Answer:
Explanation:
Given that
initial velocity ,[tex]v= 3 m/s[/tex]
[tex]a=-1.1v^{\dfrac{1}{2}}[/tex]
We know that
[tex]a=v\dfrac{dv}{dx}[/tex]
Lets take x is the distance before coming to the rest.
The final speed of the particle = 0 m/s
[tex]v\dfrac{dv}{dx}=-1.1v^{\dfrac{1}{2}}[/tex]
[tex]\dfrac{dv}{dx}=-1.1v^{-\dfrac{1}{2}}[/tex]
[tex]v^{\dfrac{1}{2}}{dv}=-1.1dx[/tex]
[tex]\int_{3}^{0}v^{\dfrac{1}{2}}{dv}=-\int_{0}^{x}1.1dx[/tex]
[tex]\left [v^{\dfrac{3}{2}}\times \dfrac{2}{3}\right]_3^0=-1.1x[/tex]
[tex]3^{\dfrac{3}{2}}\times \dfrac{2}{3}=1.1x[/tex]
[tex]x=\dfrac{3.46}{1.1}\ m\\x=3.14\ m[/tex]
(b)time taken by it
[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}=-1.1\sqrt{v}[/tex]
[tex]\int_{3}^{0}\frac{dv}{\sqrt{v}}=-1.1\int_{0}^{t}dt[/tex]
[tex]\int_{0}^{3}\frac{dv}{\sqrt{v}}=1.1\int_{0}^{t}dt[/tex]
[tex]2\times 3\sqrt{3}=1.1t[/tex]
[tex]t=9.44\ s[/tex]
Consider as a system the gas in a vertical cylinder; the cylinder is fitted with a piston onwhich a number of small weights are placed. The initial pressure is 200 kPa, and the initial volume ofthe gas is 0.04 m^3. Let a bunsen burner be placed under the cylinder, and let thevolume of the gas increase 0.1m^3 while the pressure remainsconstant.
A. Calculate the work done by the system during this process.
Answer:
Work done = 12,000J = 12KJ
Explanation:
This process is an Isobaric one i.e a process at constant pressure
P = 200KPa = 200 x 1000 Pa = 200,000PaVi = 0.04m3Vf = 0.1m3From the formula for Workdone = Integral (PdV)
Work done = P ( Vf - Vi)
Plugging the values in the equation,
Work done = 200,000 ( 0.1 - 0.04)
Work done = 12,000J = 12KJ
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
5) A 80-kg man has a total foot imprint area of 480 cm2. Determine the pressure this man exerts on the ground if (a) he stands on both feet and (b) he stands on one foot.
Answer:
The pressure exerted by this man on ground
(a) if he stands on both feet is 8.17 KPa
(b) if he stands on one foot is 16.33 KPa
Explanation:
(a)
When the man stand on both feet, the weight of his body is uniformly distributed around the foot imprint of both feet. Thus, total area in this case will be:
Area = A = 2 x 480 cm²
A = 960 cm²
A = 0.096 m²
The force exerted by man on his area will be equal to his weight.
Force = F = Weight
F = mg
F = (80 kg)(9.8 m/s²)
F = 784 N
Now, the pressure exerted by man on ground will be:
Pressure = P = F/A
P = 784 N/0.096 m²
P = 8166.67 Pa = 8.17 KPa
(b)
When the man stand on one foot, the weight of his body is uniformly distributed around the foot imprint of that foot only. Thus, total area in this case will be:
Area = A = 480 cm²
A = 0.048 m²
The force exerted by man on his area will be equal to his weight, in this case, as well.
Force = F = Weight
F = mg
F = (80 kg)(9.8 m/s²)
F = 784 N
Now, the pressure exerted by man on ground will be:
Pressure = P = F/A
P = 784 N/0.048 m²
P = 16333.33 Pa = 16.33 KPa
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]
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)
Given a Pane object appPane and a TextField object nameField, write a statement that adds nameField to the pane.
Answer:
appPane.getChildren().add(nameField);
Explanation:
The question is related to JavaFx which is a set of packages used for making interactive graphical user interface that contains graphical components for better user experience.
