14^2 = \alpha \beta\times \beta[/tex]
[tex]14^2 = 28\times \beta[/tex]
→ [tex]196 = 28\times \beta[/tex]
[tex]\beta = 7[/tex]
[tex]\alpha = 4[/tex]
(a) P(14 and 28)
→ [tex]P(14< X< 28 ) = F(\frac{x}{\beta}, \alpha )- F (\frac{x}{\beta}, d )[/tex]
[tex]= F(\frac{28}{7}, 4 )- F(\frac{14}{7} ,4)[/tex]
[tex]= F(4,4)-F(2,4)[/tex]
By using gamma tables, we get
[tex]= 0.5670-0.1430[/tex]
[tex]= 0.4240[/tex]
(b) P(almost 28)
→ [tex]P(x \leq 28) = F(\frac{x}{\beta}, \alpha )[/tex]
[tex]= F (\frac{28}{7} ,4)[/tex]
[tex]= F(4,4)[/tex]
[tex]= 0.5670[/tex]
(c) 99 percentile
[tex]F(X \leq x, \alpha) = 0.99[/tex][tex]F(x, \alpha) = 0.99[/tex][tex]F(x:4)=0.99[/tex]at x = 10,
99th percentile,
→ [tex]x \beta = 10\times 7[/tex]
[tex]= 70[/tex]
(d)
[tex]F(x;4) = 1-0.005[/tex][tex]F(x;4) = 0.995[/tex]x = 11
→ [tex]t = x \beta[/tex]
[tex]= 11\times 7[/tex]
[tex]= 77[/tex]
Thus the above answers are correct.
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2. If you have 10 ft3 of an ideal gas stored at 75 F and at 12 psia that is expanded to 15 ft3 , what is the resulting pressure? 3. The atmospheric pressure is 14.3 psi. How many inches of mercury (s=13.6) does the National Weather Service report this pressure as? Pressure = __________________
Answers
Answer:
2. The decrease in pressure will be 8psia
3. The NWS will report the pressure to be 0.0657inchHg
Explanation:
2. This is an isothermal system because it was only the volume of the gas that was increased, and temperature was constant. To find the pressure, use the Boyle's law equation below.
P1V1 ÷ P2V2..............(1)
make P2 the subject of the formula
P2 = P1V1 ÷ V2 ..............(2)
P1 = 12psia
V1 = 10ft3
V2 = 15ft3
Using equation 2
P2 = (12 × 10) ÷ 15 = 8psia
Therefore, the resulting pressure of the gas will decrease to 8psia, which obeys Boyle's law that states that pressure is inversely proportion to volume in a given mass of gas. That means, as volume increases pressure should decrease.
3. The National weather Service report, are the national agency that forecast the weather, to report how the climate will change. To report the pressure of a weather, their use a pressure barometer, and reports the reading in the pressure barometer, which is the hight the Mercury in the barometer increased to.
Using a pressure barometer.
Patm = pgh..............(3)
Make h the subject of formula
h = Patm ÷ pg ..............(4)
Patm is the Atmospheric pressure = 4.3psi
p is the density of Mercury = 13.6
g is the force of gravity = 9.8
h is the hight of the Mercury in the barometer, which is the pressure that is measured.
Using equation 4
h = 4.3 ÷ (13.6 × 9.8) = 0.0322629psi
To convert Pound- force per square inch to inch of mecury (that's is psi to inchHg)
1psi = 2.03602inchHg
Therefore
0.0322629psi = 0.065689399inchHg
The NWS will report the pressure to be 0.0657inchHg
You are designing a spherical tank to hold water for a small village in a developing country. The volume of liquid it can hold can be computed as V = πh2[3R − h]3 where V = volume (m3), h = depth of water in tank (m), and R = the tank radius (m).
Answers
Answer:
A. [tex]9\pi h^2 - \pi h^3 -90 = 0\\\\[/tex]
B. h1 = 2.0813772719
h2 = 2.0272465228
h3 = 2.0269057423
Explanation:
V = πh^2 x [3R − h]/3
Given v=30, R=3
[tex]30 = \pi h^2 [\frac{9-h}{3} ]\\\\9\pi h^2 - \pi h^3 -90 = 0\\\\[/tex]
B. An initial guess within the interval [0; 6] is selected,
initial guess h0 = 1.5
Newton method x₂ = x₁ - f'(x₁)/f(x₁)
h₂ = h₁ - f'(x₁)/f(x₁)
[tex]h_2 = h_1 - f'(x_1)/f(x_1)\\\\h_2 = h_1 - \frac{18\pi h-3\pi h^2}{9\pi h^2 - \pi h^3 -90}[/tex]
h1 = 2.0813772719
h2 = 2.0272465228
h3 = 2.0269057423
piston cylinder has 5kg water at 500kPa and 300c. Heated by 100V and 4A that operates for 500 secs. It's cooled to a saturated liquid. what is the boundary work and heat lost?
Answers
Answer: boundary work is 1.33kJ and heat lost is 200kJ
Explanation: mass of water m = 5kg
Molar mass of water mm = 18kg/mol
No of moles n = 5/18 = 0.28
Pressure P = 500x10^3
Temperature T = 300°c = 573K
R = 8.314pa.m3/mol/k a constant.
Using PV = nRT
V = nRT/P
V = (0.28x8.314x573)÷(500x10^3)
V = 0.00266m3
Work on boundary w = PV
W = 500x10^3 x 0.00266 = 1330 J
= 1.33kJ
Heat lost on cooling is equal to electrical heat Q used to heat the water initially
Q =Ivt
I = current = 4A
v = voltage = 100
T = time = 500sec
Q = 4x100x500 = 200000 = 200kJ
A 2.0-in-thick slab is 10.0 in wide and 12.0 ft long. Thickness is to be reduced in three steps in a hot rolling operation. Each step will reduce the slab to 75% of its previous thickness. It is expected that for this metal and reduction, the slab will widen by 3% in each step. If the entry speed of the slab in the first step is 40 ft/min, and roll speed is the same for the three steps.
