Answer:
$151094
Explanation:
Solution
Recall that:
Acme Chemical in 2002 purchased a large pump worth of = 112,000
The estimation for the pump to the industrial pump index is =100
The index in 2002= 212
The current index is = 286
k = is the reference year for which cost or price is known.
n = the year for which cost or price is to be estimated (n>k).
Cn = the estimated cost or price of item in year n.
Ck = the cost or price of item in reference year k.
Cn = Ck * (In / Ik )
Now,
We find the estimated cost of the new pump which is stated as follows:
Cn = (112,000 * 286) /212
=32032000/212
=$151094
Therefore, the estimated cost of the new pump is $151094
Tin atoms are introduced into an FCC copper ,producing an alloy with a lattice parameter of 4.7589×10-8cm and a density of 8.772g/cm3 .Cal the atomic percentage of tin present in the alloy
Answer:
atomic percentage = 143 %
Explanation:
Let x be the number of tin atoms and there are 4 atoms / cell in the FCC structure , then 4 -x be the number of copper atoms . Therefore, the value of x can be determined by using the density equation as shown below:
[tex]\mathbf{density (\rho) = \dfrac{(no \ of \ atoms/cell)(atomic \ mass )}{(lattice \ parameter )^3(6.022*10^{23} atoms/ mol)} }[/tex]
where;
the lattice parameter is given as : 4.7589 × 10⁻⁸ cm
The atomic mass of tin is 118.69 g/mol
The atomic mass of copper is 63.54 g/mol
The density is 8.772 g/cm³
[tex]\mathbf{8.772 g/cm^3 = \dfrac{(x)(118.69 \ g/mol) +(4-x)(63.54 \ g/mol)}{(4.7589*10^{-8} cm )^3(6.022*10^{23} atoms/ mol)} }[/tex]
569.32 = 118.69x + 254.16-63.54x
569.32 - 254.16 = 118.69x - 63.54 x
315.16 = 55.15x
x = 315.16/55.15
x = 5.72 atoms/cell
As there are four atoms per cell in FCC structure for the metal, thus, the atomic percentage of the tin is calculated as follows :
atomic % = [tex]\frac{no \ of \ atoms \ per \ cell \ in \ tin }{no \ of \ atoms \ per \ cell \ in \ the \ metal}*100[/tex]
atomic % = [tex]\frac{5.72 \ atoms / cell}{4 \ atoms/ cell} *100[/tex]
atomic % = 143 %
Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8, the compressor inlet temperature is 300 K, and the turbine inlet temperature is 1800 K. The effectiveness of the regenerator is 75 percent. Determine the thermal efficiency and the required mass flow rate of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of (a) 100 percent and (b) 80 percent. The properties of Helium are cp = 5.1926 kJ/kg.K and k = 1.667.
Answer:
Explanation:
Find the temperature at exit of compressor
[tex]T_2=300 \times 8^{\frac{1.667-1}{1.667} }\\=689.3k[/tex]
Find the work done by the compressor
[tex]\frac{W}{m} =c_p(T_2-T_1)\\\\=5.19(689.3-300)\\=2020.4kJ/kg[/tex]
Find the actual workdone by the compressor
[tex]\frac{W}{m} =n_c(\frac{W}{m} )\\\\=1 \times 2020.4kJ/kg[/tex]
Find the temperature at exit of the turbine
[tex]T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k[/tex]
Find the actual workdone by the turbine
[tex]1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg[/tex]
Find the temperature of the regeneration
[tex]\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k[/tex]
Find the heat supplied
[tex]Q_i_n=c_p(T_3-T_5)\\\\=5.19(1800-759.8)\\\\=5388.2kJ/kg[/tex]
Find the thermal efficiency
[tex]n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4[/tex]
60.4%
Find the mass flow rate
[tex]m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42[/tex]
Find the actual workdone by the compressor
[tex]\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg[/tex]
Find the actual workdone by the turbine
[tex]\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg[/tex]
Find the temperature of the compressor exit
[tex]\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k[/tex]
Find the temperature at the turbine exit
[tex]4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k[/tex]
Find the temperature of regeneration
[tex]\epsilon =\frac{T_5-T_2}{T_4-T_2}\\\\0.75=\frac{T_5-787.5}{985-787.5}\\\\T_5=935.5k[/tex]
Answer:
a) 60.4%; 18.42 kg/s
b) 37.8% ; 35.4 kg/s
Explanation:
a) at an isentropic efficiency of 100%.
Let's first find the exit temperature of the compressor T2, using the formula:
[tex](r_p) ^k^-^1^/^k = \frac{T_2}{T_1}[/tex]
Solving for T2, we have:
[tex] T_2 = 300 * (8)^1^.^6^6^7^-^1^/^1^.^6^6^7 = 689.3 K [/tex]
Let's now find the work dine by the compressor.
[tex] \frac{W_c}{m} = c_p(T_2 - T_1) [/tex]
[tex] \frac{W_c}{m} = 5.19(689.3 - 300) = 2020.4 KJ/kg[/tex]
The actual work done by the compressor =
[tex] W_c = 1 * 2020.4 = 2020.4 KJ/kg [/tex]
Let's find the temperature at the exit of the turbine, T4
[tex](r_p) ^k^-^1^/^k = \frac{T_3}{T_4}[/tex]
Solving for T4, we have:
[tex]T_4 = \frac{1800}{(8)^1^.^6^6^7^-^1^/^1^.^6^6^7} = 783.3 K[/tex]
Let's find the work done by the turbine.
[tex]\frac{W_t}{m} = c_p(T_3 - T_4)[/tex]
[tex]\frac{W_t}{m} = 5.19(1800 - 783.3) = 5276.6 KJ/kg[/tex]
The actual work done by the turbine:
= 1 * 5276.6 = 5276.6 KJ/kg
Let's find the regeneration temperature, using the formula:
[tex] e = \frac{T_r - T_2}{T_4 - T_2}[/tex]
Substituting figures, we have:
[tex] 0.75 = \frac{T_r - 689.3}{783.3 - 689.3} [/tex]
[tex] T_r = [0.75(783.3 - 689.3)] + 689.3 = 759.8 [/tex]
Let's calculate the heat supplied.