Pane is a JavaFx component which is used for adjusting the position and size of a graphical component for a given scene.
We can create a pane object by appPane = new Pane(); command.
TextField is a JavaFx object that is used for displaying a line of text. We have various functions available to set properties of TextField object.
We can create a TextField object by nameField = new TextField(); command.
A pane can have multiple set of graphical components for various parts of an application. These are called children.
Hence we can write the following statement to add a TextField object nameField to a Pane object appPane.
appPane.getChildren().add(nameField);
What is the peak runoff discharge for a 1.3 in/hr storm event from a 9.4-acre concrete-paved parking lot (C
Answer:
331809.5gallon/hr or 92.16gallon/s
Explanation:
What is the peak runoff discharge for a 1.3 in/hr storm event from a 9.4-acre concrete-paved parking lot (C
convert 9.4 acre to inches we have=5.896*10^7
How to calculate Peak runoff discharge
1. take the dimension of the roof
2. multiply the dimension by the n umber of inches of rainfall
3. Divide by 231 to get gallon equivalence (because 1 gallon = 231 cubic inches)
5.896*10^7*1.3
7.66*10^7 cubic inches/hr
1 gallon=231 cubic inches
7.66*10^7 cubic inches=331809.5gallon/hr or 92.16gallon/s
this is gotten by converting 1 hr to seconds
331809.5gallon/hr /3600s=92.16gallon/s
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.
A 4.60 L vessel contains 24.5 g of PCl3 and 3.00 g of O2 at 15.0 ⁰C. The vessel is heated to 195 ⁰C, and the contents react to give POCl3.What is the final pressure in the vessel,assuming that the reaction goes to completion and that all reactants and products are in the gas phase?
Answer:
The final pressure in the vessel is 2.270atm
Explanation:
Mass of PCl3 = 24.5g, MW of PCl3 = 137.5g/mole, number of moles of PCl3 = 24.5/137.5 = 0.178mole
Mass of O2 = 3g, MW of O2 = 32g/mole, number of moles of O2 = 3/32 = 0.094moles
Total moles of rectant (n) = 0.178mole + 0.094mole = 0.272mole
From ideal gas equation
PV = nRT
P (pressure of reactants) = nRT/V
n = 0.272mole, R = 82.057cm^3.atm/gmol.K, T = 15°C = 15+273K = 288K, V = 4.6L = 4.6×1000cm^3 = 4600cm^3
P = 0.272×82.057×288/4600 = 1.397atm
From pressure law
P1/T1 = P2/T2
P2 (final pressure in the vessel) = P1T2/T1
P1 = 1.397atm, T1 = 288K, T2 = 195°C = 195+273K = 468K
P2 = 1.397×468/288 = 2.270atm
A six- lane freeway ( three lanes in each direction) in a scenic area has a measured free- flow speed of 55 mi/ h. The peak- hour factor is 0.80, and there are 8% large trucks and buses and 6% recreational vehicles in the traffic stream. One upgrade is 5% and 0.5 mi long. An analyst has determined that the freeway is operating at capacity on this upgrade during the peak hour. If the peak- hour traffic volume is 3900 vehicles, what value for the driver population factor was used?
Answer:
0.867
Explanation:
The driver population factor ([tex]f_{p}[/tex])can be estimated using the equation below:
[tex]f_{p} = \frac{V}{PHF*N*f_{HV}*v_{p}}[/tex]
The value of the heavy vehicle factor ([tex]f_{HV}[/tex]) is determined below:
The values of the [tex]E_{T}[/tex] = 2 and [tex]E_{R}[/tex] = 3 are gotten from the tables for the RVs, trucks and buses upgrades for passenger-car equivalents. Therefore:
[tex]f_{HV}[/tex] = 1/[1+0.08(2-1)+0.06(3-1)] = 1/[1+0.08+0.12] = 1/1.2 = 0.833
Furthermore, the vp is taken as 2250 pc/(h*In) from the table of LOS criteria for lane freeway using the 15 minutes flow rate. Therefore:
[tex]f_{p}[/tex] = 3900/[0.8*3*0.833*2250] = 3900/4498.2 = 0.867
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]
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]
For H2O, determine the specified property at the indicated state.