Calculate:
a) lenghtb) exit velocity of the final slab
Answers
Answer:
L_f = 26.025 ft
v_f = 51.77 ft/min
Explanation:
Given:-
- The thickness of the slab initially, t_o = 2 in
- The width of the slab initially, w_o = 10 in
- The Length of the slab initially, L_o = 12.0 ft
- The reduction in thickness in each of three steps, r = 75%
- The widening of the slab in each of three steps , m = 3%
- The entry speed vi = 40 ft/min
- The roll speed remains the same
Find:-
a) length
b) exit velocity
Of the final slab
Solution:-
- The final thickness (t_f) after three passes is as follows:
t_f = ( r / 100 )^n * t_o
Where, n = number of passes.
t_f = ( 75 / 100 ) ^3 * ( 2.0 )
t_f = 0.844 in
- The final width (w_f) after three passes is as follows:
w_f = ( m / 100 + 1 )^n * w_o
Where, n = number of passes.
w_f = ( 3 / 100 + 1 ) ^3 * ( 10.0 )
w_f = 10.927 in
- Assuming the Volume of the slab remains the same. Zero material Loss. The final length of slab can be determined:
t_o*w_o*L_o = t_f*w_f*L_f
L_f = ( 2 * 10 * 12 ) / ( 0.844 * 10.927 )
L_f = 26.025 ft
- We can use the volume rate equation as the roll speed remains constant i.e change in rate of volume is zero. Hence, we can write the before and after the 3rd step formulation:
t_i*w_i*v_i = t_f*w_f*v_f
Where, v_i : The entry step speed
v_f : Third step exit speed.
(0.75)^2 * 2 * (1.03)^2 * 10 * 40 = (0.844)*(10.927)*v_f
v_f = 51.77 ft/min
Find the value of R2 required if vs = 18 V , R1 = 50 kΩ , and the desired no-load output voltage is vo = 6 VPart b)Several different loads are going to be used with the voltage divider from Part A. If the load resistances are 300 kΩ , 200 kΩ , and 100 kΩ , what is the output voltage that is the most different from the design output voltage vo = 6 V ?part c)The circuit designer wants to change the values of R1 and R2 so that the design output voltage vo = 6 V is achieved when the load resistance is RL = 200 kΩ rather than at no-load. The actual output voltage must not drop below 5.4 V when RL = 100 kΩ . What is the smallest resistor value that can be used forR1?
Answers
Answer:
Explanation:
Check attachment for solution
a. The value of R2 required for the desired no-load output voltage R2 = 10 kΩ
b. The most different output voltage is 5.8 V, which occurs when the load resistance is 300 kΩ.
c. The smallest resistor value that can be used for R1 is 45 kΩ.
Part A:
To find the value of R2 required for the desired no-load output voltage, we can use the following voltage divider equation:
vo = vs * R2 / (R1 + R2)
Substituting in the known values, we get:
6 V = 18 V * R2 / (50 kΩ + R2)
Solving for R2, we get:
R2 = 18 V * 6 V - 6 V * 50 kΩ / 6 V - 18 V
R2 = 10 kΩ
Part B:
To find the output voltage that is the most different from the design output voltage, we need to calculate the output voltage for each load resistance. We can use the following voltage divider equation:
vo = vs * (R2 / (R1 + R2) || RL)
Substituting in the known values for each load resistance, we get:
vo (RL = 300 kΩ) = 18 V * (10 kΩ / (50 kΩ + 10 kΩ) || 300 kΩ)
vo (RL = 300 kΩ) = 5.8 V
vo (RL = 200 kΩ) = 18 V * (10 kΩ / (50 kΩ + 10 kΩ) || 200 kΩ)
vo (RL = 200 kΩ) = 7.2 V
vo (RL = 100 kΩ) = 18 V * (10 kΩ / (50 kΩ + 10 kΩ) || 100 kΩ)
vo (RL = 100 kΩ) = 8.6 V
Therefore, the output voltage that is the most different from the design output voltage is 5.8 V, which occurs when the load resistance is 300 kΩ.
Part C:
To change the values of R1 and R2 so that the design output voltage vo = 6 V is achieved when the load resistance is RL = 200 kΩ rather than at no-load, we can use the following voltage divider equation:
vo = vs * R2 / (R1 + R2)
Substituting in the known values, we get:
6 V = 18 V * R2 / (R1 + 200 kΩ)
Solving for R2, we get:
R2 = 30 kΩ
Next, we need to calculate the value of R1 to ensure that the actual output voltage does not drop below 5.4 V when RL = 100 kΩ. We can use the following voltage divider equation:
vo = vs * (R2 / (R1 + R2) || RL)
Substituting in the known values, we get:
5.4 V = 18 V * (30 kΩ / (R1 + 30 kΩ) || 100 kΩ)
Solving for R1, we get:
R1 = 45 kΩ
Therefore, the smallest resistor value that can be used for R1 is 45 kΩ.
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Air flows steadily between two sections in a long, straight portion of 10-cm inside diameter pipe. The uniformly distributed temperature and pressure at each section are given. The average air velocity at Sections (1) and (2) are 66 m/s and 300 m/s, respectively. Assume uniform velocity distributions. Determine the frictional force (Rx) exerted by the pipe wall on the airflow between the sections. At section (1), p1 = 7 MPa, T1 = 25°C, and V1 = 66 m/s. At section 2, p2 = 1.3 MPa, T1 = -20°C, and V2 = 300 m/s. (5 pt)
Answers
Answer:
solution attached below
Explanation:
Water at 158C (r 5 999.1 kg/m3 and m 5 1.138 3 1023 kg/m·s) is flowing steadily in a 30-m-long and 5-cm-diameter horizontal pipe made of stainless steel at a rate of 9 L/s. Determine (a) the pressure drop, (b) the head loss, and (c) the pumping power requirement to overcome this pressure drop.