[tex]Q = c_p(T_3 - T_r)[/tex]
[tex] Q = 5.19(1800 - 759.8) [/tex]
Q = 5388.2 kJ/kg
For thermal efficiency, we have:
[tex] n = \frac{W_t - W_c}{Q} [/tex]
Substituting figures, we have:
[tex] n = \frac{5276.6 - 2020.4}{5388.2} = 0.604 [/tex]
0.604 * 100 = 60.4%
For mass flow rate:
Let's use the formula:
[tex] m = \frac{W_n_e_t}{P} [/tex]
Wnet = 60MW = 60*1000
[tex] m = \frac{60*10^3}{5276.6 - 2020.4} = 18.42 [/tex]
b) at an isentropic efficiency of 80%.
Let's now find the work done by the compressor.
[tex] \frac{W_c}{m} = c_p(T_2 - T_1) [/tex]
[tex] \frac{W_c}{m} = 5.19(689.3 - 300) = 2020.4 KJ/kg[/tex]
The actual work done by the compressor =
[tex] W_c = \frac{2020.4}{0.8}= 2525.5 KJ/kg [/tex]
Let's find the work done by the turbine.
[tex] \frac{W_t}{m} = c_p(T_3 - T_4) [/tex]
[tex] \frac{W_t}{m} = 5.19(1800 - 787.5) = 5276.6 KJ/kg[/tex]
The actual work done by the turbine:
= 0.8 * 5276.6 = 4221.2 KJ/kg
Let's find the exit temperature of the compressor T2, using the formula:
[tex]\frac{W_c}{m} = c_p(T_2 - T_1) [/tex]
[tex] 2525.5 = 5.19(T_2 - 300) [/tex]
Solving for T2, we have:
[tex] T_2 = \frac{2525.5 + 300}{5.19} = 787.5 [/tex]
Let's find the temperature at the exit of the turbine, T4
[tex] \frac{W_t}{m} = c_p(T_3 - T_4) [/tex]
[tex] 4221.2 = 5.19(1800 - T_4) [/tex]
Solving for T4 we have:
[tex] T_4 = 958 K[/tex]
Let's find the regeneration temperature, using the formula:
[tex] e = \frac{T_r - T_2}{T_4 - T_2}[/tex]
Substituting figures, we have:
[tex] 0.75 = \frac{T_r - 787.5}{985 - 787.5} [/tex]
[tex] T_r = [0.75(958 - 787.5)] + 787.5 = 935.5 K [/tex]
Let's calculate the heat supplied.
[tex]Q = c_p(T_3 - T_r)[/tex]
[tex] Q = 5.19(1800 - 935.5) [/tex]
Q = 4486.2 kJ/kg
For thermal efficiency, we have:
[tex] n = \frac{W_t - W_c}{Q} [/tex]
Substituting figures, we have:
[tex] n = \frac{4221.2 - 2525.2}{4486.2} = 0.378 [/tex]
0.378 * 100 = 37.8%
For mass flow rate:
Let's use the formula:
[tex] m = \frac{W_n_e_t}{P} [/tex]
Wnet = 60MW = 60*1000
[tex] m = \frac{60*10^3}{4221.2 - 2525.2} = 35.4 kg/s [/tex]
A circuit-switching scenario in whichNcs users, each requiring a bandwidth of 25 Mbps, must share a link of capacity 150 Mbps.
A packet-switching scenario withNps users sharing a 150 Mbps link, where each user again requires 25 Mbps when transmitting, but only needs to transmit 10 percent of the time.
What is the probability that a given (specific) user is transmitting, and the remaining users are not transmitting?
Answer:
0.09
Explanation:
Packet switching involves breaking a message into packets and sending them independently. Since the user only needs to transmit 10 percent of the time, the probability that a given (specific) user is transmitting = 10% = 0.1
The probability that a user is not transmitting = 100% - 10% = 90% = 0.9
Therefore, the probability that a given (specific) user is transmitting, and the remaining users are not transmitting = 0.1 * 0.9 = 0.09
Cryogenic liquid storage. Liquid oxygen is stored in a thin-walled spherical container, 96 cm in diameter, which is further enclosed in a concentric container 100 cm in diameter. The surfaces facing each other at coated with an emittance of only 0.05. The inner surface is at 95 K, and the outer surface is at 280 K.
a. Draw an equivalent electrical circuit.
b. What is the heat exchange [W] between the two surfaces?
Answer:
The answer is "26.55 V"
Explanation:
Given values:
[tex]d_i= 0.96m\\d_o= 1m\\\epsilon = 0.05\\T_0= 280k\\T_i= 95k\\[/tex]
For Answer (a) please find the attachment.
Answer (b):
[tex]q_{i-0}= \frac{\sigma (T_{0}^4)-(T_{i}^4)}{\frac{1-\epsilon i }{\epsilon_{i} A_{i}}+ \frac{1 }{\ f_{i o} A_{i}} +\frac{1-\epsilon_{0}}{\epsilon_{0} A_{0}}}[/tex]
[tex]f_{i0}= 1 \ it \ is \ fully \ inside \ the \ large \ sphero \\[/tex]
[tex]q_{i-0}= \frac{\sigma A_i (T_{0}^4)-(T_{i}^4)}{\frac{1 }{\epsilon_{i}} - 1+ 1 +\frac{1-\epsilon_{0}}{\epsilon_{0}} \times \frac{A_i}{A_0}}\\\\q_{i-0}= \frac{\sigma A_i (T_{0}^4)-(T_{i}^4)}{\frac{1 }{\epsilon_{i}} +\frac{1-\epsilon_{0}}{\epsilon_{0}} \times (\frac{d_i}{d_0})^2}\\\\q_{i-0}= \frac{\sigma (\pi d^2_i) (T_{0}^4)-(T_{i}^4)}{\frac{1 }{\epsilon_{i}} +\frac{1-\epsilon_{0}}{\epsilon_{0}} \times (\frac{r_i}{r_0})^2}\\\\[/tex]
[tex]q_{i-0}= \frac{5.67 \times 10^{-8} \times 3.14 \times 9.62 \times 9.62 \times (280^4-94^4)}{\frac{1 }{0.05} +\frac{1-0.05}{0.05} \times (\frac{0.96}{1})^2}\\\\\ After \ solve the \ equation \ the \ answer \ is:\\\\q_{i-0} = 26.55 \ V[/tex]
Q#3:(A)Supose we extend the circular flow mode to add imports and export copy the circular flow digram onto a sheet paper and then add a foreign country as athird agent.Draw a through sketch of the flows of imports exports and the payment for each on your digrams?
