To determine the final volume and work done during the heating of water in a piston-cylinder assembly at constant pressure, tabulated data like steam tables are required since water is not an ideal gas. The work done is calculated using the formula W = PΔV.
Explanation:The student has been asked to find the final volume and the work done during a constant pressure process in which 5 kg of water is heated in a piston-cylinder assembly from an initial state of 5 bar and 300°C. To solve for the final volume and work done, one would typically use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat input minus the work output. However, since water at this state is not an ideal gas, tabulated data from steam tables or software would be used to determine the specific volume at the final state and then multiplied by the mass to find the total volume. The work done in a constant pressure process is equal to the pressure times the change in volume (W = PΔV). Without the final specific volume from the tables, we cannot compute the final volume or work directly.
BJP4 Self-Check 7.16a: countStrings Language/Type: Java arrays Strings Author:Marty Stepp (on 2016/09/08) Write a method countStrings that takes an array of Strings and a target String and returns the number of occurences target appears in the array.
Answer
//countStrings Method
public class countStrings {
public int Arraycount(String[] arrray, String target) {
int count = 0;
for(String elem : arrray) {
if (elem.equals(target)) {
count++;
}
}
return count;
}
// Body
public static void main(String args [] ) {
countStrings ccount = new countStrings();
int kount = ccount.Arraycount(new String[]{"Sick", "Health", "Hospital","Strength","Health"}, " Health"
System.out.println(kount);
}
}
The Program above is written in Java programming language.
It's Divided into two parts
The first is the method countStrings
While the second part of the program is the main method for the program execution
Assume that the number of seeds a plant produces is proportional to its aboveground biomass. Find an equation that relates number of seeds and above ground biomass if a plant that weighs 225 g has 26 seeds. Use the variables s for number of seeds and w for weight in grams.
Answer:
s= 0.1156 * w or
s= 0.115*(q+p) in terms of Top Biomass and Root Biomass
Explanation:
Since s (number of seeds) is proportional to biomass (w), and above ground biomass also increases with total plant biomass.
s α w
s= 26
w= 225
k= constant
Thus s = k * w
s/w= k
26/225= k
0.1156= k
The equation showing the relationship between seeds and plant biomass is:
s= 0.1156 * w
Assume q= Top Biomass, and p= Root Biomass
w= q+p
Our equation now becomes
s= 0.115*(q+p)
Water is the working fluid in an ideal Rankine cycle. The condenser pressure is 8 kPa, and saturated vapor enters the turbine at (a) 18 MPa and (b) 4 MPa. The net power output of the cycle is 100 MW.
Determine for each case a) the mass flow rate of steam, in kg/h, b) the heat transfer rates for the working fluid passing through the boiler and condenser, each in kW, and c) the thermal efficiency.
Explanation:
The obtained data from water properties tables are:
Point 1 (condenser exit) @ 8 KPa, saturated fluid
[tex]h_{f} = 173.358 \\h_{fg} = 2402.522[/tex]
Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid
[tex]h_{2a} = 489.752\\h_{2b} = 313.2[/tex]
Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam
[tex]h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876[/tex]
Point 4 (Turbine exit) @ 8 KPa, mixed fluid
[tex]x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119[/tex]
Calculate mass flow rates
Part a) @ 18 MPa
mass flow
[tex]\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a} - h_{4a}) - (h_{2a} - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26 - 2241.448938 ) - (489.752 - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}[/tex]
Heat transfer rate through boiler
[tex]Q_{in} = mass flow * (h_{3a} - h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W[/tex]
Heat transfer rate through condenser
[tex]Q_{out} = mass flow * (h_{4a} - h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W[/tex]
Thermal Efficiency
[tex]n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.06485[/tex]
Part b) @ 4 MPa
mass flow
[tex]\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b} - h_{4b}) - (h_{2b} - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14 - 2405.54119 ) - (313.12 - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}[/tex]
Heat transfer rate through boiler
[tex]Q_{in} = mass flow * (h_{3b} - h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W[/tex]
Heat transfer rate through condenser
[tex]Q_{out} = mass flow * (h_{4b} - h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W[/tex]
Thermal Efficiency
[tex]n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.038275[/tex]
For (a) 18 MPa:
- The mass flow rate of steam is approximately 238,050 kg/h.
- The heat transfer rates are approximately 674,970 kW (boiler) and 481,360 kW (condenser).
- The thermal efficiency is approximately 14.81%.
For (b) 4 MPa:
- The mass flow rate of steam is approximately 187,320 kg/h.
- The heat transfer rates are approximately 480,040 kW (boiler) and 380,450 kW (condenser).
- The thermal efficiency is approximately 20.83%.
Explanation and Calculation
To analyze the ideal Rankine cycle, we need to follow these steps for each case:
Case (a): 18 MPa
1. Determine Specific Enthalpies
- Condenser exit: Saturated liquid at 8 kPa.
From steam tables: [tex]\( h_1 \approx 191.81 \, \text{kJ/kg} \)[/tex]
-Pump exit: Compressed liquid at 18 MPa.
Using [tex]\( h_2 = h_1 + v_1 \Delta P \)[/tex]
[tex]\[ v_1 \approx 0.001 \, \text{m}^3/\text{kg} \][/tex]
[tex]\[ h_2 \approx 191.81 + 0.001 \times (18000 - 8) \approx 209.81 \, \text{kJ/kg} \][/tex]
- Boiler exit: Saturated vapor at 18 MPa.