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
(a) the pressure drop is [tex]\Delta P_L = 100.185\ kPa[/tex]
(b) the head loss is the [tex]h_L =10.22\ m[/tex]
(c) the pumping power requirement to overcome this pressure drop [tex]W_p = 901.665\ W[/tex]
Explanation:
So the question we are told that the water is at a temperature of 15°C
And also we are told that the density of the water is [tex]\rho=999.1\ kg/m^3[/tex]
We are also told that the dynamic viscosity of the water is [tex]\mu = 1.138 *10^{-3} kg/m \cdot s[/tex]
From the diagram the length of the pipe is [tex]L=[/tex] 30 m
The diameter is given as [tex]D=[/tex] 5 cm [tex]\frac{5}{100} m = 0.05m[/tex]
The volumetric flow rate is given as [tex]Q=[/tex] 9 L/s [tex]= \frac{9}{1000} m^3/ s = 0.009\ m^3/s[/tex]
Now the objective of this solution is to obtain
i the pressure drop, ii the head loss iii the pumping power requirement to overcome this pressure drop
to obtain this we need to get the cross-sectional area of the pipe which will help when looking at the flow analysis
So mathematically the cross-sectional area is
[tex]A_s =\frac{\pi}{4} D^2[/tex]
[tex]= \frac{\pi}{4} (5*10^{-2})[/tex]
[tex]=1.963*10^{-3} m^2[/tex]
Next thing to do is to obtain average flow velocity which is mathematically represented as
[tex]v = \frac{Q}{A_s}[/tex]
[tex]v =\frac{9*10^{-3}}{1.963*10^{-3}}[/tex]
[tex]=4.585 \ m/s[/tex]
Now to determine the type of flow we have i.e to know whether it a laminar flow , a turbulent flow or an intermediary flow
We use the Reynolds number
if it is below [tex]4000[/tex] then it is a laminar flow but if it is higher then it is a turbulent flow ,now when it is exactly the value then it is an intermediary flow
This Reynolds number is mathematically represented as
[tex]Re = \frac{\rho vD}{\mu}[/tex]
[tex]=\frac{(999.1)(4.585)(0.05)}{1.138*10^{-3}}[/tex]
[tex]=2.0127*10^5[/tex]
Now since this number is greater than 4000 the flow is turbulent
So we are going to be analyse the flow using the Colebrook's equation which is mathematically represented as
[tex]\frac{1}{\sqrt{f} } =-2.0\ log [\frac{\epsilon/D_h}{3.7} +\frac{2.51}{Re\sqrt{f} } ][/tex]
Where f is the friction factor , [tex]\epsilon[/tex] is the surface roughness ,
Now generally the surface roughness for stainless steel is
[tex]\epsilon = 0.002 mm = 2*10^{-6} m[/tex]
Now substituting the values into the equation we have
[tex]\frac{1}{\sqrt{f} } =-2.0\ log [\frac{2*10^{-6}/5*10^{-2}}{3.7} +\frac{2.51}{(2.0127*10^{5})\sqrt{f} } ][/tex]
So solving to obtain f we have
[tex]\frac{1}{\sqrt{f} } =-2.0\ log [\frac{4*10^{-5}}{3.7} +\frac{2.51}{(2.0127*10^{5})\sqrt{f} } ][/tex]
[tex]= -2.0log[1.0811 *10^{-5} +\frac{1.2421*10^{-5}}{\sqrt{f} } ][/tex]
[tex]f = 0.0159[/tex]
Generally the pressure drop is mathematically represented as
[tex]\Delta P_L =f\ \frac{L}{D} \ \frac{\rho v^2}{2}[/tex]
Now substituting values into the equation
[tex]= 0.0159 [\frac{30}{5*10^{-5}} ][\frac{(999.1)(4.585)}{2} ][/tex]
[tex]=100.185 *10^3 Pa[/tex]
[tex]=100.185 kPa[/tex]
Generally the head loss in the pipe is mathematically represented as
[tex]h_L =\frac{\Delta P_L}{\rho g}[/tex]
[tex]= \frac{100.185 *10^3}{(999.1)(9.81)}[/tex]
[tex]=10.22m[/tex]
Generally the power input required to overcome this pressure drop is mathematically represented as
[tex]W_p = Q \Delta P_L[/tex]
[tex]=(9*10^{-3}(100.185*10^3))[/tex]
[tex]= 901.665\ W[/tex]
The 1000-lb elevator is hoisted by the pulley system and motor M. The motor exerts a constant force of 500 lb on the cable. The motor has an efficiency of ε = 0.65. Determine the power that must be supplied to the motor at the instant the load has been hoisted s = 27 ft starting from rest.
Answers
The power that must be supplied to the motor is 136 hp
Explanation:
Given-
weight of the elevator, m = 1000 lb
Force on the table, F = 500 lb
Distance, s = 27 ft
Efficiency, ε = 0.65
Power = ?
According to the equation of motion:
F = ma
[tex]3(500) - 1000 = \frac{1000}{32.2} * a[/tex]
a = 16.1 ft/s²
We know,
[tex]v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (27-0)\\\\v = 29.48m/s[/tex]
To calculate the output power:
Pout = F. v
Pout = 3 (500) * 29.48
Pout = 44220 lb.ft/s
As efficiency is given and output power is known, we can calculate the input power.
ε = Pout / Pin
0.65 = 44220 / Pin
Pin = 68030.8 lb.ft/s
Pin = 68030.8 / 500 hp
= 136 hp
Therefore, the power that must be supplied to the motor is 136 hp
To calculate the power supplied to the motor, we use the work done by the motor and the motor efficiency. However, without the time, it takes to lift the elevator, a crucial piece of information is missing, thus preventing an exact calculation.