Answer & Explanation:
Circular Flow model denotes how goods & services, factor incomes & prices move within sectors of economy.
A closed economy has two sectors - households & firms, having following features of circular flow between them:
Households provide factor services to firms , & get factor payments from firms in returnFirms provide goods & services to households, & get prices for households in returnIn case of open economy - with rest of world & foreign country, exports & imports also come in circular flow.
Firms export to foreign ROW, receive export payments from them. Households, firms import from foreign ROW, pay their import payments to them.g Replacing incandescent lights with energy-efficient fluorescent lights can reduce the lighting energy consumption to one-fourth of what it was before. The energy consumed by the lamps is eventually converted to heat, and thus switching to energy-efficient lighting also reduces the cooling load in summer but increases the heating load in winter. Consider a building that is heated by a natural gas furnace with an efficiency of 80 percent and cooled by an air conditioner with a COP of 3.5. The electricity costs $0.12/kWh and natural gas costs $1.40/therm (1 therm = 105,500 kJ).
Answer:
For a 1 kWh, heat supplied for the cost of furnace is = $1.66 *1 0^-5 /kJ * 3600 =$0.06/kWh.
Therefore, supplied heat for natural gas furnace is $0.06/kWh which is lower than $0.12/kWh cost of electricity of refrigerator.
Explanation:
Solution
Given that:
The thermal efficiency of furnace = 0.8 or 80%
The cop of refrigerant COP = 3.5
The unit cost of electricity for refrigerant is =$ 0.12kWh
The unit cost of natural gas of furnace is = $1.40/therm
1 therm = 105,500 kJ
Now,
In the summer the usage or utilization of lightening in the house will be lesser than winter, hence the total cost of energy used in the household will be decreased.
Thus,
In winter, the utilization of lightening will be higher with regards to electricity
For a natural gas furnace it is given below:
The unit cost =$1.40/ηth (I therm/105,500)
= 1.40/0.8 (I therm/105,500)
= $1.66 *1 0^-5 /kJ
For a 1 kWh, heat supplied for the cost of furnace is = $1.66 *1 0^-5 /kJ * 3600 =$0.06/kWh.
So,for the heat supplied for natural gas furnace is $0.06/kWh which is less er than $0.12/kWh cost of electricity of refrigerator.
Therefore, the efficiency of energy lightning will reduce the total energy cost of the building both in winter and summer.
A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print: 1. The printer's electronic circuits must be energized. 2. Paper must be loaded and ready to advance. 3. The printer must be "on line" with the microprocessor. As each of the above conditions is satisfied, a HIGH is generated and applied to a 3-input logic gate. When all three conditions are met, the logic gate produces a HIGH output indicating readiness to print. The basic logic gate used in this circuit would be an): A) NOR gate. B) NOT gate. C) OR gate. D) AND gate.
Answer:
D) AND gate.
Explanation:
Given that:
A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print
These conditions are:
1. The printer's electronic circuits must be energized.
2. Paper must be loaded and ready to advance.
3. The printer must be "on line" with the microprocessor.
Now; if these conditions are met the logic gate produces a HIGH output indicating readiness to print.
The objective here is to determine the basic logic gate used in this circuit.
Now;
For NOR gate;
NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.
For NOT gate.
NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.
Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".
Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.
If someone told you that a certain AC circuit was a capacitive, you would know that in that circuit the current
A) current and voltage are zero
B) leads the voltage
C) and voltage are in phrase
D) lags the voltage
Answer:
B) leads the voltage
Explanation:
One way to think about it is that the current causes charge to be accumulated on the capacitor, changing its voltage. The current must be non-zero before the voltage can change. Hence current leads voltage.
Find the requested quantities for the circuit. We used the mesh-current method to identify the meshes. We then identified the mesh currents and wrote a KVL equation for each mesh and a constraint equation for the dependent source that defines its controlling variable in terms of the mesh currents. We solved these equations simultaneously for the unknown mesh currents and constrained current, and we checked the solution by verifying that the power in the circuit balances Now use the mesh-current values to calculate the voltage v0 and the total power generated in the circuit. Enter your answers directly on the figure.
Answer:
Explanation:
The image that is supposed to be attached to the question is displayed in the diagram below.