From steam tables: [tex]\( h_3 \approx 2821.6 \, \text{kJ/kg} \)[/tex]
-Turbine exit: Expanding to 8 kPa.
From steam tables: [tex]\( h_4 \approx 2201.4 \, \text{kJ/kg} \)[/tex]
2. Mass Flow Rate of Steam
Using the power output:
[tex]\[ \dot{W}_{\text{net}} = \dot{m} (h_3 - h_4 - (h_2 - h_1)) \][/tex]
[tex]\[ 100 \times 10^6 = \dot{m} (2821.6 - 2201.4 - (209.81 - 191.81)) \][/tex]
[tex]\[ 100 \times 10^6 = \dot{m} \times 420.2 \][/tex]
[tex]\[ \dot{m} \approx 238050 \, \text{kg/h} \][/tex]
3. Heat Transfer Rates
- Boiler:
[tex]\[ \dot{Q}_{\text{in}} = \dot{m} (h_3 - h_2) \][/tex]
[tex]\[ \dot{Q}_{\text{in}} = 238050 \times (2821.6 - 209.81) \approx 674.97 \times 10^3 \, \text{kW} \][/tex]
- Condenser:
[tex]\[ \dot{Q}_{\text{out}} = \dot{m} (h_4 - h_1) \][/tex]
[tex]\[ \dot{Q}_{\text{out}} = 238050 \times (2201.4 - 191.81) \approx 481.36 \times 10^3 \, \text{kW} \][/tex]
4. Thermal Efficiency
[tex]\[ \eta = \frac{\dot{W}_{\text{net}}}{\dot{Q}_{\text{in}}} \][/tex]
[tex]\[ \eta = \frac{100 \times 10^3}{674.97 \times 10^3} \approx 14.81\% \][/tex]
Case (b): 4 MPa
1. Determine Specific Enthalpies
- Condenser exit: Saturated liquid at 8 kPa.
From steam tables: [tex]\( h_1 \approx 191.81 \, \text{kJ/kg} \)[/tex]
- Pump exit: Compressed liquid at 4 MPa.
Using [tex]\( h_2 = h_1 + v_1 \Delta P \)[/tex]
[tex]\[ v_1 \approx 0.001 \, \text{m}^3/\text{kg} \][/tex]
[tex]\[ h_2 \approx 191.81 + 0.001 \times (4000 - 8) \approx 195.81 \, \text{kJ/kg} \][/tex]
- Boiler exit: Saturated vapor at 4 MPa.
From steam tables: [tex]\( h_3 \approx 2749.7 \, \text{kJ/kg} \)[/tex]
- Turbine exit: Expanding to 8 kPa.
From steam tables: [tex]\( h_4 \approx 2222.0 \, \text{kJ/kg} \)[/tex]
2. Mass Flow Rate of Steam
Using the power output:
[tex]\[ \dot{W}_{\text{net}} = \dot{m} (h_3 - h_4 - (h_2 - h_1)) \][/tex]
[tex]\[ 100 \times 10^6 = \dot{m} (2749.7 - 2222.0 - (195.81 - 191.81)) \][/tex]
[tex]\[ 100 \times 10^6 = \dot{m} \times 531.9 \][/tex]
[tex]\[ \dot{m} \approx 187320 \, \text{kg/h} \][/tex]
3. Heat Transfer Rates
- Boiler:
[tex]\[ \dot{Q}_{\text{in}} = \dot{m} (h_3 - h_2) \][/tex]
[tex]\[ \dot{Q}_{\text{in}} = 187320 \times (2749.7 - 195.81) \approx 480.04 \times 10^3 \, \text{kW} \][/tex]
- Condenser:
[tex]\[ \dot{Q}_{\text{out}} = \dot{m} (h_4 - h_1) \][/tex]
[tex]\[ \dot{Q}_{\text{out}} = 187320 \times (2222.0 - 191.81) \approx 380.45 \times 10^3 \, \text{kW} \][/tex]
4. Thermal Efficiency
[tex]\[ \eta = \frac{\dot{W}_{\text{net}}}{\dot{Q}_{\text{in}}} \][/tex]
[tex]\[ \eta = \frac{100 \times 10^3}{480.04 \times 10^3} \approx 20.83\% \][/tex]
1. A group of 45 tests on a given type of concrete had a mean strength of 4780 psi and a standard deviation of 525 psi. Does this concrete satisfy the requirements of ACI code for 4000 psi concrete
Answer:
Yes, because both design compressive stress is greater than 4000 psi
Explanation:
To design a concrete that satisfy the requirements of ACI code for 4000 psi concrete
Step 1: design the compressive stress, using the two equations below
[tex]f_{c} =f_{cr}-1.34*s[/tex] -------equation 1
where;
[tex]f_{c}[/tex] is the design compressive stress
[tex]f_{cr}[/tex] is the critical stress = 4780 psi mean strength
s is the standard deviation = 525 psi
[tex]f_{c} = 4780 -1.34*525[/tex] = 4076.5 psi
Step 2: design the compressive stress, using the second design equation
[tex]f_{c} =f_{cr}-2.33*s + 500[/tex] -------equation 2
[tex]f_{c} =4780-2.33*525 + 500[/tex] = 4056.75 psi
Therefore, since both compressive stress is greater than 4000 psi, the concrete satisfies the requirements of ACI code for 4000 psi concrete
The speed of an aircraft is given to be 260 m/s in air. If the speed of sound at that location is 330 m/s, the flight of the aircraft is _____.
