Answer:
L = 10.32 m
Explanation:
The resistance of a copper-nickel alloy wire is given by the following formula:
[tex]R = \frac{\rho L}{A}[/tex]
where,
R = Resistance = 0.5 Ω
ρ = resistivity of copper-nickel alloy = 3.8 x 10⁻⁸ Ωm
L = Length of wire = ?
A = cross-sectional area of wire = [tex]\frac{\pi d^2}{4} = \frac{\pi(1\ x\ 10^{-3}\ m)^2}{4}[/tex] = 7.85 x 10⁻⁷ m²
Therefore,
[tex]0.5\ \Omega = \frac{(3.8\ x\ 10^{-8}\ \Omega.m)L}{7.85\ x\ 10^{-7}\ m^2}[/tex]
L = 10.32 m
Select the correct answer. The most frequent maintenance task for a car is: A. Oil changes B. Tire replacements C. Coolant changes D. Brake replacements
Answer:
A. Oil changes
Explanation:
It depends on the car and its usage and environment. Usually oil is supposed to be changed every few months, more often if the car is driven a lot. Coolant changes may be indicated as seasons change, so will generally occur less frequently than oil changes.
Tire and brake replacement depend on usage and driving habits. Some owners may never have to replace either one, if they trade their car every year or two. Folks who drive with their foot on the brake pedal may have to replace brakes relatively often.
The most frequent task is generally oil changes.
Answer:
A. Oil changesthe most frequent maintenance task for a carSteam enters a turbine with a pressure of 30 bar, a temperature of 400 oC, and a velocity of 160 m/s. Saturated vapor at 100 oC exits with a velocity of 100 m/s. At steady state, the turbine develops work equal to 540 kJ per kg of steam flowing through the turbine. Heat transfer between the turbine and its surroundings occurs at an average outer surface temperature of 350 K. Determine the rate at which entropy is produced within the turbine per kg of steam flo
Answer:
The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.
Explanation:
By either the Principle of Mass Conservation and First and Second Laws of Thermodynamics, we model the steam turbine by the two equations described below:
Principle of Mass Conservation
[tex]\dot m_{in} - \dot m_{out} = 0[/tex] (1)
First Law of Thermodynamics
[tex]-\dot Q_{out} + \dot m \cdot \left[h_{in}-h_{out}+ \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} \right] = 0[/tex] (2)
Second Law of Thermodynamics
[tex]-\frac{\dot Q_{out}}{T_{out}} + \dot m\cdot (s_{in}-s_{out}) + \dot S_{gen} = 0[/tex] (3)
By dividing each each expression by [tex]\dot m[/tex], we have the following system of equations:
[tex]-q_{out} + h_{in}-h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} = 0[/tex] (2b)
[tex]-\frac{q_{out}}{T_{out}} + s_{in}-s_{out} + s_{gen} = 0[/tex] (3b)
Where:
[tex]\dot Q_{out}[/tex] - Heat transfer rate between the turbine and its surroundings, in kilowatts.
[tex]q_{out}[/tex] - Specific heat transfer between the turbine and its surroundings, in kilojoules per kilogram.
[tex]T_{out}[/tex] - Outer surface temperature of the turbine, in Kelvin.
[tex]\dot m[/tex] - Mass flow rate through the turbine, in kilograms per second.
[tex]h_{in}[/tex], [tex]h_{out}[/tex] - Specific enthalpy of water at inlet and outlet, in kilojoules per kilogram.
[tex]v_{in}[/tex], [tex]v_{out}[/tex] - Speed of water at inlet and outlet, in meters per second.
[tex]w_{out}[/tex] - Specific work of the turbine, in kilojoules per kilogram.
[tex]s_{in}[/tex], [tex]s_{out}[/tex] - Specific entropy of water at inlet and outlet, in kilojoules per kilogram-Kelvin.
[tex]s_{gen}[/tex] - Specific generated entropy, in kilojoules per kilogram-Kelvin.