Applying Nodal Analysis at node 1;
[tex]\dfrac{V_o -50}{12.5*10^{-3}} + \dfrac{V_o}{50*10^3}+\dfrac{V_o-7500 \ in}{10*10^3}=0[/tex]
where;
[tex]in = \dfrac{V_o}{50*10^3}[/tex] (from the circuit)
= [tex]\dfrac{V_o-50}{12.5}+\dfrac{V_o}{50} + V_o -\dfrac{7500 *V_o }{\frac{50*10^3}{10}}=0[/tex]
= [tex]V_o [ \dfrac{1}{12.5}+\dfrac{1}{50}+\dfrac{1}{10}-\dfrac{75}{500}] = \dfrac{50}{12.5}[/tex]
= [tex]V_o[ \dfrac{500*500+12.5*5000+12.5*5000*5-75*12.5*500}{12.5*50*10*500}]= \dfrac{50}{12.5}[/tex]
= [tex]V_o = 80 \ volts[/tex]
[tex]in = \frac{80}{50*10^3}= 1.6 mA \\ \\ 7500*in = 120 volts \\ \\ I = \frac{120-80}{10(10^3} =4*10^{-3} Amps \\ \\ \\ \\ P_{generated} = 75000*in*I \\ \\ P_{generated} = 120*4*10^{-3} \\ \\ P_{generated} = 480 \ MW[/tex]
Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radius of the inner sheave is 125 mm and the radius of the outer sheave is 250 mm. Knowing that when the system is at rest, the tension is 100 N in both portions of belt B and 160 N in both portions of belt C, determine the reactions at A and D. Assume that the bearing at D does not exert any axial thrust.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
When subject to an unknown torque, the shear stress in a 2 mm thick rectangular tube of dimension 100 mm x 200 mm was found to be 50 MPa. Using the same amount of material, if the dimensions are changed to 50 mm x 250 mm, what will be the shear stress (in MPa)? The breadth and depth of the section are given along the centreline of the wall.
Answer:
The shear stress will be 80 MPa
Explanation:
Here we have;
τ = (T·r)/J
For rectangular tube, we have;
Average shear stress given as follows;
Where;
[tex]\tau_{ave} = \frac{T}{2tA_{m}}[/tex]
[tex]A_m[/tex] = 100 mm × 200 mm = 20000 mm² = 0.02 m²
t = Thickness of the shaft in question = 2 mm = 0.002 m
T = Applied torque
Therefore, 50 MPa = T/(2×0.002×0.02)
T = 50 MPa × 0.00008 m³ = 4000 N·m
Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m
Therefore, [tex]A_m[/tex] = 0.05 m × 0.25 m = 0.0125 m².
Therefore, from the following average shear stress formula, we have;
[tex]\tau_{ave} = \frac{T}{2tA_{m}}[/tex]
Plugging in then values, gives;
[tex]\tau_{ave} = \frac{4000}{2\times 0.002 \times 0.0125} = 80,000,000 Pa[/tex]
The shear stress will be 80,000,000 Pa or 80 MPa.
Because A-B=A+ (-B), the subtraction of signed numbers can be accomplished by adding the complement. Subtract each of the following pairs of 5-bit binary numbers by adding the complement of the subtrahend to the minuend. Indicate when an overflow occurs. Assume that negative number are represented in 1’s complement. Then repeat using 2’s complement.
a) 01001-11010b) 11010-11001c) 10110-01101d) 11011-00111e) 11100-10101
Answer:
Using 1's complement
a)
Therefore the difference is -10001
b)
Therefore the difference is 00001
c)
Therefore the difference is 01001
d)
Therefore the difference is 10100
e)
Therefore the difference is 00111
Explanation:
Using 1's complement
a) The 1's complement of the subtrahend 11010 = 00101.
Therefore 01001-11010 = 01001 + 00101 = 01110
Since no overflow, we take the 1's complement of the result and it is negative.
Therefore the difference is -10001
b) The 1's complement of the subtrahend 11001 = 00110.
Therefore 11010-11001 = 11010 + 00110 =1 00000
Since there is an overflow, we add the overflow to the result
Therefore the difference is 00001
c) The 1's complement of the subtrahend 01101 = 10010
Therefore 10110-01101 = 10110 + 10010 = 1 01000
Since there is an overflow, we add the overflow to the result
Therefore the difference is 01001
d) The 1's complement of the subtrahend 00111 = 11000
Therefore 11011-00111= 11011 + 11000 = 1 10011
Since there is an overflow, we add the overflow to the result
Therefore the difference is 10100
e) The 1's complement of the subtrahend 10101 = 01010
Therefore 11100-10101= 11100 + 01010 = 1 00110
Since there is an overflow, we add the overflow to the result
Therefore the difference is 00111
Using 2's complement
a) The 2's complement of the subtrahend 11010 = 00110.
Therefore 01001-11010 = 01001 + 00110 = 01111
Since no overflow, we take the 2's complement of the result and it is negative.
Therefore the difference is -10001
b) The 2's complement of the subtrahend 11001 = 00111.
Therefore 11010-11001 = 11010 + 00111 =1 00001
Since there is an overflow, we drop the overflow
Therefore the difference is 00001
c) The 1's complement of the subtrahend 01101 = 10011
Therefore 10110-01101 = 10110 + 10011 = 1 01001
Since there is an overflow, we drop the overflow
Therefore the difference is 01001
d) The 1's complement of the subtrahend 00111 = 11001
Therefore 11011-00111= 11011 + 11001 = 1 10100
Since there is an overflow, we drop the overflow
Therefore the difference is 10100
e) The 1's complement of the subtrahend 10101 = 01011
Therefore 11100-10101= 11100 + 01011 = 1 00111
Since there is an overflow, we drop the overflow
Therefore the difference is 00111
A Newtonian liquid flows in the annular space between to fixed horizontal concentric cylinders. The radius of the inner cylinder is ri and the outer cylinder is ro. Two static pressure taps separated by a distance L along the outer pipe are connected to a manometer with reading of h. Develop an expression for the shear stress on the inner and outer cylinder walls as a function of h. Assume the flow is fully developed and laminar.