Answer:
[tex]Ma=\frac{260}{330} \\Ma=0.7878[/tex]
So, Ma < 1 Flow is Subsonic
Explanation:
Mach Number:
Mach Number is the ratio of speed of the object to the speed of the sound. It is used to categorize the speed of the object on the basis of mach number as sonic, supersonic and hyper sonic. (It is a unit less quantity)
Mach < 1 Subsonic
Mach > 1 Supersonic
Ma= Speed of the object/Speed of the sound
[tex]Ma=\frac{260}{330} \\Ma=0.7878[/tex]
So, Ma < 1 Flow is Subsonic
A signalized intersection approach has an upgrade of 4%. The total width of the cross street at this intersection is 60 feet. The average vehicle length of approaching traffic is 16 feet. The speed of approaching traffic is 40 mi/h. Determine the sum of the minimum necessary change and clearance intervals.
Answer:
change interval is 3.93 sec
clearance interval is 1.477 sec
Explanation:
Given data:
upgrade of intersection 4%
street total width at intersection is 60 ft
vehicle length of approaching traffic = 16 ft
speed of approaching traffic =40 mi/hr
85th percentile speed is calculated as
S_{85} = S +5
S_{ 85} = 40 + 5 = 45 mi/h
15th Percentile speed
[tex]S_{15} =S-5[/tex]
= 40 - 5 = 35 mi/hr
change in interval is calculated as
[tex]y = t + \frac{1.47 S_{85}}{2a +(64.4\times 0.01 G}[/tex]
t is reaction time is 1.0,
deceleration rate is given as 10 ft/s^2
[tex]y = 1.0 +\frac{1.47\times 45}{2a +(64.4\times 0.01\times 4}[/tex]
y = 3.93 s
clearance interval is calculated as
[tex]a_r = \frac{W+ L}{1.47\times S_{15}}[/tex]
[tex]a_r = \frac{60+16}{1.47\times 35} = 1.477 s[/tex]
The real power delivered by a source to two impedances, ????1=4+????5Ω and ????2=10Ω connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current.
Answer:
The question is incomplete, below is the complete question
"The real power delivered by a source to two impedance, Z1=4+j5Ω and Z2=10Ω connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current."
answer:
a. 615W, 384.4W
b. 17.4A
Explanation:
To determine the real power absorbed by the impedance, we need to find first the equivalent admittance for each impedance.
recall that the symbol for admittance is Y and express as
[tex]Y=\frac{1}{Z}[/tex]
Hence for each we have,
[tex]Y_{1} =1/Zx_{1}\\Y_{1} =\frac{1}{4+j5}\\converting to polar \\ Y_{1} =\frac{1}{6.4\leq 51.3}\\ Y_{1} =(0.16 \leq -51.3)S[/tex]
for the second impedance we have
[tex]Y_{2}=\frac{1}{10}\\Y_{2}=0.1S[/tex]
we also determine the voltage cross the impedance,
P=V^2(Y1 +Y2)
[tex]V=\sqrt{\frac{P}{Y_{1}+Y_{2}}}\\[/tex]
[tex]V=\sqrt{\frac{1000}{0.16+0.1}}\\ V=62v[/tex]
The real power in the impedance is calculated as
[tex]P_{1}=v^{2}G_{1}\\P_{1}=62*62*0.16\\ P_{1}=615W[/tex]
for the second impedance
[tex]P_{2}=v^{2}*G_{2}\\ P_{2}=62*62*0.1\\384.4w[/tex]
b. We determine the equivalent admittance
[tex]Y_{total}=Y_{1}+Y_{2}\\Y_{total}=(0.16\leq -51.3 )+0.1\\Y_{total}=(0.16-j1.0)+0.1\\Y_{total}=0.26-J1.0\\[/tex]
We convert the equivalent admittance back into the polar form
[tex]Y_{total}=0.28\leq -19.65\\[/tex]
the source current flows is
[tex]I_{s}=VY_{total}\\I_{s}=62*0.28\\I_{s}=17.4A[/tex]
Because assembly language is so close in nature to machine language, it is referred to as a ____________.
low-level language
Answer:
symbolic machine code.
Explanation:
The instructions in the language are closely linked to the machine's architecture.
Using Pascal’s Law and a hydraulic jack, you want to lift a 4,000 lbm rock. The large cylinder has a diameter of 6 inches.