Explanation:To determine the power that must be supplied to the motor at the instant the 1000-lb elevator has been hoisted 27 ft from rest, we must first calculate the useful power output. Since the motor has an efficiency of 0.65 and exerts a constant force of 500 lb, we can use the work-energy principle along with the efficiency to find the required power.
The work done by the motor on the elevator is the force applied times the distance moved. The useful work is W = force × distance = 500 lb × 27 ft.
The useful power output, or work done per unit time, can be found by P = W / t. However, the time is not provided directly in this scenario; we assume this process occurs instantaneously, which is not realistic for real-world scenarios.
Since power supplied is equal to the useful power output divided by the efficiency, we have Power supplied = Useful power / ε.
Assumptions in the CalculationWe would normally need time to calculate power, which is work done or energy converted per unit time.The efficiency of the motor is used to calculate the actual power supplied versus the useful power output.The actual lifting process takes time, and an instantaneous lift is not realistic.Due to missing time information, the exact numerical answer cannot be provided without assumptions or additional data concerning how quickly the elevator is lifted.
A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 19.636 and 19.661 mm, respectively, and its final length is 75.9 mm, calculate its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 108 GPa and 37.1 GPa, respectively.
Answers
Answer:
The original length of the specimen is found to be 76.093 mm.
Explanation:
From the conservation of mass principal, we know that the volume of the specimen must remain constant. Therefore, comparing the volumes of both initial and final state as state 1 and state 2:
Initial Volume = Final Volume
πd1²L1/4 = πd2²L2/4
d1²L1 = d2²L2
L1 = d2²L2/d1²
where,
d1 = initial diameter = 19.636 mm
d2 = final diameter = 19.661 mm
L1 = Initial Length = Original Length = ?
L2 = Final Length = 75.9 mm
Therefore, using values:
L1 = (19.661 mm)²(75.9 mm)/(19.636 mm)²
L1 = 76.093 mm
A cylindrical specimen of cold-worked steel has a Brinell hardness of 250.(a) Estimate its ductility in percent elongation.(b) If the specimen remained cylindrical during deformation and its original radius was 5 mm (0.20 in.), determine its radius after deformation.
Answers
Answer:
A) Ductility = 11% EL
B) Radius after deformation = 4.27 mm
Explanation:
A) From equations in steel test,
Tensile Strength (Ts) = 3.45 x HB
Where HB is brinell hardness;
Thus, Ts = 3.45 x 250 = 862MPa
From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.
Also, from image 2,at CW of 27%,
Ductility is approximately, 11% EL
B) Now we know that formula for %CW is;
%CW = (Ao - Ad)/(Ao)
Where Ao is area with initial radius and Ad is area deformation.
Thus;
%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100
%CW = [1 - (rd)²/(ro)²]
1 - (%CW/100) = (rd)²/(ro)²
So;
(rd)²[1 - (%CW/100)] = (ro)²
So putting the values as gotten initially ;
(ro)² = 5²([1 - (27/100)]
(ro)² = 25 - 6.75
(ro) ² = 18.25
ro = √18.25
So ro = 4.27 mm
An equal-tangent vertical curve is to be constructed between grades of -2% (initial) and 1% (final). The PVI is at station 110 00 and at elevation 420 ft. Due to a street crossing the roadway, the elevation of the roadway at station 112 00 must be at 424.5 ft. Design the curve, and determine the elevations and stations for PVC and PVI.
Answers
Answer:
The curve length (L) will be = 1218 ft
The elevations and stations for PVC and PVI
a. station of PVC = 103 + 91.00
b. station of PVI = 116 + 09.00
c. elevation of PVC = 432.18ft
d. elevation of PVI = 426.09ft
Explanation:
First calculate for the length (L)
To calculate the length, use the formula of "elevation at any point".
where, elevation at any point = 424.5.
and ∴ PVC Elevation = (420 + 0.01L)
Then, calculate for Station of PVC and PVI and elevation of PVC and PVI
An online music platform, S record, is planning to implement a database to enhance its data management practice and ultimately advance its business operations. The initial planning analysis phases have revealed the following system requirements:
Each album has a unique Album ID as well as the following attributes: Album Title, Album Price, and Release Date. An album contains at least one song or more songs. Songs are identified by Song ID. Each song can be contained in more than one album or not contained in any of them at all and has a Song Title and Play Time. Each song belongs to at least one genre or multiple genres. Songs are written by at least an artist or multiple artists. Each artist has a unique Artist ID, and an artist writes at least one song or multiple songs, to be recorded in the database. Data held by each artist includes Artist Name and Debut Date.
Each customer must sign up as a member to make a purchase on the platform. The customer membership information includes Customer ID, Customer Name, Address (consisting of City, State, Postal Code), Phone Number, Birthday, Registration Date. Customers place orders to purchase at least one album or more albums. They can purchase multiple quantities of the same album, which should be recorded as Quantities Ordered. Each order is identified by an Order IDand has Order Date, Total Price, Payment Method, and Delivery Option.
Q1. Draw an ER diagram for Statement 1 (You can add notes to your diagram to explain additional assumptions, if necessary).
File format: asgmt1_q1_ lastname_firstname.pdf
Answers
Answer:
Explanation:
We would be taking a breakdown of the following entities.
online music platform database can have the following entities:
1. Artist
has the following attributes:
Artists Name
Artist ID
Debut Date
2. Albums
has the following attributes
:
Album ID
Title
Release Date
Price
3. Song
has the following attributes
:
Song ID
Play Time
Title
genres
4. Items
This represents the items the customers purchased with several albums and quantity.
this shows the following attributes
Album ID
Qty
5. Orders
has the following attributes
The Order ID
The Order Date
Total Price
Payment Method
Delivery Option
6. Customer
has the following attributes
Customer ID
Customer Name
Address - City, State, Postal code
Phone Number
Birthday
Registration Date
NB. the uploaded image shows the ER Diagram.
cheers i hope this helps.