Answer:
See explaination
Explanation:
please kindly see attachment for the step by step solution of the given problem
Durante el segundo trimestre de 2001, Tiger Woods fue el golfista que más dinero ganó en el PGATour. Sus ganancias sumaron un total de $5517777. De los 10 principales golfistas mejor remunerados, siete usaron pelotas de golf de la marca Titleist (sitio web de PGATour). Suponga que seleccionan al azar a dos de los 10 principales golfistas que ganan más dinero. Determine:
Answer: a. 0.4667
b. 0.4667 and C 0.0667
Explanation:
Given Data:
N = population size (10)
n = random selection (2)
r = number of observations = 7
Therefore
f(y) = ( r/y ) ( N - r / n - y ) / ( N /n )
When y = 1
f(1) = ( 7/1 ) ( 10 - 7 / 2 -1 ) / ( 10/2 )
= 7 / 15
= 0.4667
When y = 2
f(2) = ( 7/2 ) ( 10 - 7 / 2 -2 ) / ( 10/2 )
= 7 / 15
= 0.4667
When y = 0
f(0) = ( 7/0 ) ( 10 - 7 / 2 -0) / ( 10/2 )
= 1 / 15
= 0.0667
Firebrick is referred to as "common brick" because it is the most commonly used
type of brick.
Answer: Firebrick is referred to as "common brick" because it is the most commonly used type of brick --- False
Explanation:
Firebrick is not the most commonly used, it is a refractory ceramic brick used in lining of furnaces and kilns, which is built basically to withstand or resist high temperature.The most commonly used type of brick, is the Building Brick called "common brick" because it is versatile and used for applications where appearance is not an important factor.
It is required to design and implement: 1. A counter which counts from 0 to 255 with seven segment display.
2. A logic function y=∑( 0,3,5,10,16,20,30 35).
3. Summing and subtracting circuit of 8 digit numbers.
4. ROM (a,b,c,d) to 60 to 75 in binary.
5. A timing module which counts time in us or ms or seconds.
Answer:
A counter which counts from 0 to 255 with seven segment display
Timer Mode Control (TMOD)
Explanation:
Why does a BJT transistor require detailed calculations for its base resistor value to operate?
Answer: Because of the role the base region play in the transistor.
Explanation:
The base region of BJT transistor - an opposite polarity charge carrier from emitter region to collector region, plays a vital role in triggering for a sufficient emiter - to - collector current.
The current received by the base region of BJT determines the effect of the continue flow of current into the collector region which will eventually determine the output current.
– A cloud customer has asked you to do a forensics analysis of data stored in on CSP’s server. The customer’s attorney explains that the CSP offers little support for data acquisition and analysis will help you with data collection for a fee. The attorney asks you to prepare a memo with detailed questions of what you need to know to perform the task .She plans to use this memo to negotiate for services you will provide in collecting and analyzing evidence .Write a one –to two page menu with questions to ask the CSP .
Answer:
A one -two pages menu was written with questions directed to the CSP which is stated below in the explanation section
Explanation:
Solution
If CSP has no team or limited staff, you will need to ask the following questions to understand how the CSP is set up:
Is detailed knowledge of cloud topology, storage devices is available ?Are there any restrictions in taking digital evidence from a cloud storage?For e-discovery demands on multi tenant cloud systems, is the data of investigation local or remote?Does the investigator have the power to make use of cloud staff conduct an investigation? What is the relationship of CSP's with cloud users?What are the SLA's and what are the guidelines to define them ? SLAs should also specify support options, penalties for services not provided, system performance,fees, provided software/hardware. CSP must explain who has the right to access the data ? and limitations for conducting acquisitions for an investigation.For guidelines of operations, digital forensics should review CSP's policies, and standards..What are the CSP's business continuity and disaster recovery plans.Are there Any plans to revise current laws ?Are there Any cases involving data commingling with other customer's data?Ask What law controls data stored in the cloud is a challenge?To access evidence in the cloud :
What is the configuration of the CSP?Is the data storage location secretly kept or it is open ?Are there any court orders, subpoenas with prior notice, search warrants etc?What are the procedures for log keeping ? so that complications we not arise in the investigations chain of evidence.What is the configuration of the CSP?What is the right key of encryption to read the data if at all the CSP has provided encryption to the data.Is there any threat from hackers so that they will not use any malware an modify the file meta data?Does CSP have a personnel trained to respond to network incidents?Who are the data owners, identity protection, users and access controls for a better role management.One of my tools stopped working, and I found that in order to fix it, I need to replace a broken wire. The broken wire has a diameter of 5mm and a length of 15 cm. When it was connected to the battery, a current of 12.5mA flowed through the wire. The only wire that I have at home is the same material, but has a diameter of 2mm.
a) If I replace the broken piece of wire with a 15cm length of my 2mm diameter wire, what will happen to the current (increase or decrease)?
What is the value of the current through the new wire (assume that it’s connected to the same battery as the original wire)? Hint: Think about the resistances of the old and new wires.
b) How long should the new wire (2mm diameter) be if I want to ensure that the current passing through it is the same as the original wire (12.5mA)?
Answer:
a) I₂ = 2 mA (The current has decreased)
b) L₂ = 2.4 cm
Explanation:
Consider the old wire as wire 1 and the new wire as wire 2. The data that we have from the question is:
Current through wire 1 = I₁ = 12.5 mA
Diameter of wire 1 = d₁ = 5 mm
Length of wire 1 = L₁ = 15 cm
Cross-Sectional Area of wire 1 = A₁ = πd₁²/4 = π(5 mm)²/4 = 19.63 mm²
Diameter of wire 2 = d₂ = 2 mm
Cross-Sectional Area of wire 2 = A₂ = πd₂²/4 = π(2 mm)²/4 = 3.14 mm²
a)
Length of wire 2 = L₂ = 15 cm
Since, the battery is same. Therefore, the voltage will be same for both wires.
V₁ = V₂
using Ohm's Law (V = IR)
I₁R₁ = I₂R₂
Since resistance of wire is given by formula: R = ρL/A
Therefore,
I₁ρ₁L₁/A₁ = I₂ρ₂L₂/A₂
where,
ρ = resistivity, and it depends upon material of wire and the material of both wires is same and the length of wires is also same.