a. What would the diameter of the small cylinder need to be if the amount of forceyou could apply was limited to your weight (120 lbf) ? (neglect the leveragegained by using a handle)
Answer:
a diameter of D₂ = 0.183 inches would be required
Explanation:
appyling pascal's law
P applied to the hydraulic jack = P required to lift the rock
F₁*A₁ = F₂*A₂
since A₁= π*D₁²/4 , A₂= π*D₂²/4
F₁*π*D₁²/4 = F₂* π*D₂²/4
F₁*D₁²=F₂*D₂²
D₂ = D₁ *√(F₁/F₂)
replacing values
D₂ = D₁ *√(F₁/F₂) = 6 in * √(120 lbf/(4000 lbm * 32.174 (lbf/lbm)) = 0.183 inches
An electric current of 237.0 mA flows for 8.0 minutes. Calculate the amount of electric charge transported. Be sure your answer has the correct unit symbol and the correct number of significant digits x10
Answer:
amount of electric charge transported = 1.13 × [tex]10^{-2}[/tex] C
Explanation:
given data
electric current = 237.0 mA = 0.237 A
time = 8 min = 8 × 60 sec = 480 sec
solution
we get here amount of electric charge transported that is express as
amount of electric charge transported = electric current × time ...........1
put here value and we get
amount of electric charge transported = 0.237 × 480
amount of electric charge transported = 113.76 C
amount of electric charge transported = 1.13 × [tex]10^{-2}[/tex] C
A hydraulic cylinder has a 125-mm diameter piston with hydraulic fluid inside the cylinder and an ambient pressure of 1 bar. Assuming standard gravity, find the piston mass that will create a pressure inside of 2500 kPa.
Answer: 3002.86kg
Explanation:
Hydraulic cylinder diameter =125mm
Ambient pressure =1bar
Pressure =2500kpa
Piston Mass (MP) =?
F|(when it moves upward )=PA=F|(when it moves downward) =PoA+Mpg
Po=1 bar=100kpa
A=(π/4)D^2=(π/4)*0.125^2=0.01227m^2
Mp=(P-Po) A/g=(2500-100)*1000*0.01227/9.80665
Mp=3002.86kg.
In a Major scale the half-steps always fall between SUPERTONIC and SUBDOMINANT, and between LEADING TONE and TONIC.
True/False
Answer:
False
Explanation: Half-steps are two keys that are adjacent. A major scale have half step between 3 and 4,7 and 8. The fourth note- subdominant, and the second note- SUPERTONIC. 7th note-leading tonic, and first note-tonic.
The student's statement is incorrect. In a Major scale, half-steps always fall between the MEDIAN and the SUBDOMINANT, and the LEADING TONE and TONIC
Explanation:The statement in your question - 'In a Major scale the half-steps always fall between SUPERTONIC and SUBDOMINANT, and between LEADING TONE and TONIC' is False. In a Major scale, the half steps always occur between the 3rd and 4th steps (i.e., MEDIAN and SUBDOMINANT) and between the 7th and 8th steps (i.e., LEADING TONE and TONIC). The SUPERTONIC is the second step of the scale, not directly involved in the half steps of a Major scale. Therefore, the correct sequence of half steps in a Major scale falls between the MEDIAN and the SUBDOMINANT, and the LEADING TONE and TONIC.
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For a short time a rocket travels up and to the left at a constant speed of v = 650 m/s along the parabolic path y=600−35x2m, where x isin m. The origin of polar coordinate system is the same as the origin of the rectangular coordinate system xy.
Part A
Determine the radial component of velocity of the rocket at the instant when its transverse coordinate θ = 60∘, where θ is measured counterclockwise from the x axis.
Express your answer to three significant figures and include the appropriate units.
Part B
Determine the transverse component of velocity of the rocket at the instant when its transverse coordinate θ = 60∘, where θ is measured counterclockwise from the x axis.
Express your answer to three significant figures and include the appropriate units.
Answer:
Detailed working is shown
Explanation:
The attached file shows a detailed step by step calculation..
The radial and transverse components of the rocket's velocity at an angle of 60 degrees from the x-axis are 325 m/s and 563 m/s, respectively.
Explanation:Using Physics principles, we know that when a rocket moves along a parabolic path, its velocity can be decomposed into two components: the radial component (the component of velocity directly in line with the radial direction) and the transverse component (the component of velocity perpendicular to the radial direction).
Given that the absolute speed |v| of the rocket is 650 m/s and the angle θ that the velocity makes with respect to x-axis (measured counterclockwise) is 60°, we can use the trigonometric definitions of sine and cosine to compute the radial and transverse components respectively.
Part A: The radial component of velocity (vr) at θ = 60° can be computed using the formula vr = v * cosθ. So, vr = 650 m/s * cos60° = 325 m/s.
Part B: The transverse component of velocity (vt) at θ = 60° can be computed using the formula vt = v * sinθ. So, vt = 650 m/s * sin60° = 563 m/s.
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The atomic radii of Mg2+ and F- ions are 0.079 and 0.120 nm, respectively.
(a) Calculate the force of attraction between these two ions at their equilibrium inter-ionic separation (i.e., when the ions just touch each other).
(b) What is the force of repulsion at this same separation distance?
Answer:
a) 1.165 × 10⁻⁸ N b)- 1.165 × 10⁻⁸ N
Explanation:
Using Coulomb's law
F(attraction) = [tex]\frac{Z1Z2qelectron}{4piER^2}[/tex]
where
R = sum of the distance between the centers of charges = sum of ionic radii = 0.079 nm + 0.120nm = 0.199 nm = 0.199 × 10⁻⁹ m
Z₁ = valency of Mg²⁺ = 2
Z₂ = valency of F ⁻ = - 1
qelectron = charge on electron =1.062 × 10⁻19 C
E = permitivity of free space = 8.85 × 10 ⁻¹² C²/ Nm²
Fa= (1×2× (1.602 × 10⁻¹⁹)²) / (4× 3.142 × 8.85 × 10⁻¹² × (0.199 × 10⁻⁹)²) = 1.165 × 10⁻⁸ N
b) At equilibrium F of repulsion = - F of attraction = - 1.165 × 10⁻⁸ N
a) A total charge Q = 23.6 μC is deposited uniformly on the surface of a hollow sphere with radius R = 26.1 cm. Use ε0 = 8.85419 X 10−12 C2/Nm2. What is the magnitude of the electric field at the center of the sphere? b) What is the magnitude of the electric field at a distance R/2 from the center of the sphere? c) What is the magnitude of the electric field at a distance 52.2 cm from the center of the sphere?