(a) T = 140°C, v = 0.5 m3/kg. Find p, in bar.
(b) p = 30 MPa, T = 80°C. Find v, in m3/kg.
(c) p = 10 MPa, T = 600°C. Find v, in m3/kg.
(d) T = 80°C, x = 0.4. Find p, in bar, and v, in m3/kg.
For H2O, determine the specific volume at the indicated state, in m3/kg.
(a) T = 440°C, p = 20 MPa.
(b) T = 160°C, p = 20 MPa.
(c) T = 40°C, p = 2 MPa.
The properties of water and steam used to do work can be determined from the steam table, given input of the temperature, pressure, specific volume, as well as other thermodynamic variables
The responses to the required values of pressure and specific volume are as follows;
Part I
(a) Given T = 140°C, v = 0.5 m³/kg, to find p
From the steam tables, we have, at 140°, [tex]v_f[/tex] = 0.00107976, [tex]v_g[/tex] = 0.508519, [tex]p_s[/tex] = 3.61501
Given that the [tex]v_f[/tex] < v < [tex]v_g[/tex], the fluid has two phases and the experienced pressure [tex]p_s[/tex] = 3.61501 bar
(b) Given that at 30 MPa, and 80°C the steam is in the superheated heated region
From the single phase region of the steam tables, we have;
v = 1.01553 × 10⁻³ m³/kg
(c) Given that at p = 10 MPa = 100 bar, and T = 600°C, we have from the single phase region of the steam table
v = 3.83775 × 10⁻² m³/kg
(d) At T = 80°C, and x = 0.4, we have;
v = 0.00102904 + 0.4 × (3.40527 - 0.00102904) ≈ 1.363
v = 1.363 m³/kg
Part II
(a) At T = 440 °C, p = 20 MPa. from the single phase region of the steam table, we have;
v = 1·22459 × 10⁻² m³/kg
(b) At T = 160°C, p = 20 MPa = 200 bar, the steam is in the superheated region, and we have;
v = 1.08865 × 10⁻³ m³/kg
(c) At T = 40°C and p = 2 MPa = 20 bar, we have
v = 0.00100700 m³/kg = 1.007 × 10⁻³ m³/kg
Learn more about the steam tables here;
https://brainly.com/question/13794937
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
The pressure in a water line is 1500 kPa. What is the line pressure in (a) lb/ft2units and (b) lbf/in2(psi) units?
Answer:
Part A:
[tex]1500\ KPa= 31328.145 \frac{lb}{ft^2}[/tex]
Part B:
[tex]1500 KPa=217.55656 \frac{lb}{in^2}[/tex](Psi)
Explanation:
Part A:
Line Pressure is 1500 KPa
We need a conversion factor which converts KPa to lb/ft^2.
[tex]20.88543 \frac{lb}{ft^2}= 1\ KPa[/tex]
In order to convert 1500 KPa to lb/ft^2, we proceed as:
[tex]1\ KPa=20.88543 \frac{lb}{ft^2} \\1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa}\\1500\ KPa= 31328.145 \frac{lb}{ft^2}[/tex]
1500 KPa is 31328.145 lb/ft^2
Part B:
We will use the same procedure we did in Part A:
1 ft= 12 in
[tex](1\ ft)^2=(12\ in)^2\\1 ft^2=144 in^2[/tex]
Converting [tex]1500 KPa\ into\ \frac{lb}{in^2}[/tex]
[tex]1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa} * \frac{ft^2}{144\ in^2}[/tex]
[tex]1500 KPa=217.55656 \frac{lb}{in^2}[/tex](Psi)
1500 KPa is 217.55656 lb/in^2 (psi)
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)
The displacement of an oscillating mass is 0.004*sin(7t) meters. What is the peak amplitude of its velocity in meters/second?
Answer:
The peak amplitude velocity is found to be 0.028 m/s.
Explanation:
Given that the displacement of an oscillating mass is:
Displacement = x = 0.004 Sin(7t)
Now, to find out the velocity of this particle, me have to take derivative of 'x' with respect to 't'.