Consider the following matrix transpose routine:
typedef int array[4] [4];
void transpose2(array dst, array src)
{
int i, j;
for (i = 0; i < 4; i++);
for (j = 0; j < 4; j++);
dst [j] [i] = src[i][j];
}
}
}
Assume this code runs on a machine with the following properties:
(a) sizeof (int) = 4
(b) The src array starts at address 0 and the dst array starts at address 64 (decimal).
(c) There is a single LI data cache that is direct-mapped, write-through, write-allocate, with a block size of 16 bytes.
(d) The cache has a total size of 32 data bytes (i.e. not considering tags, etc.), and the cache is initially empty.
(e) Accesses to the src and dst arrays are the only sources of read and write misses, respectively.
For each row and col, indicate whether the access to src [row] [col] and dst [row] [col] is a hit (h) or a miss (m). For example, reading src [0] [0] is a miss and writing dst [0] [0] is also a miss.
Answers
The question is about determining cache hits and misses when performing a matrix transposition, taking into account the cache's specifications. Given the setup, each new cache block accessed will result in a miss, followed by hits if subsequent accesses fall within the same block. The non-sequential access pattern of matrix transposition is likely to incur more cache misses compared to sequential access.
Explanation:The student is asking about cache behavior when transposing a matrix, which involves changing the positions of each element in the array so that rows become columns and vice versa. In a cache memory system that is direct-mapped, block size determines how many elements will fit within one block of cache. Because the array starts at a known memory address and each integer occupies 4 bytes (given by the property that sizeof(int) is 4), we can predict how the cache will behave during the execution of the transpose routine.
Given the cache specifics - direct-mapped, write-through, write-allocate, and block size of 16 bytes (which can hold 4 integers), and size of 32 bytes total - the access pattern will largely depend on the congruence of array indices and cache block boundaries. For example, when accessing the source array src at address 0 (src[0][0]), it will be a cache miss as the cache is initially empty. However, subsequent accesses like src[0][1], src[0][2], and src[0][3] may hit as they reside in the same cache block if the block can accommodate them. Similarly, writes to the destination array dst will initially miss and then follow a similar pattern of hits and misses based on cache boundaries and block allocation behavior after write misses (due to write-allocate policy).
Matrix transposition involves swapping elements such that element (i, j) becomes element (j, i). As a result of accessing elements in this non-sequential manner, there could be more cache misses compared to sequential access because the transpose operation accesses elements across different rows and hence, potentially different cache lines.
Consider the animation code from the end of Chapter 4. Modify the ShapeIcon class so that it aggregates a collection of objects implementing the MovableShape interface and its paintIcon() method takes care of painting all the given MovableShape objects. Modify the animation program (AnimationTester.main) to display 5 cars moving horizontally, each starting from a different vertical coordinate, so that the car with index i moves twice as fast as the car with index i - 1.
Answers
Answer:
import java.awt.*;
import java.util.*;
import javax.swing.*;
/**
An icon that contains a moveable shape.
*/
public class ShapeIcon implements Icon
{
private int width;
private int height;
private MoveableShape shape;
private ArrayList <MoveableShape> cars;
public ShapeIcon(MoveableShape shape,
int width, int height)
{
this.shape = shape;
this.width = width;
this.height = height;
cars = new ArrayList<MoveableShape>();
}
public int getIconWidth()
{
return width;
}
public int getIconHeight()
{
return height;
}
public void paintIcon(Component c, Graphics g, int x, int y)
{
Graphics2D g2 = (Graphics2D) g;
shape.draw(g2);
}
}
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;
/**
This program implements an animation that moves
a car shape.
*/
public class AnimationTester
{
private static final int ICON_WIDTH = 400;
private static final int ICON_HEIGHT = 100;
private static final int CAR_WIDTH = 100;
public static void main(String[] args)
{
JFrame frame = new JFrame();
final MoveableShape shape
= new CarShape(0, 0, CAR_WIDTH);
ShapeIcon icon = new ShapeIcon(shape,
ICON_WIDTH, ICON_HEIGHT);
final JLabel label = new JLabel(icon);
frame.setLayout(new FlowLayout());
frame.add(label);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.pack();
frame.setVisible(true);
final int DELAY = 10; //100
// Milliseconds between timer ticks
Timer t = new Timer(DELAY, new
ActionListener()
{
public void actionPerformed(ActionEvent event)
{
shape.translate(1, 0);
label.repaint();
}
});
t.start();
}
}
import java.awt.*;
import java.awt.geom.*;
import java.util.*;
/**
A car that can be moved around.
*/
public class CarShape implements MoveableShape
{
private int x;
private int y;
private int width;
/**
Constructs a car item.