Answer:
(a) E = 0 N/C
(b) E = 0 N/C
(c) E = 7.78 x10^5 N/C
Explanation:
We are given a hollow sphere with following parameters:
Q = total charge on its surface = 23.6 μC = 23.6 x 10^-6 C
R = radius of sphere = 26.1 cm = 0.261 m
Permittivity of free space = ε0 = 8.85419 X 10−12 C²/Nm²
The formula for the electric field intensity is:
E = (1/4πεo)(Q/r²)
where, r = the distance from center of sphere where the intensity is to be found.
(a)
At the center of the sphere r = 0. Also, there is no charge inside the sphere to produce an electric field. Thus the electric field at center is zero.
E = 0 N/C
(b)
Since, the distance R/2 from center lies inside the sphere. Therefore, the intensity at that point will be zero, due to absence of charge inside the sphere (q = 0 C).
E = 0 N/C
(c)
Since, the distance of 52.2 cm is outside the circle. So, now we use the formula to calculate the Electric Field:
E = (1/4πεo)[(23.6 x 10^-6 C)/(0.522m)²]
E = 7.78 x10^5 N/C
Many car companies are performing research on collision avoidance systems. A small prototype applies engine braking that decelerates the vehicle according to the relationship a = − k √ t , where a and t are expressed in m/s² and seconds, respectively.
The vehicle is traveling at 20 m/s when its radar sensors detect a stationary obstacle. Knowing that it takes the prototype vehicle 4 seconds to stop, determine; (a) expressions for its velocity and position as a function of time, (b) how far the vehicle traveled before it stopped.
Answer:
[tex]v(t)=-\frac{5}{2}\sqrt{t^3}+20\\s(t)=-\sqrt{t^5}+20t[/tex]
[tex]s(t=4)=48\text{ m}[/tex]
Explanation:
In this case acceleration is defined as:
[tex]a(t)=-k\sqrt{t}[/tex] ,
where k is a constant to be found.
To find the expressions for velocity and position as a function of time you must integrate the expression above for acceleration two times.
Initial conditions and boundary conditions are defined with the rest of the data as:
[tex]v(t=0)=20\text{ m/s}\\v(t=4)=0\text{ m/s}\\s(t=0)=0\text{ m}[/tex]
First integration is equal to:
[tex]a'(t)=v(t)=-k\int\sqrt{t}dt=-\frac{2}{3}k\sqrt{t^3}+C_1[/tex]
The boundary condition and initial condition can be used to calculate [tex]k[/tex] and [tex]C_1[/tex]:
[tex]C_1=20\\k=\frac{15}{4}[/tex]
With this expression for velocity is defined as:
[tex]v(t)=-\frac{5}{2}\sqrt{t^3}+20[/tex]
The same can be done to get to expression for position:
[tex]s(t)=-\sqrt{t^5}+20[/tex]
To get the total distance traveled you can integrate the velocity expression from time=0 sec to time=4 sec:
[tex]s_{tot}=\int_0^4(-\frac{5}{2}\sqrt{t^3}+20)dt=48\text{ m}[/tex]
Derive the following conversion factors:
(a) Convert a volume flow rate in cubic inches per minute to cubic millimeters per minute.
(b) Convert a volume flow rate in cubic meters per second to gallons per minute (gpm).
(c) Convert a volume flow rate in liters per minute to gpm.
(d) Convert a volume flow rate of air in standard cubic feet per minute (SCFM) to cubic meters per hour.
A standard cubic foot of gas occupies one cubic foot at standard temperature and pressure (T = 15∘ C and p= 101:3 kPa absolute).
Answer:
A. 0.0283 mm3/min
B. 15850.2 gal/min
C. 0.2642 gal/min
D. 1.7 m3/hour
Explanation:
A.
[(1 in)3/min *(25.4mm)3/(1 in)]
= 0.02832 mm3/min
B.
[(1m)3/sec*(264.173gal)/(1m)3]*(60secs)/1min
= 15850.2 gal/min
C.
[(Liter/min)*(0.264172gal/liter)]
=0.2642 gal/min
D.
[(1ft)3/min*(0.3048m)3/(1ft)3*(60mins/1hour)]
=1.7 m3/hour
Below is an attachment that should help.
DMZ stands for "data-mining zombie," and it is a type of zombie that uses targeted algorithms to steal important, private data from computers connected to the botnet.
True/False
Answer:
False
Explanation:
DMZ stand for DeMilitarized Zone(perimeter network).
It is a sub-network(physical or logical) that contains external facing services to untrusted network. Not only that it contains it also exposes this kind of services.