Hence, ρ₁ = ρ₂
and L₁ = L₂
and the equation becomes:
I₁/A₁ = I₂/A₂
I₂ = I₁A₂/A₁
I₂ = (12.5 mA)(3.14 mm²)/(19.63 mm²)
I₂ = 2 mA
Thus, the current has decreased.
b)
In order to have same current the resistance of both wires must be same:
R₁ = R₂
ρ₁L₁/A₁ = ρ₂L₂/A₂
Since, ρ₁ = ρ₂
Therefore,
L₁/A₁ = L₂/A₂
L₂ = L₁A₂/A₁
L₂ = (15 cm)(3.14 mm²)/(19.63 mm²)
L₂ = 2.4 cm
Describe the process of sowing of maize by a seed drill.
Answer:
Find explanation below.
Explanation:
Seed drills make it possible for seeds to be planted at equal spacing or distances.
The process begins with the plowing of the field. This is is done to open up the soil and enable it to easily accept the seeds. The farmer then sets the drill in a manner as to accommodate the size of the maize seeds. The maize seeds are then loaded on the hopper located above the drill. The drill can then start planting the maize seeds while giving them equal spacing and depth. Depth of 7 to 8 cm, and spacing of 10-13 cm are recommended for optimum results.Seed drills are important because they allow seeds to be planted at specific depths and distances. This allows for increased crop yield. They also help to curb the growth of weeds.
An application demands that a sinusoidal pressure variation of 250 Hz be measured with no more than 2% dynamic error. In selecting a suitable pressure transducer from a vendor catalog, you note that a desirable line of transducers has a fixed natural frequency of 600 Hz but that you have a choice of transducer damping ratios of between 0.5 and 1.5 in increments of 0.05. Select a suitable transducer.
Answer:
Explanation:
Given that:
f = 250 Hz
[tex]\delta[/tex]= 2%
[tex]f_n[/tex]= 600 Hz
[tex]\zeta[/tex] = 0.5 to 1.5 increment by 0.05
[tex]F = A sin (Xt)[/tex]
For 250 Hz = 250 cycle/sec
[tex]X = 2 \pi t 250[/tex]
[tex]X = 500 \pi t[/tex]
[tex]X = Asin (500 \pi t)[/tex]
[tex]\omega = 250 \\ \\ \omega_n = 600[/tex]
M = 0.98 , 1.02
[tex]M_{(w)}} = \sqrt{[1-(\frac{w}{w_n})^2 + ( 2\zeta \frac{w}{w_n})^2}[/tex]
[tex]\frac{1}{M} = [1-(\frac{w}{w_n})^2]+(2 \zeta \frac{w}{w_n})^2[/tex]
[tex]\zeta = \frac{w_n}{2w}\sqrt{\frac{1}{M^2}-(1-(\frac{w}{w_n})^2)^2}[/tex]
[tex]\zeta = \frac{600}{2(250)}\sqrt{\frac{1}{0.98^2}-(1-(\frac{250}{600})^2)^2}[/tex]
[tex]\zeta = 0.7183[/tex]
At 0.7183 value of damping ratio the error value was 2% at 0.98 value of M
[tex]\frac{1}{M} = [1-(\frac{w}{w_n})^2]+(2 \zeta \frac{w}{w_n})^2[/tex]
[tex]\zeta = \frac{w_n}{2w}\sqrt{\frac{1}{M^2}-(1-(\frac{w}{w_n})^2)^2}[/tex]
[tex]\zeta = \frac{600}{2(250)}\sqrt{\frac{1}{1.02^2}-(1-(\frac{250}{600})^2)^2}[/tex]
[tex]\zeta = 0.6330[/tex]
At 0.6330 value of damping ratio the error value was 2% at 1.02 value of M.
Hence, the damping ratio [tex]\zeta[/tex] of the transducer must be placed between 0.6330 to 0.7183
Wheels A and B have weights of 150 lb and 100 lb , respectively. Initially, wheel A rotates clockwise with a constant angular velocity of 100 / A rad s and wheel B is at rest. If A is brought into contact with B, determine the time required for both wheels to attain the same angular velocity. The coefficient of kinetic friction between the two wheels is 0.3 k and the radii of gyration of A and B about their respective centers of mass are 1 A k ft and 0.75 B k ft . Neglect the weight of link AC.
The image attached that is supposed to be attached to the question is shown in the first file below.
Answer:
t = 2.19 seconds
Explanation:
The free body diagram showing the center of mass A and B is attached in the second diagram below.
NOTE : that from the second diagram; Mass A and B do not have any acceleration
Taking the moment about wheel A:
[tex]\sum M_A = I_A \alpha _A[/tex]
[tex]-f(r_A) = I_A \alpha _A ----- (1)[/tex]
The equilibrium forces in the y-direction is 0
i.e
[tex]F_y = 0[/tex]
So;
[tex]N +T sin 30^0 -W_A = 0 ----- (2)[/tex]
The equilibrium forces in the x-direction is as follows:
[tex]\sum F_x = 0[/tex]
[tex]Tcos 30^0 + f= 0 -----(3)[/tex]
The kinetic friction f can be expressed as :
[tex]f = \mu _k N[/tex]
From above equation (2) and equation (3);
[tex]N + [\dfrac{-f}{cos 30^0}]sin 30^0 -150 =0[/tex]
[tex]N - \mu _k N \ tan 30^0 -150 =0[/tex]
[tex]N = \dfrac{150}{1-0.3 \ tan 30^0}[/tex]
N = 181.423 lb
Similarly; from equation(1)
[tex]\alpha_A = - \dfrac{f(r_A)}{I_A}[/tex]
[tex]\alpha _A = \dfrac{-\mu_k N(r_A)}{I_A}[/tex]
[tex]\alpha _A = \dfrac{-0.3*181.423*1.25}{\frac{150}{32.2}*I^2}[/tex]
[tex]\alpha _A =-14.6045 \ rad/s^2[/tex]
However; from the kinematics ; as moments are constant ; so is the angular acceleration is constant )
Thus;
[tex]\omega _A - \omega_o^A = \alpha_A t[/tex]
[tex]\omega _A = \omega_o^A + \alpha_A t[/tex]
[tex]\omega _A = 100 -14.6045 \ t ---- (4)[/tex]
Let's take a look at wheel B now;
Taking the moment about wheel B from the equation of motion:
[tex]\sum M_B = I_B \alpha _B[/tex]
[tex]f(r_B) = I_B \alpha _B[/tex]
[tex]\mu_k N (r_B) = I_B \alpha_B[/tex]
[tex]\mu_k N (r_B) = \dfrac{W_B}{g}* k^2_B \alpha_B[/tex]
[tex]\alpha_B = \dfrac{0.3*181.423*1}{\frac{100}{32.2}*0.75^2}[/tex]
[tex]\alpha = 31.1563 \ rad/s^2[/tex]
Again; from the kinematics; as the moments are constant which lead to the angular accleration;
[tex]\omega _B = \omega _o^B + \alpha _B \ t[/tex]
[tex]\omega _B =0 + 31.156 \ t-----(5)[/tex]
From equation 4 and 5 which attain the same angular velocity; we have;
[tex]\omega^A = \omega^B[/tex]
100 - 14.6045 t = 31.1563 t
100 = 31.1563 t + 14.6045 t
100 = 45.761 t
t = 100/45.761
t = 2.19 seconds
Superheated water vapor at a pressure of 20 MPa, a temperature of 500oC, and a flow rate of 10 kg/s is to be brought to a saturated vapor state at 10 MPa in an open feedwater heater. This process is accomplished by mixing this stream with a stream of liquid water at 20oC and 10 MPa. What flow rate is needed for the liquid stream?
Answer:
1.96 kg/s.
Explanation:
So, we are given the following data or parameters or information which we are going to use in solving this question effectively and these data are;
=> Superheated water vapor at a pressure = 20 MPa,
=> temperature = 500°C,
=> " flow rate of 10 kg/s is to be brought to a saturated vapor state at 10 MPa in an open feedwater heater."
=> "mixing this stream with a stream of liquid water at 20°C and 10 MPa."
K1 = 3241.18, k2 = 93.28 and 2725.47.
Therefore, m1 + m2= m3.
10(3241.18) + m2 (93.28) = (10 + m3) 2725.47.
=> 1.96 kg/s.
A. A solidified lava flow containing zircon mineral crystals is present in a sequence of rock layers that are exposed in a hillside.A mass spectrometer analysis was used to count the atoms of uranium-235 and lead-207 isotopes in zircon samples from thelava flow. The analysis revealed that 71% of the atoms were uranium-235, and 29% of the atoms were lead-207. Refer toFIGURE 8.11 to help you answer the following questions.1. About how many half-lives of the uranium-235to lead-207 decay pair have elapsed in the zircon crystals? ______________2. What is the absolute age of the lava flow based onits zircon crystals? Show your calculations.3. What is the age of the rocklayers above the lava flow? _______________4. What is the age of the rocklayers beneath the lava flow? _______________B. Astronomers think that Earth probably formed at the same time as
Answer:
1. 0.494
2. = 352.299 million years
3. It is between 0 years to 352.299 million years
4. greater than the 352.299 million years.
Explanation:
Given
Uranium 235 ———— lead 207( zircon sample)
t= 0 100%. 0%
t= t. 71%. 29%
1)- half lives elapsed = n
(1/2)n = 0.71
By taking log and solving n = 0.494
No of half lives = 0.494
2) Calculating the age of the larva
- age of lava flow = half life of uranium 235 x n
= 713 x 0.494
= 352.299 million years
3)- The rock layers was created above this lava flow is of later occurrence so the age will be lower than that of the age of the lava flow so the age of rock over the lava flow will lower than the 352.299 million years it will be in between 0 years to 352.299 million years.
4)- The rock layers created underneath this lava flow is of earlier occurrence so its age is more than that of the age of the lava flow so the age of rock will more than the lava flow is greater than the 352.299 million years.
B. The astronomers think the earth was created as all of the other rock materials in our solar system, including the oldest meteorites. The oldest meteorites ever found on Earth contain nearly equal amounts of both Uranium-238 and Lead-206.
The number of half-lives of the uranium-235to lead-207 decay pair have elapsed in the zircon crystals is; 0.494
What is the number of half lives?We are given;
The atoms of Uranium-235 and lead-207 which made up the zircon sample.
At t = 0; Uranium-235 is 100% while lead-207 is 0%
At t = t; Uranium-235 is 71% while lead-207 is 29%
1) Let the half lives that elapsed be n. Thus;
(¹/₂)ⁿ = 0.71
n*log0.5 = log 0.71
n = (log 0.71)/(log 0.5)
n = 0.494
Thus;
Number of half lives = 0.494
2) Formula to get the absolute age of the larva is;
Absolute age age of lava flow = half life of uranium-235 * n
The half life of uranium-235 is 713 million years. Thus;
Absolute age age of lava flow = 713 * 0.494
Absolute age of lava flow = 352.22 million years
3) The rock layers above the lava flow were created after the lava flow and so the age will be lower than that of the age of the lava flow. Thus, it's age will be between 0 years and 352.299 million years.
4) A: The rock layers beneath the lava flow were in existence earlier than the lava flow and as such, the age of rock layers beneath the lava flow will be greater than 352.22 million years.
B; The astronomers think the earth was created as all of the other rock materials in our solar system.
Read more about Half life at; https://brainly.com/question/26148784
A small family home in Tucson, Arizona, has a rooftop area of 1967 square feet, and it is possible to capture rain falling on about 56% of the roof. A typical annual rainfall is about 14 inches. If the family wanted to install a tank to capture the rain for an entire year, without using any of it, what would be the required volume of the tank in m3 and in gallons? How much would the water weigh when the tank was full (in N and in lbf)?
Answer:
V = 36.4 m³ = 4.86 gallons
W = 80193.88 lbf = 356720 N = 356.72 KN
Explanation:
We have the following data given in the question:
At = Total area of roof = 1967 ft²
h = Annual Rainfall = 14 inches = 1.17 ft
V = Volume of tank in m³ and gallons = ?