Velocity = V = dx/dt = (7)(0.004)Cos (7t)
V = 0.028 Cos (7t)
Now, for the maximum displacement, the value of the Cos function must be maximum. And we know that the maximum value of Cos function is 1. Thus, to get maximum displacement, we set the value of Cos (7t) to be equal to 1.
Vmax = 0.028(1)
Vmax = 0.028 m/s (Peak Amplitude of Velocity)
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]
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.
Why is the contractor normally required to submit a bid bond when making a proposal to an owner on a competitively bid contract?
Answer:
It serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.
Explanation:
A bid bond is a type of construction bond that protects the obligee in a construction bidding process.
A bid bond typically involves three parties:
The obligee; the owner or developer of the construction project under bid. The principal; the bidder or proposed contractor.
The surety; the agency that issues the bid bond to the principal example insurance company or bank.
A bid bond generally serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.
Contractors are required to submit a bid bond to ensure they are financially able and willing to complete the project they bid on, preserving the integrity of the bidding process by preventing cost overruns, favoritism, and corruption.
Explanation:A contractor is normally required to submit a bid bond when making a proposal to an owner on a competitively bid contract to ensure that the contractor is serious and financially able to carry out the project. The bid bond is a form of guarantee that the contractor will enter into the contract at the bid price if awarded the project and provides a financial penalty if the contractor fails to do so. Municipalities and other entities require this bond to prevent situations where a contractor might win a bid and then refuse to complete the work or ask for additional funds to finish the project (“cost overruns”). This is significant in a competitive bidding environment where fairness and the avoidance of favoritism or corruption are essential, as it stops anyone from being favored over others and ensures the project will be completed at the bid price.
Tenders, which are bid documents that outline the project plan and cost, play a crucial role in the construction industry. A well-prepared tender demonstrates that the projected cost reflects a balance between safety, environmental friendliness, and profitability. In contrast, a sealed-bid auction often used in these processes, requires strategic bidding and is susceptible to collusion or bribery to uncover competitor's bid details. Hence, a bid bond safeguards the process by providing a financial deterrent against such unethical practices.
The concentration of carbon monoxide (co) in an exhaust gas is 1x10^4 ppmv. what is the concentration in mg/m^3 at 25 and 1 atm
Answer:
1.1451 x 104 (11451.13)mg/m3
Explanation:
1 ppmv is defined as one volume of a contaminant or solid(CO)(mL) in 1 x 106 volume of solvent/water.
1ppmv = 1mL/m3
Concentration in mg/m3 = volume in ppm x molecular weight x pressure(kPa)/( gas constant x temperature(K)
Molecular weight of CO = 12 + 16
= 28g/mol
Temperature = 273.15 + 25
= 298.15K
Pressure = 1 x 101.325kPa
= 101.325kPa
Ppmv = 1 x 10-4ppmv
Gas constant, R = 8.3144 L.kPa/mol.K
Concentration in mg/m3 = (1 x 104 * 28 * 101.325)/(8.3144 * 298)
= 1.1451 x 104mg/m3
= 11451.13 mg/m3
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.
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]
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
The motion of a particle is defined by the relation x = t3 – 6t2 + 9t + 3, where x and t are expressed in feet and seconds, respectively. Determine When the velocity is zero The position, acceleration and total distance traveled when t= 5 sec
The displacement of a particle when its velocity is zero can be found by differentiating the displacement function to get the velocity function, then determining when this velocity is zero. By substituting the corresponding time back into the displacement equation, we calculate the displacement at that time. Additionally, for acceleration, we can integrate to find velocity and displacement with given initial conditions.
Explanation:Finding the Displacement When Velocity is ZeroThe question involves determining the displacement of a particle when its velocity is zero. Since velocity is the derivative of displacement with respect to time, we first need to find the velocity function by differentiating the displacement function x = t3
- 6t2 + 9t + 3. Once we have the velocity function, we can locate the time t at which the velocity is zero. The displacement at this time will be our required value. Similarly, if given an acceleration function, a(t) = t - 1, and initial conditions for velocity and position, we integrate the acceleration to find the velocity, and again to find the displacement.