Adding another layer of security is the main purpose of DMZ
DMZ is positioned in between the Internet and private network
Consider the base plate of an 800-W household iron with a thickness of L 5 0.6 cm, base area of A 5 160 cm2, and thermal conductivity of k 5 60 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be 112°C. Disregarding any heat loss through the upper part of the iron, (a) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate, (b) obtain a relation for the variation of temperature in the base plate by solving the differential equation, and (c) evaluate the inner surface temperature. Answer: (c) 117°C
Answer:
a. [tex]\frac{-kdT(0)}{dx} =q_{0}[/tex]=5000W/m^2
b.833.3(0.006-x)+112
c. 117 deg C
Explanation:
Consider the base plate of an 800-W household iron with a thickness of L 5 0.6 cm, base area of A 5 160 cm2, and thermal conductivity of k 5 60 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be 112°C. Disregarding any heat loss through the upper part of the iron,
Assumption
Heat conduction is steady state and unidimensional 2. thermal conductivity is constant. Heat supplied is not in the plate
4. we disregard heat loss
Heat flux=heat/area
[tex]\alpha[/tex]/A=800W/160*10^-4
with direction to the surface been in the x direction,
the mathematical expression will be
[tex]\frac{d^2T}{dx^2}[/tex]=0..............1
and [tex]\frac{-kdT(0)}{dx} =q_{0}[/tex]=5000W/m^2
from fourier law, for conductivity
T(L)=T2=112C
b. integrating equation 1 twice we have\dT/dx=c1
T(x)=C1x+C2
C1 and C2 are arbitrary constant
at x=0 the boundary conditions become
-kC1=qo
C1=-(qo/k)
at x=L
=T(L)=C1L+C2=T2
C2=T2-cL1
C2=T2+qoL/k
Juxtaposing C1 and C2 into the general equation , we have
T(x)=-qo/k+T2+qoL/k=qo(L-k)/k+T2
50000*(0.006-x)/60+112
833.3(0.006-x)+112
c. inner surface plate temperature is
T(0)=833.33(0.006-0)+112 ( using the derivation in answer b)
117 deg C
The density of a fluid is given by the empirical equation rho 70:5 exp 8:27 107 P where rho is density (lbm/ft3 ) and P is pressure (lbf/in2 ). (a) What are the units of 70:5 and 8:27 107?
Answer:
The unit of 70.5 is lbm/ft^3
The unit of 8.27×10^7 is in^2/lbf
Explanation:
The unit of 70.5 has the same unit as density which is lbm/ft^3 because exponential is found of constant values (unitless values)
The unit of 8.27×10^7 (in^2/lbf) is the inverse of the unit of pressure P (lbf/in^2) because the units have to cancel out so a unitless value can be obtained. Exponential is found of figures with no unit
A 1000 KVA three phase transformer has a secondary voltage of 208/120. What is the secondary full load amperage?
Answer:
The three phase full load secondary amperage is 2775.7 A
Explanation:
Following data is given,
S = Apparent Power = 1000 kVA
No. of phases = 3
Secondary Voltage: 208 V/120 V (Here 208 V is three phase voltage and 120 V is single phase voltage)
Since,
[tex]V_{1ph} =\frac{ V_{3ph}}{\sqrt{3} }\\V_{1ph) = \frac{208}{\sqrt{3} }\\[/tex]
[tex]V_{1ph} = 120 V[/tex]
The formula for apparent power in three phase system is given as:
[tex]S = \sqrt{3} VI[/tex]
Where:
S = Apparent Power
V = Line Voltage
I = Line Current
In order to calculate the Current on Secondary Side, substituting values in above formula,
[tex]1000 kVA = \sqrt{3} * (208) * (I)\\1000 * 1000 = \sqrt{3} * (208) * (I)\\I = \frac{1000 * 1000}{\sqrt{3} * (208) }\\ I = 2775.7 A[/tex]
Consider a pond that initially contains 10 million gallons of fresh water. Water containing a chemical pollutant flows into the pond at the rate of 5 million gallons per year (gal/yr), and the mixture in the pond flows out at the same rate. The concentration c=c(t) of the chemical in the incoming water varies periodically with time to the expression c(t) = 2 + sin(2t) grams per gallon (g/gal).
Construct a mathematical model of this flow process and determine the amount of chemical in the pond at any time t. Then, plot the solution using Maple and describe in words the effect of the variation in the incoming chemical.
Answer:
kindly find attachment for detailed answer
Explanation
Consider a pond that initially contains 10 million gallons of fresh water. Water containing a chemical pollutant flows into the pond at the rate of 5 million gallons per year (gal/yr), and the mixture in the pond flows out at the same rate. The concentration c=c(t) of the chemical in the incoming water varies periodically with time to the expression c(t) = 2 + sin(2t) grams per gallon (g/gal).
Construct a mathematical model of this flow process and determine the amount of chemical in the pond at any time t. Then, plot the solution using Maple and describe in words the effect of the variation in the incoming chemical.
Use the laws of propositional logic to prove that each statement is a tautology. (p n q) rightarrow (p V r) p rightarrow (r rightarrow p) (8 points each for a total of 16, zyBook section 1.5, exercise 1.5.3(a, b))
Answer:
See explanation below.