@param x the left of the bounding rectangle
@param y the top of the bounding rectangle
@param width the width of the bounding rectangle
*/
public CarShape(int x, int y, int width)
{
this.x = x;
this.y = y;
this.width = width;
}
public void translate(int dx, int dy)
{
// x += dx;
// y += dy;
if (x <= 400)
x += dx;
else
x = 0;
y += dy;
}
public void draw(Graphics2D g2)
{
Rectangle2D.Double body
= new Rectangle2D.Double(x, y + width / 6,
width - 1, width / 6);
Ellipse2D.Double frontTire
= new Ellipse2D.Double(x + width / 6, y + width / 3,
width / 6, width / 6);
Ellipse2D.Double rearTire
= new Ellipse2D.Double(x + width * 2 / 3, y + width / 3,
width / 6, width / 6);
// The bottom of the front windshield
Point2D.Double r1
= new Point2D.Double(x + width / 6, y + width / 6);
// The front of the roof
Point2D.Double r2
= new Point2D.Double(x + width / 3, y);
// The rear of the roof
Point2D.Double r3
= new Point2D.Double(x + width * 2 / 3, y);
// The bottom of the rear windshield
Point2D.Double r4
= new Point2D.Double(x + width * 5 / 6, y + width / 6);
Line2D.Double frontWindshield
= new Line2D.Double(r1, r2);
Line2D.Double roofTop
= new Line2D.Double(r2, r3);
Line2D.Double rearWindshield
= new Line2D.Double(r3, r4);
g2.draw(body);
g2.draw(frontTire);
g2.draw(rearTire);
g2.draw(frontWindshield);
g2.draw(roofTop);
g2.draw(rearWindshield);
Explanation:
Define the Artist class in Artist.py with a constructor to initialize an artist's information. The constructor should by default initialize the artist's name to "None" and the years of birth and death to 0. Define the Artwork class in Artwork.py with a constructor to initialize an artwork's information. The constructor should by default initialize the title to "None", the year created to 0, and the artist to use the Artist default constructor parameter values. Add an import statement to import the Artist class. Add import statements to main.py to import the Artist and Artwork classes.
Answers
The codes for the files are:
File: Artist.py
This file will contain the Artist class with a constructor initializing the artist's name, birth year, and death year.
# Artist. py
class Artist:
def __init__(self, name="None", birth_year=0, death_year=0):
self.name = name
self. birth_year = birth_year
self. death_year = death_year
File: Artwork.py
This file will define the Artwork class and import the Artist class from Artist.py. It initializes the artwork's title, year of creation, and artist.
# Artwork. py
from Artist import Artist
class Artwork:
def __init__(self, title="None", year_created=0, artist=Artist()):
self. title = title
self. year_created = year_created
self. artist = artist
File: main. py
This file will import both Artist and Artwork classes and create instances of them to demonstrate their usage.
# main. py
from Artist import Artist
from Artwork import Artwork
# Creating an instance of Artist
artist1 = Artist("Vincent van Gogh", 1853, 1890)
# Creating an instance of Artwork using the artist instance
artwork1 = Artwork("Starry Night", 1889, artist1)
# Printing artist and artwork details to verify
print(f"Artist: {artwork1. artist. name}, Born: {artwork1. artist. birth_year}, Died: {artwork1. artist. death_year}")
print(f"Artwork: {artwork1. title}, Created in: {artwork1. year_created}")
Execution and Output
When you run the main.py script, it will create an Artist object for Vincent van Gogh and an Artwork object for one of his most famous pieces, "Starry Night". The script will then print the details of both the artist and the artwork.
Expected Output:
Artist: Vincent van Gogh, Born: 1853, Died: 1890
Artwork: Starry Night, Created in: 1889
In many problems where the potential energy is considered between two objects, the potential energy is defined as zero when the particles are in infinite distance apart. Using the definition of potential energy, explain why the potential energy would be positive if the force between the particles is repulsive and negative if the force between the particles is attractive for noninfinite distances.
Answers
Answer:
potential energy=ε Qq/r. Putting values of charge along with sign and calculating the potential energy will tell whether the resultant force is attractive or repulsive.
Explanation:
potential energy=ε Qq/r
q and Q are charges on each particle
So in case the particles have attractive force between them, one particle will be negative and the other will be positiive. The result will be negative.
In case the particles have repulsive forces, the charge on both particles will be either positive or negative. The result will be positive
The structure supports a distributed load of w. The limiting stress in rod (1) is 370 MPa, and the limiting stress in each pin is 220 MPa. If the minimum factor of safety for the structure is 2.10, determine the maximum distributed load magnitude w that may be applied to the structure plus the stresses in the rod and pins at the maximum w.
Answers
In engineering, the maximum load a structure can withstand while maintaining a factor of safety can be calculated based on material properties and stress limits. Analyzing stress distributions in rods and pins under maximum loads is crucial for ensuring the structural integrity of a system.
Explanation:The maximum distributed load magnitude w that may be applied to the structure can be determined using the concept of factor of safety, which is the ratio of the materials' strength to the maximum stress it is subjected to. The factor of safety is given as 2.10, limiting stress in rod (1) as 370 MPa, and limiting stress in each pin as 220 MPa. By setting up equations based on these data, you can calculate the maximum load w.
To calculate the stresses in the rod and each pin at the maximum w, you would need to consider the equilibrium of forces acting on the structure with the maximum load applied. By analyzing the forces acting on the rod and pins, you can determine the stresses within them when the structure is subjected to the maximum load.
Example: Assuming the structure consists of multiple rods and pins, each experiencing different loads, you can analyze the stress distribution within the structure by considering the individual material properties and load distributions to ensure structural integrity.
If block A of the pulley system is moving downward at 6 ft>s while block C is moving down at 18 ft>s, determine the relative velocity of block B with respect to C.
Answers
Answer:
Explanation:
The detailed steps and appropriate calculation with analysis is as shown in the attachment.
Determine the string's length.
[tex]S_A +2S_B +2S_C = Constant[/tex]
Calculate the equation in terms of time.
[tex]v_A+2v_B +2v_c =0 \\\\6 +2v_B +2(18)=0 \\\\2v_B = -42 \\\\V_B =-21 \frac{ft}{s}[/tex]
Calculate the relative velocity of B with respect to C.[tex]V_B = v_c + V_{\frac{B}{C}} \\\\-21=18+ V_{\frac{B}{C}} \\\\ V_{\frac{B}{C}}= -39 \ \frac{ft}{s} = 39 \ \frac{ft}{s} \uparrow[/tex]
Find out more information about the velocity here:
brainly.com/question/15088467
A civil engineer is studying a left-turn lane that is long enough to hold seven cars. Let X be the number of cars in the line at the end of a randomly chosen red light. The engineer believes that the probability that X = x is proportional to (x + 1)(8 − x) for x = 0, . . . , 7 (the possible values of X), i.e. P(X = x) = ( C(x + 1)(8 − x) for x = 0, 1, . . . , 7 0 otherwise 1. Determine the value of C. 2. Find the probability that X will be at least 5.