By property charts for steam, we get the following information:
Inlet
[tex]T = 400\,^{\circ}C[/tex], [tex]p = 3000\,kPa[/tex], [tex]h = 3231.7\,\frac{kJ}{kg}[/tex], [tex]s = 6.9235\,\frac{kJ}{kg\cdot K}[/tex]
Outlet
[tex]T = 100\,^{\circ}C[/tex], [tex]p = 101.42\,kPa[/tex], [tex]h = 2675.6\,\frac{kJ}{kg}[/tex], [tex]s = 7.3542\,\frac{kJ}{kg\cdot K}[/tex]
If we know that [tex]h_{in} = 3231.7\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 2675.6\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 160\,\frac{m}{s}[/tex], [tex]v_{out} = 100\,\frac{m}{s}[/tex], [tex]w_{out} = 540\,\frac{kJ}{kg}[/tex], [tex]T_{out} = 350\,K[/tex], [tex]s_{in} = 6.9235\,\frac{kJ}{kg\cdot K}[/tex] and [tex]s_{out} = 7.3542\,\frac{kJ}{kg\cdot K}[/tex], then the rate at which entropy is produced withing the turbine is:
[tex]q_{out} = h_{in} - h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2})-w_{out}[/tex]
[tex]q_{out} = 3231.7\,\frac{kJ}{kg} - 2675.6\,\frac{kJ}{kg} + \frac{1}{2}\cdot \left[\left(160\,\frac{m}{s} \right)^{2}-\left(100\,\frac{m}{s} \right)^{2}\right] - 540\,\frac{kJ}{kg}[/tex]
[tex]q_{out} = 7816.1\,\frac{kJ}{kg}[/tex]
[tex]s_{gen} = \frac{q_{out}}{T_{out}}+s_{out}-s_{in}[/tex]
[tex]s_{gen} = \frac{7816.1\,\frac{kJ}{kg} }{350\,K} + 7.3542\,\frac{kJ}{kg\cdot K} - 6.9235\,\frac{kJ}{kg\cdot K}[/tex]
[tex]s_{gen} = 22.762\,\frac{kJ}{kg\cdot K}[/tex]
The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.
A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)
Answer:
Hydrostatic force = 41168 N
Explanation:
Complete question
A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)
Let "x" be the side length submerged in water.
Then
w(x)/base = (4+3-x)/altitude
w(x)/5 = (4+3-x)/3
w(x) = 5* (7-x)/3
Hydrostatic force = 62.5 integration of x * 4 * (10-x)/3 with limits from 4 to 7
HF = integration of 40x - 4x^2/3
HF = 20x^2 - 4x^3/9 with limit 4 to 7
HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))
HF = 658.69 N *62.5 = 41168 N
The hydrodynamic entrance length for laminar flow in a tube is linearly proportional to Reynolds number, Re, while turbulent flow is approximately independent of Re.
a. True
b. False
Answer:
B. False
Explanation:
In laminar flow, the Reynolds number (RE) is less than 2100 while in turbulent flow, it is greater than 4000.
Now,the entrance lengths for both laminar and turbulent flow is usually dependent on Reynolds number.
Thus is because the hydrodynamic entrance length for laminar flow is given by the formula;
Le = 0.05 DRe
Where;
Le is hydrodynamic entrance length
D is diameter of tube
Re is Reynolds number
Meanwhile, for turbulent flow, entrance length is given by;
Le = 1.359D(Re)^(1/4)
In both cases, we can see that the hydrodynamic entry length is directly proportional to the Reynolds number and not independent. Thus, we can say that the statement in the question is false.
Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 21 m3/min and exits at 12 bar, 400 K. Heat transfer occurs at a rate of 3.5 kW from the compressor to its surroundings.
Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.