Explanation:
If the statement is a tautology is true for all the possible combinations
Part a
[tex] (p \land q) \Rightarrow (p \lor r)[/tex] lets call this condition (1)
[tex](p \land q) [/tex] condition (2) and [tex](p \lor r)[/tex] condition (3)
We can create a table like this one:
p q r (2) (3) (1)
T T T T T T
T T F T T T
T F T F T T
T F F F T T
F T T F T T
F T F F F T
F F T F T T
F F F F F T
So as we can see we have a tautology.
Part b
[tex] p \Rightarrow (r \Rightarrow p)[/tex] lt's call this condition 1
And [tex] (r \Rightarrow p)[/tex] condition 2
We can create the following table:
p r (2) (1)
T T T T
T F T T
F T F T
F F T T
So is also a tautology.
A car starts at rest and moves along a perfectly straight highway with an acceleration of α1 = 10 m/s2 for a certain amount of time t1. It then moves with constant speed (zero acceleration) for a time t2 and finally decelerates with an acceleration α2= -10 m/s2 for a time t3 until it comes to a complete stop. The total time of motion is t1 +t2+t3=25 s. The total distance travelled by the car is 1 km. Find t2 Hints: (i) Recognize that each segment of the journey is at constant acceleration! (ii) What is the relationship between the quantities t1, t2, and t3? Use this to help simplify the set of equations that you obtain during the solution process
Answer:
t1 = t3 = 5 seconds
t2 = 15 seconds
Explanation:
For t = t1
a = 10 m/s^2
v(t) = 10*t
s(t) = 5*t^2
Distance traveled = 5*t1^2
For t = t2
a = 0 m/s^2
v(t) = 10*t1
s(t) = 10*t1*t
Distance traveled = 10*t1*t2
For t = t3
a = -10 m/s^2
v(t) = -10*t
s(t) = - 5t^2
Distance traveled = 5t3^2
Sum of all distances = 5*t1^2 + 10*t1*(t2) + 5t3^2
1000 = 5t1^2 + 10t1t2 + 5t3^2 + 10*t1*t2 ....... Eq 1
Distance traveled in first and last segments are the same:
t1 = t3 ..... Eq 2
Given: t1+t2+t3 = 25 .... Eq 3
Solving Equations simultaneously:
Subs Eq 2 into Eq 3 & Eq 1
1000 = 5t1^2 + 10*t1*t2 + 5*t1^2
100 = t1^2 + t1*t2 ..... Eq 4
2t1 + t2 = 25
t2 = 25 - 2t1 .... Eq 5
Subs Eq 5 into Eq 4
100 = t1^2 + t1*(25 - 2t1)
t1^2 -25t1 + 100 = 0
Solve for t1
t1 = 5 , 20 Hence, t1 = 5 sec is selected
t1 = t3 = 5 sec
t2 = 15 sec
Explain the concept of an electric field as if you were addressing a friend or relative.
Answer: Electric Field Formula An electric charge produces an electric field, which is a region of space around an electrically charged particle or object in which an electric charge would feel force. The electric field exists at all points in space and can be observed by bringing another charge into the electric field.
Explanation:
A particle is moving along a circular path having a radius of 6 in. such that its position as a function of time is given by θ=cos2t, where θ is in radians and t is in seconds.
Determine the magnitude of the acceleration of the particle when θ= 35 ∘
Answer:
The angular acceleration is -2.44 rad/s², while the linear acceleration is -14.66 in/s².
Explanation:
First we need to find the time, at the given position. W e are given the position of particle to be:
θ = 35°
Converting it to radians because, the given equation is in radians:
θ = (35°) (π radians/180°)
θ = 0.611 radians
Now, we have the equation:
θ = Cos(2t)
2t = Arc Cos (θ)
2t = Arc Cos (0.611 radians)
t = 0.91/2
t = 0.457 sec
Now, to determine angular acceleration of the particle, we must derivate the equation twice with respect to 't'
Angular Velocity = ω = dθ/dt = -2Sin(2t)
Angular Acceleration = α = -4Cos(2t)
Now, we use the value of t:
α = -4Cos(2 x 0.457)
α = -2.44 rad/s² (negative sign shows decceleration)
Now for linear acceleration, we know that:
a = rα
a = (6 in)(-2.44 rad/s²)
a = -14.66 in/s² (negative sign shows decceleration)
Real resistors can only be manufactured to a specific tolerance, so that in effect the value of the resistance is uncertain. For example, a 1Ω resistor specified as 5% tolerance could in practice be found to have a value anywhere in the range of 0.95 to 1.05Ω. Calculate the potential voltage range across a 2.2 kΩ 10% tolerance resistor if the current flowing through the element is 4 sin 44t mA.
Answer:
The potential voltage range across a 2.2 kΩ 10% tolerance resistor when current of 4 sin 44t mA is flowing through the element is between a range of 7.92sin44t and 9.68sin44t volts.