Answers
Answer:
A) C = 1/120
B) P( x ≥ 5) = 1/3
Explanation:
I've attached explanations to this.
The value of C is: C = 1/126
The probability that X will be at least 5 is 1/21.
To determine the value of C, we need to make sure that the sum of the probabilities of all possible values of X is equal to 1. This means that the following equation must hold:
\sum_{x=0}^7 P(X = x) = 1
Substituting in the given expression for P(X = x), we get:
\sum_{x=0}^7 C(x + 1)(8 - x) = 1
Expanding the summation and simplifying, we get:
7C + 21C + 35C + 35C + 21C + 7C = 1
126C = 1
Therefore, the value of C is:
C = 1/126
To find the probability that X will be at least 5, we need to calculate the following sum:
P(X \ge 5) = P(X = 5) + P(X = 6) + P(X = 7)
Substituting in the expression for P(X = x), we get:
P(X \ge 5) = C(5 + 1)(8 - 5) + C(6 + 1)(8 - 6) + C(7 + 1)(8 - 7)
P(X \ge 5) = C(6)(3) + C(7)(2) + C(8)(1)
Substituting in the value of C, we get:
P(X \ge 5) = (1/126)(6)(3) + (1/126)(7)(2) + (1/126)(8)(1)
P(X \ge 5) = 1/21
Therefore, the probability that X will be at least 5 is 1/21.
For such more question on probability
https://brainly.com/question/31064097
#SPJ3
To accomplish the design project's Outback Challenge mission, your third design topic to research is a Flight Controller (i.e. manual RC transmitter) and GPS Module. The Flight Controller is not an autopilot, but rather a manual handheld flight control transmitter. If the autonomous operations fail, the Ground Base Control will take control of the UAS to manually release the rescue package (water bottle) to Outback Joe. Using the criteria found in the Outback Challenge, research and select the type of flight controller (i.e. manual handheld RC transmitter) and GPS module you feel would best accomplish the two-fold mission of searching for and delivering a rescue package to Outback Joe.
Answers
Answer:
Explanation: see attachment below
Air enters the compressor of an ideal gas refrigeration cycle at 7∘C and 35 kPa and the turbine at 37∘C and 160 kPa. The mass flow rate of air through the cycle is 0.2 kg/s. Assuming variable specific heats for air, determine
(a) the rate of refrigeration, (b) the net power input, and (c) the coefficient of performance.
Answers
It appears that your answer contains either a link or inappropriate words. Please correct and submit again! error
Had to screenshot the solution check attached
The surface energy of a single crystal depends on crystallographic orientation. Does this surface energy increase or decrease with an increase in planar density
Answers
Answer:
The surface energy of a single crystal depends on the planar density (i.e., degree of atomic packing) of the exposed surface plane because of the number of unsatisfied bonds. As the planar density increases, the number of nearest atoms in the plane increases, which results in an increase in the number of satisfied atomic bonds in the plane, and a decrease in the number of unsatisfied bonds. Since the number of unsatisfied bonds diminishes, so also does the surface energy decrease. (That is, surface energy decreases with an increase in planar density.)
Explanation:
Tanya Pierce, President and owner of Florida Now Real Estate is seeking your assistance in designing a database for her business. One of her employees has experience in developing and implementing Access-based systems, but has no experience in conceptual or logical data modeling. So, at this point Tanya only wants you to develop a conceptual data model for her system. You are to use our entity-data diagramming notation - Crows foot symbols.
Tanya has some very specific needs for her system. There are several aspects of the business that need to be represented in the data model. Of central interest are properties that are listed or sold by the company. Note that a separate division of Florida Now handles raw land, so your system only has to deal with developed property. For all types of properties, Tanya wants to keep track of the owner (client), the listing and selling dates, the asking and selling prices, the address of the property, the Multiple Listing Service (MLS) number, and any general comments. In addition, the database should store the client that purchases a property. There are some specialized data that need to be stored, depending on the type of property. For single family houses she wants to store the area (for example UCF or Conway), the size of the house in square feet, the number of bedrooms and baths, the size of the garage (for example, 2 car), and the number of stories. For condominiums, the database should track the name of the complex, the unit number, the size in square feet, the number of bedrooms and baths, and the type of community (coop or condo). For commercial properties, she needs to know the zoning, size in square feet, type (industrial, retail, or office), and the general condition of the property.
Tanya also wants the database to store information about her real estate agents, such as their name, home address, home phone number, mobile phone number, email address, and their real estate license number. Also, the database should track which agent lists and sells each piece of property. Note that she has a separate system that tracks selling agents and listings from outside brokerages, so you don’t have to worry about external agents. However, sometimes a property will be listed and sold by the same Florida Now agent, but other times one Florida Now agent will list a property and another will sell it.
Of course, Tanya thinks that it would be good to have the database track information about Florida Now’s clients, such as their name, phone number, and street and email addresses. Also, she wants to be able to record comments about the client. Florida Now offers referral fees to clients who refer potential customers to the brokerage. The database should store these referral relationships between clients, including the amount and date of the referral payment. Note that only one client can be paid for a referral. In other words, it is not possible for two clients to be paid for referring the same client.
Finally, Tanya wants to be able to use the database to examine the effectiveness of various advertising outlets. For each outlet, the database should store the name of the outlet (for example, realestate.com or the Orlando Sentinel), the main contact person and their phone number. She also wants to know how much it cost to advertise each property on any outlet used, and when a property was advertised on each outlet used for that property. Keep in mind that a property may be advertised on multiple outlets, and that the cost of advertising on an outlet might vary from property to property.
Create an entity relationship diagram that captures Tanya’s database requirements. The ERD should indicate all entities, attributes, and relationships (including maximum and minimum cardinalities). Also be sure to indicate primary key attributes.
You must draw the ERD using computer software such as Visio, PowerPoint or Word. In addition, you must resolve all M:N (many- to-many) relationships and multi-valued attributes. State any assumptions you make. Remember that your assumptions must be reasonable and must not violate any stated requirements. Think through the entities, and attributes for each entity.
Answers
Answer:
the answer is attributes for each entity
Using the Distortion-Energy failure theory: 8. (5 pts) Calculate the hydrostatic and distortional components of the stress 9. (10 pts) Calculate the von Mises stress and the factor of safety. 10. (10 pts) Of the two factors of safety computed, which one is more realistic? What failure theory should you use if you want to be conservative? 11. (10 pts) Suppose all the principal stresses are equal in magnitude and sign, and larger than Sy. What are the predicted safety factors by the maximum shear stress and distortion energy failure theories? Calculate your results and explain them. What do you think would happen in reality?
Answers
Answer:
Detailed solution is given below:
Calculate the diffusion coefficient for magnesium in aluminum at 500oC. Pre-exponential and activation energy values for this system are 1.2 x 10-4m2/s and 144,000 J/mole respectively. R = 8.314 J/mol K
Answers
Answer:
The diffusion coefficient for magnesium in aluminum at the given conditions is 2.22x10⁻¹⁴m²/s.
Explanation:
The diffusion coefficient represents how easily a substance (solute, e.g.: magnesium) can move through another substance (solvent, e.g.: aluminum).
The formula to calculate the diffusion coefficient is:
D=D₀.exp(- EA/R.T)
Each term of the formula means:
D: diffusion coefficient
D₀: Pre-exponential factor (it depends on each particular system, in this case the magnesium-aluminium system)
EA: activation energy
R: gas constant
T: temperature
1st) It is necessary to make a unit conversion for the temperature from °C to K, because we need to solve the problem according to the units of the gas constant (R) to assure the units consistency, so:
Temperature conversion from °C to K: 500 + 273 = 773 K
2nd) Replace the terms in the formula and solve:
D=D₀.exp(- EA/R.T)
D=1.2x10⁻⁴m²/s . exp[- 144,000 J/mole / (8.314 J/mol K . 773K)]
D=1.2x10⁻⁴m²/s . exp(-22.41)
D=1.2x10⁻⁴m²/s . 1.85x10⁻¹⁰
D= 2.22x10⁻¹⁴m²/s
Finally, the result for the diffusion coefficient is 2.22x10⁻¹⁴m²/s.
A specimen of some metal having a rectangular cross section 10.5 mm x 13.7 mm is pulled in tension with a force of 2650 N, which produces only elastic deformation. Given that the elastic modulus of this metal is 79 GPa, calculate the resulting strain.
Answers
The calculated resulting strain of the metal is 0.0002329
How to determine the resulting strain
To calculate the resulting strain [tex](\(\epsilon\))[/tex], we can use Hooke's Law
This is represented as
[tex]\[\epsilon = \frac{\sigma}{E}\][/tex]
Where:
-[tex]\(\sigma\)[/tex] is the stress
- E is the elastic modulus
From the question, we have
Force F = 2650 N
The cross-sectional area is calclated as
[tex]\(A\) = 10.5 mm x 13.7 mm[/tex]
This gives
[tex]A = \(143.85 \times 10^{-6}\) m\(^2\)[/tex] (converted to [tex]m\(^2\)[/tex])
Calculating the stress [tex](\(\sigma\))[/tex] using the formula:
[tex]\[\sigma = \frac{F}{A}\][/tex]
Substitute the given values:
[tex]\[\sigma = \frac{2650}{143.85 \times 10^{-6}} = \frac{2650}{143.85 \times 10^{-6}} = 18422118.06 \text{ Pa}\][/tex]
Next, we have
[tex]\[\epsilon = \frac{\sigma}{E} = \frac{18422118.06}{79 \times 10^9} = \frac{18422118.06}{79 \times 10^9} = 0.0002329\][/tex]
Hence, the resulting strain is 0.0002329
The contents of a tank are to be mixed with a turbine impeller that has six flat blades. The diameter of the impeller is 3 m. If the temperature is 20°Cand the impeller is rotated at 30 rpm (rev/min), what will be the power consumption? Use power number (Np) of 3.5
Answers
Answer:
P=3.31 hp (2.47 kW).
Explanation:
Solution
Curve A in Fig1. applies under the conditions of this problem.
S1 = Da / Dt ; S2 = E / Dt ; S3 = L / Da ; S4 = W / Da ; S5 = J / Dt and S6 = H / Dt
The above notations are with reference to the diagram below against the dimensions noted. The notations are valid for other examples following also.
32.2
Fig. 32.2 Dimension of turbine agitator
The Reynolds number is calculated. The quantities for substitution are, in consistent units,
D a =2⋅ft
n= 90/ 60 =1.5 r/s
μ = 12 x 6.72 x 10-4 = 8.06 x 10-3 lb/ft-s
ρ = 93.5 lb/ft3 g= 32.17 ft/s2
NRc = (( D a) 2 n ρ)/ μ = 2 2 ×1.5×93.5 8.06× 10 −3 =69,600
From curve A (Fig.1) , for NRc = 69,600 , N P = 5.8, and from Eq. P= N P × (n) 3 × ( D a )5 × ρ g c
The power P= 5.8×93.5× (1.5) 3 × (2) 5 / 32.17 =1821⋅ft−lb f/s requirement is 1821/550 = 3.31 hp (2.47 kW).