Answer:
- 46.5171kW
Explanation:
FIrst, the value given:
P1 = 1.05 bar (Initial pressure)
P2 = 12 bar (final pressure)
Heat transfer, Q = - 3.5 kW (It is negative because the compressor losses heat to the surroundings)
Mgaseous nitrogen = Mair = 28.0134 Kg/mol (constant)
Universal gas constant, Ru = 8.3143 Kj/Kgmolk
Specific gas constant, R = 0.28699 Kj/KgK
Initial temperature, T1 = 300 K
Final temperature, T2 = 400 K
Finding the volume:
P1V1 = RT1
V1 = RT1 ÷ P1
= (0.28699 Kj/KgK X 300k) ÷ 105
Note convert bar to Kj/Nm by multiply it by 100
V1 = 0.81997 m3/Kg
To get the mass flow rate:
m = volumetric flow rate / V1
= (21 m3/min x 1/60seconds) ÷ 0.81997 m3/Kg
= 0.4268Kg/s
Using tables for the enthalpy,
hT1 = 300.19 KJ/Kg
hT2 = 400.98 KJ/Kg
The enthalpy change = hT2 - hT1
= 100.79 KJ/Kg
Power, P = Q - (m X enthalpy change)
= - 3.5 - (0.4268 X 100.79)
= - 46.5171kW
A six-lane divided multilane highway (three lanes in each direction) has a measured free-flow speed of 50 mi/h. It is on mountainous terrain with a traffic stream consisting of 7% large trucks and buses and 3% recreational vehicles. The driver population adjustment in 0.92. One direction of the highway currently operates at maximum LOS C conditions and it is known that the highway has PHF = 0.90.
Required:
How many vehicles can be added to this highway before capacity is reached, assuming the proportion of vehicle types remains the same but the peak-hour factor increases to 0.95?
Answer:
2901 vehicles
Explanation:
We are given;
Percentage of large trucks & buses; p_t = 7% = 0.07
Percentage of recreational vehicles; p_r = 3% = 0.03
PHF = 0.90
Driver population adjustment; f_p = 0.92
First of all, let's Calculate the heavy vehicle factor from the formula;
f_hv = 1/[1 + p_t(e_t - 1) + p_r(e_r - 1)]
Where;
e_t = passenger car equivalents for trucks and buses
e_r = passenger car equivalents for recreational vehicles
From the table attached, for a mountainous terrain, e_t = 6 and e_r = 4. Thus;
f_hv = 1/[1 + 0.07(6 - 1) + 0.03(4 - 1)]
f_hv = 1.44
Let's now calculate the initial hourly volume from the formula;
v_p = V1/(PHF × N × f_hv × f_p)
Where;
v_p = 15-minute passenger-car equivalent flow rate
V1 = hourly volume
N = number of lanes in each direction
From online tables of LOS criteria for multilane freeway segments, v_p = 1300 pc/hr/ln
Thus;
1300 = V1/(0.9 × 3 × 1.44 × 0.92)
V1 = 1300 × (0.9 × 3 × 1.44 × 0.92)
V1 = 4650 veh/hr
Now, let's Calculate the final hourly volume;
From online sources, the maximum capacity of a 6 lane highway with free-flow speed of 50 mi/h is 2000 pc/hr/ln.
We are told the online peak-hour factor increases to 0.95 and so PHF = 0.95.
Thus;
2000 = V2/(0.95 × 3 × 1.44 × 0.92)
V2 = 2000(0.95 × 3 × 1.44 × 0.92)
V2 = 7551 veh/hr
Number of vehicles added to the highway = V2 - V1 = 7551 - 4650 = 2901 vehicles
Outline how the technological innovations of the Renaissance led to the Industrial Age.
Answer:
Printing press , lenses etc,
Explanation:
Almost all of the Renaissance innovations that influenced the world, printing press was one of them. The printing press is widely regarded as the Renaissance's most significant innovation. The press was invented by Johannes Gutenberg, and Germany's culture was altered forever.