Explanation:
Given that there is 10% tolerance for the 2.2 kΩ resistor, this implies that the resistance would range between 2,200 — 10% of 2,200 and 2,200 + 10% of 2,200, which is:
(i) 2,200 — 10% of 2,200 = 2,200 — 220 = 1,980 Ω, and
(ii) 2,200 + 10% of 2,200 = 2,200 + 220 = 2,420 Ω
Therefore, we will calculate the potential voltage for 1,980 Ω and 2,420 Ω if the current flowing through the element is 4sin44t mA:
(a) The potential voltage for a resistance of 1,980 Ω: we will use the formula: potential voltage v = i × R
Where i = 4sin44t mA = 0.004sin44t A, and R = 1,980 Ω
The potential voltage = v = 1,980 × 0.004sin44t = 7.92sin44t (in volts)
(b) The potential voltage for a resistance of 2,420 Ω: we will use the formula: potential voltage v = i × R
Where i = 4sin44t mA = 0.004sin44t A, and R = 2,420 Ω
The potential voltage = v = 2,420 × 0.004sin44t = 9.68sin44t (in volts)
Calculate the resulting power factor if a synchronous motor rated at 500 hp with 90% efficiency operating at rated load and at unity power factor is added to the plant instead of the capacitor.
Assume constant voltage (1 hp = 0.746 kW).
Answer:
The question is incomplete. Below is the complete question
"An industrial plant consisting primarily of induction motor loads absorbs 500 kW at 0.6 power factor lagging. (a) Compute the required kVA rating of a shunt capacitor to improve the power factor to 0.9 lagging. (b) Calculate the resulting power factor if a synchronous motor rated 500 hp with 90% efficiency operating at rated load and at unity power factor is added to the plant instead of the capacitor. Assume constant voltage. (1 hp = 0.746 kW)"
Answer:
a. 424.5KVA
b. 0.808 lagging.
Explanation:
Let's first determine the real power, reactive and the apparent power delivered.
Ql= Ptan(the real power angle)
Ql=500*tan53.13°
Ql=666.7 Kvar
The reactive angle is
Arccos(0.9)=25.84°
Now we calculate the reactive power
Qs=Ptan25. 84°
Qs=242.4KVAR
Hence the apparent power is
Qc=Ql-Qs
Qc=666.7-242.2
Qc=424.5KVAR
The required KVA rating of the shunt capacitor is 424.5KVA
b. To calculate the resulting power factor we first determine the power absorbs by the synchronous motor.
Pm=(500*0.746)/0.9
Pm=414.4KW
The total reactive power is
Ps=P+Pm
Ps=414.4+500=914.4KW
Hence we compute the source power factor as
PF=cos[arctan(Qs/Ps)]
PF=cos[arctan(666.7/914.4)
From careful calculation, we arrive at
PF=0.808.
Note this power factor will be lagging.
Holmes owns two suits: one black and one tweed. He always wears either a tweed suit or sandals. Whenever he wears his tweed suit and a purple shirt, he chooses to not wear a tie. He never wears the tweed suit unless he is also wearing either a purple shirt or sandals. Whenever he wears sandals, he also wears a purple shirt. Yesterday, Holmes wore a bow tie. What else did he wear?
Answer:
He wore his black suit, another color of shirt (not purple) and shoes
Explanation:
Holmes owns two suits: one black and one tweed.
Whenever he wears his tweed suit and a purple shirt, he chooses not to wear a tie and whenever he wears sandals, he always wears a purple shirt.
So, if he wore a bow tie yesterday, it means he wore his black suit, another color of shirt (not purple) and shoes because the shirt color is not purple
Consider a cylinder of height h, diameter d, and wall thickness t pressurized to an internal pressure P_0 (gauge pressure, relative to the external atmospheric pressure). The cylinder consists of material with Young's modulus E, Poisson's ratio v, and density rho. Derive expressions for the axial and hoop strains of the cylinder wall in terms of the can dimensions, properties, and internal pressure. You may assume plane stress conditions.
Continuing on Problem 1, assume a strain gage is bonded to the cylinder wall surface in the direction of the axial strain. The strain gage has nominal resistance R_0 and a Gage Factor GF. It is connected in a Wheatstone bridge configuration where all resistors have the same nominal resistance: the bridge has an input voltage V_in. (The strain gage is bonded and the Wheatstone bridge balanced with the vessel already pressurized.) Develop an expression for the voltage change delta V across the bridge if the cylinder pressure changes by delta P.
Repeat Problem 2, but now assuming the strain gage is bonded to the cylinder wall surface in the direction of the hoop strain. Does the voltage change more when the strain gage is oriented in the axial or hoop direction?
Continuing on Problem 3 (strain gage in the hoop direction), calculate the voltage change delta V across the Wheatstone bridge when the cylinder pressure increases by 1 atm. Assume the vessel is made of aluminum 3004 with height h = 10.5 cm, diameter d = 5.5 cm, and thickness t = 50 mu m. The Gage Factor is GF = 2 and the Wheatstone bridge has V_in = 6 V. The strain gage has nominal resistance R_0 = R_4 = 120 ohm.
Explanation:
Note: For equations refer the attached document!
The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:
Eq 1
Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:
Eq 2
For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:
Eq 3
Finally, solving for unknown pressure as a function of hoop strain:
Eq 4
Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is
Eq 5
Consequently, a small differential change in ΔR/R can be expressed as
Eq 6
Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves
Eq 7
Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF
Eq 8
For Wheat stone bridge:
Eq 9
Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:
Eq9
Substituting e with respective stress-strain relation
Eq 10
part b
Since, axial strain(1-2v) < hoop strain (2-v). V out axial < V out hoop.
Hence, dV hoop < dV axial.
part c
Given data:
P = 253313 Pa
D = d + 2t = 0.09013 m
t = 65 um
GF = 2
E = 75 GPa
v = 0.33
Use the data above and compute Vout using Eq3
Eq 11