W = Weight of water in N and lbf = ?
So, for volume we know that the area of roof that receives rainfall is 56% of total area and 14 inches of annual rainfall means that there is a standing height of 14 inches of rain water for a given area, for 1 year.
Area to receive rain = A = 0.56*1967 ft² = 1101.52 ft²
Now,
Volume = V = A * h = 1101.52 ft²)(1.17 ft)
V = 1285.11 ft³
Converting to m³:
V = (1285.11 ft³)(1 m³/35.3147 ft³)
V = 36.4 m³
Converting to gallons:
V = (1285.11 ft³)(1 m³/264.172 gal)
V = 4.86 gal
Now, for the weight of water, we use formula:
W = ρVg
where,
W = weight of water = ?
ρ = Density of water = 1000 kg/m³
V = Volume of tank = 36.4 m³
g = 9.8 m/s²
Therefore,
W = (1000 kg/m³)(36.4 m³)(9.8 m/s²)
W = 356720 N = 356.72 KN
Converting to lbf:
W = (356720 N)(1 lbf/4.44822 N)
W = 80193.88 lbf
. (20 pts) A horizontal cylindrical pipe (k = 10 W/m·K) has an outer diameter of 15 cm and a wall thickness of 5 cm. The pipe is situated in a stationary air, where the air and surrounding temperature is 27°C. The outer surface temperature of the pipe is 127°C, and the pipe surface has an emissivity of 0.5. Determine the inner surface temperature of the pipe. Use the following air properties for the analysis: k = 0.03 W/m∙K, ν = 20.92 × 10−6 m2 /s, α = 29.90 × 10−6 m2 /s, Pr = 0.70
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Last autumn our office received complaints of a large fish kill along the Ohio River, indicating that someone had discharged highly toxic material into the river. Our water monitoring stations at Cincinnati and Portsmouth, Ohio (119 miles apart) report that a large slug of phenol is moving down the river and we strongly suspect that this is the cause of the pollution. The slug took 9 hours to pass the Portsmouth monitoring station, and its concentration peaked at 8:00 A.M. Monday. About 24 hours later the slug peaked at Cincinnati, taking 12 hours to pass this monitoring station. Phenol is used at a number of locations on the Ohio River, and their distance upriver from Cincinnati are as follows:_______.Ashland, KY-150 miles upstream Marietta, OH-303Huntington, WV-168 Wheeling, WV-385Pomeroy, OH-222 Steubenville, OH-425Parkersburg, WV-290 Pittsburgh, PA-500What can you say about the probable pollution source?
Answer:
Explanation:
Base on the scenario been described in the question, this solution, I have assumed that the river has the exact flow all through and therefore, dispersion characteristics will is also the same
This implies that the river doesn't separate or any other water body does not mixed with Ohio river
For us to begin we can assume that all phenol was injected at a time at upstream from Cincinnati.
Please check the image below
For a p-n-p BJT with NE 7 NB 7 NC, show the dominant current components, with proper arrows, for directions in the normal active mode. If IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA, calculate the base transport factor, emitter injection efficiency, common-base current gain, common-emitter current gain, and ICBO. If the minority stored base charge is 4.9 * 10-11 C, calculate the base transit time and lifetime.
Answer:
=> base transport factor = 0.98.
=> emitter injection efficiency = 0.99.
=> common-base current gain = 0.97.
=> common-emitter current gain = 32.34.
=> ICBO = 1 × 10^-6 A.
=> base transit time = 0.325.
=> lifetime = 1.875.
Explanation:
(Kindly check the attachment for the diagram showing the dominant current components, with proper arrows, for directions in the normal active mode).
The following parameters or data are given for a p-n-p BJT with NE 7 NB 7 NC and they are: IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA.
(1). The base transport factor = ICp/IEp=9.8/10 = 0.98.
(2). emitter injection efficiency =IEp/ IEp + ICn = 10/10 + 0.1 = 0.99.
(3).common-base current gain = 0.98 × 0.99 = 0.9702.
(4).common-emitter current gain =0.97 / 1- 0.97 = 32.34.
(5). Icbo = Ico = 1 × 10^-6 A.
(6). base transit time = 1248 × 10^-2 × (1.38× 10^-23/1.603 × 10^-19). = 0.325.
(7).lifetime;
= > 2 = √0.325 + √ lifetime.
= Lifetime = 2.875.
Calibrations on a recent version of an operating system showed that on the client side, there is a delay of at least 0.5 ms for a packet to get from an application to the network interface and a delay of 1.4 ms for the opposite path (network interface to application buffer). The corresponding minimum delays for the server are 0.20 ms and 0.30 ms, respectively.
What would be the accuracy of a run of the Cristian's algorithm between a client and server, both running this version of Linux, if the round trip time measured at the client is 6.6 ms?
Answer:
4.2ms
Explanation:
Calibrated time= 0.3+0.2+0.5+1.4= 2.4
Measured time= 6.6ms
Accuracy is closeness of measurement to an observed or true value
Accuracy= 6.6-2.4= 4.2ms
Suppose that the material to be used in a fuse melts when the current density rises to 459 A/cm2. What diameter of cylindrical wire should be used to limit the current to 0.53 A?
Answer:
The required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm² .
Explanation:
We are given current density of 459 A/cm² and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.
Recall that current density is given by
j = I/A
where I is the current flowing through the wire and A is the area of the wire
A = πr²
but r = d/2 so
A = π(d/2)²
A = πd²/4
so the equation of current density becomes
j = I/πd²/4
j = 4I/πd²
Re-arrange the equation for d
d² = 4I/jπ
d = √4I/jπ
d = √(4*0.53)/(459π)
d = 0.0383 cm
Therefore, the required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm² .
Answer:
diameter is 1 cm
Explanation:
diameter is 1 cm