Astrophysics, humanist psychology, the printed word, vernacular vocabulary of poetry, drawing, and sculpture practice are only a few of the Renaissance's main inventions.
What is the force applied to a bone with a cross-sectional area of 0.0004 m2 if it is being stressed with 5,000 Pa?
Answer:
[tex]2\ \text{N}[/tex]
Explanation:
[tex]A[/tex] = Cross sectional area of bone = [tex]0.0004\ \text{m}^2[/tex]
[tex]P[/tex] = Pressure = [tex]5000\ \text{Pa}[/tex]
[tex]F[/tex] = Force applied to the bone
Force is given by
[tex]F=PA[/tex]
[tex]\Rightarrow F=5000\times 0.0004[/tex]
[tex]\Rightarrow F=2\ \text{N}[/tex]
The force applied to the bone is [tex]2\ \text{N}[/tex].
We need to use the Cr-Au galvanic cell to power a 75.0-watt light bulb. Please determine how long (in min) the galvanic cell will power the light bulb before running out power. The galvanic cell uses the following solutions: 0.810 L 1.50 M Cr2(SO4)3 solution and 1.20 L 0.860 M Au(NO3)3 solution. power(W) = electric potential(V) x current(A); charge(C)= current(A) x time(s); 1 mol of electrons = 96485 C
Answer:
148.6 mins
Explanation:
Aim: using Cr-Au galvanic cell to power a 75 watt light bulb
Calculate how long the Galvanic cell will power the light bulb
power ( W ) = electric potential * current ---- ( 1 )
charge = current * time ------ ( 2 )
first step : determine the electric potential
E⁰cell = E⁰cathode - E⁰anode ( values gotten from balancing reaction )
= ( 1.5 - ( -0.74 )) = 2.24 V
Given that amount of AU^+3 is limited the
Ecell ( cell voltage ) ≈ E⁰cell = 2.24 V
back to equation 1
Power = 2.24 * current
∴ current = 75 / 2.24 = 33.48 A
back to equation 2
Charge = current * time
Time = Charge / current = ( 3 * 1.032 * 96485 ) / (33.48 * 60 )
= 148.6 mins
Note : 1 mol AU^+3 will give 3 e^-
1.032 mol AU^+3 = 3 * 1.032 mol e^-
hence charge = 3 * 1.032 * 96485
There is an AC series circuit that is constructed of a 150.0-ohm resistor along with 300.0 ohm inductive reactance and 200.0 ohm capacitive reactance. What is the difference in phase between the current and resistor voltage of the circuit
Answer:
0°
Explanation:
The resistor voltage has the same phase as the circuit current. There is no phase difference.
Answer:
0° (zero degree)Explanation:
the difference in pjase between the current and resistor voltage of the given 150.0 ohm, 300.0 ohm and 200.0 ohmHey answr this sajida Yusof
Answer:
that not even a question
If the old radiator is replaced with a new one that has longer tubes made of the same material and same thickness as those in the old unit, what should the total surface area available for heat exchange be in the new radiator to achieve the desired cooling temperature gradient
Answer: hello some parts of your question is missing attached below is the missing information
The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube
answer : Total surface area = 3/2 * area of old radiator
Explanation:
we will use this relation
K = [tex]\frac{Qd }{A* change in T }[/tex]
change in T = ΔT
therefore New Area ( A ) = 3/2 * area of old radiator
Given that the thermal conductivity is the same in the new and old radiators
A 50 mm 45 mm 20 mm cell phone charger has a surface temperature of Ts 33 C when plugged into an electrical wall outlet but not in use. The surface of the charger is of emissivity 0.92 and is subject to a free convection heat transfer coefficient of h 4.5 W/m2 K. The room air and wall temperatures are T 22 C and Tsur 20 C, respectively. If electricity costs C $0.18/kW h, determine the daily cost of leaving the charger plugged in when not in use.