To address the specific question of the student regarding the particle's position, acceleration, and total distance traveled at t = 5 seconds, we would substitute t into our established functions for position and acceleration accordingly.
If the initial acceleration of a particle is to be calculated from the provided differential equation dv(t)/dt = 6.0 - 3v(t), with the initial condition of the particle being at rest, we would substitute t = 0 into this equation to find the initial acceleration.
You are provided with a fuel cell that is designed to operate at j = 3 A∕cm2 and P = 1.5 W∕cm2. How much fuel cell active area (in cm2) is required to deliver 2 kW of electrical power? (This is approximately enough to provide power to the average American home.)
(a) 296.3 cm2
(b) 1333.3 cm2
(c) 444.4cm2
(d) 666.6 cm2
To provide 2 kW of electrical power, the fuel cell requires an active area of approximately 1333.3 cm2, calculated by dividing the total power required by the fuel cell's power density.
Explanation:The question involves calculating the area of a fuel cell required to deliver a specific power output, given its operational parameters. This problem is grounded in the principles of electrochemical engineering and requires an understanding of how power density influences the design and performance of fuel cells. Given that the fuel cell operates at a current density (j) of 3 A/cm2 and a power density (P) of 1.5 W/cm2, we are tasked with determining the active area needed to supply 2 kW of electrical power.
The formula to calculate the area is straightforward:
Start by identifying the total power output required, which is 2 kW or 2000 W.Use the provided power density of the fuel cell, 1.5 W/cm2.Calculate the area by dividing the total power by the power density: Area = 2000 W / 1.5 W/cm2 = 1333.3 cm2.Therefore, to deliver 2 kW of electrical power, a fuel cell with an active area of approximately 1333.3 cm2 is required.
Assess the capabilities of a hydroelectric power plant from the following field data: Estimated water flow rate, 40 m3/s River inlet at 1 atm, 10 oC Discharge at 1 atm, 10.2 oC, 200 m below the intake?
Answer:
This hydroelectric power plant has the capability of producing around 78.4 MW of electric energy.
Explanation:
First, we calculate the pressure the water exerts or gains when it travels 200 m below the intake. W have the data:
Density of water = ρ = 1000 kg/m^3
Acceleration due to gravity = g = 9.8 m/s²
Height lost = h = 200 m
Volume flow rate = 40 m^3/s
The pressure difference, is now given as:
ΔP = ρgh
ΔP = (1000 kg/m^3)(9.8 m/s²)(200 m)
ΔP = 1960000 Pa = 1.96 MPa
Now, we calculate the power capacity of this flow:
Power = ΔP(Volume flow rate)
Power = (1960000 N/m²)(40 m^3/s)
Power = 78400000 Watt
Power = 78.4 MW
Thus, this plant can produce at max 78.4 MW. The actual power produced will be slightly less than this due of the losses and efficiency of turbine system used.
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.
A branched molecular structure is stronger in the solid-state and more viscous in the molten state than a linear structure for the same polymer.
True or False?
Answer:
True
Explanation:
Write a program that adds the numbers 1 through 5 into the numbers ArrayList and then prints out the first element in the list.
Answer:
Answer code is given below along with comments to explain each step
Explanation:
Method 1:
// size of the array numbers to store 1 to 5 integers
int Size = 5;
// here we have initialized our array numbers, the type of array is integers and size is 5, right now it is empty.
ArrayList<Integer> numbers = new ArrayList<Integer>(Size);
// lets store values into the array numbers one by one
numbers.add(1)
numbers.add(2)
numbers.add(3)
numbers.add(4)
numbers.add(5)
// here we are finding the first element in the array numbers by get() function, the first element is at the 0th position
int firstElement = numbers.get(0);
// to print the first element in the array numbers
System.out.println(firstElement);
Method 2:
int Size = 5;
ArrayList<Integer> numbers = new ArrayList<Integer>(Size);
// here we are using a for loop instead of adding manually one by one to store the values from 1 to 5
for (int i = 0; i < Size; i++)
{
// adding the values 1 to 5 into array numbers
numbers.add(i);
}
//here we are finding the first element in the array numbers
int firstElement = numbers.get(0);
// to print the first element in the array
System.out.println(firstElement);