Answer:
C = $0.0032 per day
Explanation:
We are given;
Dimension of cell phone; 50 mm × 45 mm × 20 mm
Temperature of charger; T1 = 33°C = 306K
Emissivity; ε = 0.92
convection heat transfer coefficient; h = 4.5 W/m².K
Room air temperature; T∞ = 22°C = 295K
Wall temperature; T2 = 20°C = 293 K
Cost of electricity; C = $0.18/kW.h
Chargers are usually in the form of a cuboid, and thus, surface Area is;
A = (50 × 45) + 2(50 × 20) + 2(45 × 20)
A = 6050 mm²
A = 6.05 × 10^(-3) m²
Formula for total heat transfer rate is;
E_t = hA(T1 - T∞) + εσA((T1)⁴ - (T2)⁴)
Where σ is Stefan Boltzmann constant with a value of; σ = 5.67 × 10^(-8) W/m².K⁴
Thus;
E_t = 4.5 × 6.05 × 10^(-3) (306 - 295) + (0.92 × 6.05 × 10^(-3) × 5.67 × 10^(-8)(306^(4) - 293^(4)))
E_t = 0.7406 W = 0.7406 × 10^(-3) KW
Now, we know C = $0.18/kW.h
Thus daily cost which has 24 hours gives;
C = 0.18 × 0.7406 × 10^(-3) × 24
C = $0.0032 per day
A body of weight 300N is lying rough
horizontal plane having
a Coefficient of friction as 0.3
Find the magnitude of the forces which can move the
body while acting at an angle of 25 with the horizonted
Answer:
Horizontal force = 89.2 N
Explanation:
The frictional force = coefficient of friction * magnitude of the force (weight of the body) * cos theta
Substituting the given values, we get -
Frictional Force = 0.3*300 * cos 25 = 89.2 N
Horizontal force = 89.2 N
How do guest room hotel smoke alarms work and differ then regular home versions?
Answer: As to the more sophisticated way of detecting "smoke" from an object a human may use in hotel rooms, this sensor called a Fresh Air Sensor does not just detect and, and but alerts the management about a smoking incident in a hotel room
identify the unit of the electrical parameters represented by L and C and prove that the resonant frequency (fr)=1÷2π√LC
Answer:
L = Henry
C = Farad
Explanation:
The electrical parameter represented as L is the inductance whose unit is Henry(H).
The electrical parameter represented as C is the inductance whose unit is Farad
Resonance frequency occurs when the applied period force is equal to the natural frequency of the system upon which the force acts :
To obtain :
At resonance, Inductive reactance = capacitive reactance
Equate the inductive and capacitive reactance
Inductive reactance(Xl) = 2πFL
Capacitive Reactance(Xc) = 1/2πFC
Inductive reactance(Xl) = Capacitive Reactance(Xc)
2πFL = 1/2πFC
Multiplying both sides by F
F * 2πFL = F * 1/2πFC
2πF²L = 1/2πC
Isolating F²
F² = 1/2πC2πL
F² = 1/4π²LC
Take the square root of both sides to make F the subject
F = √1 / √4π²LC
F = 1 /2π√LC
Hence, the proof.
state the degree of the homogeneity (1)sin() (2) (x+y+1) ². (3)√x+y(4x+3y).
Answer:
(2) ( (x+y)⁴)³(3)+(3)x(4x+y)
Explanation:
correct me if I'm wrong^_^
Describe the invention process. What is involved in inventing, and what must you consider during the invention process?
The invention process can be regarded as the process involving the creation of an idea as well as development of this idea.
It should be noted that invention process involves;
Creation of ideaDevelopment of the idea.Market the idea.Things that must consider during the invention process are;
documention of the invention protecting confidentialitysearching and applying for patents. selling the invention.Therefore, invention process involves creation of an idea and this idea must be unique.
Learn more about invention process at;
https://brainly.com/question/4459550
Forget it because I almost have it
Answer:
hahahaha
Explanation: