Answer:
Product Teardown 28 pieces (1) Plastic packaging: protect and display product for purchase. (4) Exterior screws: hold case halves together. (1) Right case half: acts as part of a handle and contains the rest of the parts. (1) Left case half: acts as part of a handle and contains the rest of the parts.
Explanation:
A product teardown process is an orderly way to know about a particular product and identify its parts, system functionality to recognize modeling improvement and identify cost reduction opportunities. Unlike the traditional costing method, tear down analysis collects information to determine product quality and price desired by the consumers.
Answer:
?
Explanation:
A magician claims that he/she has invented a novel, super-fantastic heat engine. This engine operates between two reservoirs of temperatures 250 K and 750 K, respectively. Please verify and explain if this is possible for any of the following QH (heat received from the hot reservoir), QL (heat rejected to the cool reservoir), and W (mechanical work produced):
a. (3a). (3p) QH= 900 kJ, Wmech = 400 kJ, QL = 600 kJ.
b. (3b). (3p) Qu= 900 kJ, W mech = 400 kJ, QL = 500 kJ.
c. (3c). (3p) Qh= 900 kJ, Wmech = 600 kJ, QL = 300 kJ.
d. (3b). (3p) Qh= 900 kJ, Wmech = 800 kJ, QL = 100 kJ.
Answer:
A) Not possible, B) Posible, C) Possible, D) Not possible.
Explanation:
The maximum theoretical efficiency for any thermal engine is defined by Carnot's cycle, whose energy efficiency ([tex]\eta[/tex]), no unit, is expressed below:
[tex]\eta = 1-\frac{T_{L}}{T_{H}}[/tex] (1)
Where:
[tex]T_{L}[/tex] - Cold reservoir temperature, in Kelvin.
[tex]T_{H}[/tex] - Hot reservoir temperature, in Kelvin.
If we know that [tex]T_{L} = 250\,K[/tex] and [tex]T_{H} = 750\,K[/tex], then the maximum theoretical efficiency for the thermal engine is:
[tex]\eta = 1-\frac{T_{L}}{T_{H}}[/tex]
[tex]\eta = 0.667[/tex]
For real thermal engines, the following inequation is observed:
[tex]0 \le \eta_{r} \le \eta[/tex] (2)
Where [tex]\eta_{r}[/tex] is the efficiency of the real heat engine, no unit.
There are two possible criteria to determine if a given heat engine is real:
Efficiency
[tex]\eta_{r} = 1 - \frac{Q_{L}}{Q_{H}}[/tex] (3)
Where:
[tex]Q_{L}[/tex] - Heat rejected to the cold reservoir, in kilojoules.
[tex]Q_{H}[/tex] - Heat received from the hot reservoir, in kilojoules.
Power output
[tex]W = Q_{H}-Q_{L}[/tex] (4)
Where [tex]W[/tex] is the power output, in kilojoules.
Now we proceed to verify each case:
A) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 600\,kJ[/tex], [tex]W_{m} = 400\,kJ[/tex]
[tex]\eta_{r} = 0.333[/tex]
[tex]0 \le \eta_{r} \le \eta[/tex]
[tex]W = 300\,kJ[/tex]
[tex]W \ne W_{m}[/tex]
This engine is not possible.
B) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 500\,kJ[/tex], [tex]W_{m} = 400\,kJ[/tex]
[tex]\eta_{r} = 0.444[/tex]
[tex]0 \le \eta_{r} \le \eta[/tex]
[tex]W = 400\,kJ[/tex]
[tex]W = W_{m}[/tex]
The engine is possible.
C) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 300\,kJ[/tex], [tex]W_{m} = 600\,kJ[/tex]
[tex]\eta_{r} = 0.667[/tex]
[tex]0 \le \eta_{r} \le \eta[/tex]
[tex]W = 600\,kJ[/tex]
[tex]W = W_{m}[/tex]
The engine is possible.
D) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 100\,kJ[/tex], [tex]W_{m} = 800\,kJ[/tex]
[tex]\eta_{r} = 0.889[/tex]
[tex]\eta_{r} > \eta[/tex]
[tex]W = 800\,kJ[/tex]
[tex]W = W_{m}[/tex]
The engine is possible.
. A fire hose with an inside diameter of 6.40 cm is connected to a water hydrant very close to the ground. The other end is attached to a nozzle with an inside diameter of 3.5 cm and is brought 8.0 m above the ground. The flow rate on the hose is 30.0 L/s. What is the pressure in the nozzle assuming the pressure inside the hose on the ground is 1.50 x 106 N/m2
Answer:
Solving
Explanation:
If we want to create a single function for insertion and deletion in the linked list, Suggest any approach by which this task can be carried out. Just verbally explain how we can accomplish this, no need to write the code
Answer:
Explanation:
One approach to this would be to create the function with two parameters. One would be the element to be added or deleted while the other would be a char variable that would indicate the process to be completed (adding or deleting). This char can have a value of '+' or '-' accordingly. Then within the function, we would use an IF/Else statement to detect whether the char variable had a value of '+' or '-'. If the value was '+' we would execute the insertion process of the code but if the value was '-' we would only execute the deletion process of the code. Therefore, allowing both processes to be available within a single function.
If it takes 35 s for the 50-Mg tugboat to increase its speed uniformly to 25 km/h, starting from rest, determine the force of the rope on the tugboat. The propeller provides the propulsion force F which gives the tugboat forward motion, whereas the barge moves freely. Also, determine F acting on the tugboat. The barge has a mass of 75 Mg.
Answer:
- the force of the rope on the tugboat is 14.87 kN
- Force acting on the tugboat is 24.79 kN
Explanation:
Given the data in the question;
tugboat increases its speed uniformly to 25 km/h
v₁ = 25 km/h = (25 × 1000) / ( 1 × 60min × 60sec )
= 25000m / 3600s
= 6.94 m/s
Now, lets determine the force on the rope using the following relation;
T = [tex]m_b[/tex]v₁ / t₁
[tex]m_b[/tex] is mass of barge( 75 Megagram = 75 × 10³ Kilogram ), time t₁ is 35 s and v₁ is 6.94 m/s
so we substitute
T = [(75 × 10³) × 6.94 ] / 35
T = 520500 / 35
T = 14871.43 N
T = 14871.43 / 1000
T = 14.87 kN
Therefore, the force of the rope on the tugboat is 14.87 kN
Now, to determine F acting on the tugboat;
[tex]Ft_1[/tex] = ( [tex]m_t[/tex] + [tex]m_b[/tex] )[tex]v_1[/tex]
we solve for F
F = ( [tex]m_t[/tex] + [tex]m_b[/tex] )[tex]v_1[/tex] / [tex]t_1[/tex]
where [tex]m_t[/tex] is mass of tugboat (50 Megagram = 50 × 10³ Kilogram )
so we substitute
F = [( (50 × 10³) + (75 × 10³) )6.94] / 35
F = [ 125000 × 6.94 ) / 35
F = 867500 / 35
F = 24785.7 N
F = 24785.7 / 1000
F = 24.79 kN
Therefore, Force acting on the tugboat is 24.79 kN
A driver younger than 18 years of age may not operate a motor vehicle with any passenger who is not an immediate family member until 6 months from the date that the person's driver's license was issued.
O True
O False
name the process by which mild steel can be converted into high carbon steel and explain it briefly ?
Answer:
please give me brainlist and follow
Explanation:
Mild steel can be converted into high carbons steel by which of the following heat treatment process? Explanation: Case hardening, also referred as carburizing increases carbon content of steel, thus, imparting hardness to steel.
True or false, Increasing the spring force of the pressure plate that clamps the clutch disc to the flywheel increases torque capacity
but takes more foot pressure to operate the clutch pedal.
Answer: True
Explanation:
What relates a landscape architect's knowledge of design to the landscape architect's knowledge of law and government?
Answer:
The landscape architect uses knowledge of legal requirements to inform design choices.
Explanation:
The landscape architect doesn’t only draw on knowledge of law and government to design grounds for public buildings, but will draw on that knowledge quite often on many projects. The landscape architect’s knowledge of design is not purely intuitive but is a product of study and might involve research. The landscape architect’s knowledge of law and government, however, does come into play in making design choices that are consistent with the law and government regulations.
Switching to first order kinetics, enzymes are immobilized in 6 mm radius (R) beads. Beads are immersed in a solution containing substrate at a concentration (CAs) of 22 gmol/m3. The observed reaction rate (rA,obs) of the heterogeneous reaction is 4.3 x 10-6 gmol/sec. The first order rate constant (k1) is 0.24 /sec. Calculate the hypothetical rate, r*As of the reaction in this system.
How do Geothermal plowerplants relate to engineering?
Design a Finite State Moore Machine (FSM) that will take an arbitrary-sized integer as input, one bit at a time (starting from most significant bit), and return the remainder after this integer is divided by 5. This is also called as MOD 5 function. For example if the input number is 34, then its binary representation will be 100010 and 34 MOD 5 will return 4 as an output.
Answer:
attached below
Explanation:
Applying the state transition Formula
state - next = ( state *2 + in ) % 5
how this works : remainder of previous cycle is doubled to enable the calculation of the new remainder.
Input of current cycle is represented as either 1 or 0
since the dividing number = 5 . possible remainders = 1,2,3,4,0
each remainder is represented as :
S0 = zero remainder , S1 = 1 remainder , S2 = 2 remainder, S3 = 3 remainder,
S4 = 4 remainder
The number of bits on a transmission line that are in the process of actively being transmitted (i.e.,the number of bits that have been transmitted but have not yet been received) is referred to as the bit length of the line. Plot the line distance versus the transmission speed for a bit length of 1000 bits.Assume a propagation velocity of 2 X 108 m/s.
Answer:
The plotted diagram of line distance versus the transmission speed is uploaded below.
Explanation:
Given the data in the question;
Bit length B = 1000 bits
propagation velocity V = 2 × 10⁸ m/s
now, we know that the bit length of a link is expressed as;
B = R × d/V
where V is propagation velocity
d is the distance
R is the transmission speed
B is bit length
so we substitute
1000 = R × d/(2 × 10⁸)
1000 = Rd/(2 × 10⁸)
2 × 10¹¹ = Rd
R = 2 × 10¹¹ / d
R = 2E+11
Hence, we plot the transmission speed versus line distance; as shown in the image BELOW.
From the plot, if the transmission speed increases, the distances between stations decreases and vise versa.
Hence, both are inversely proportional.
The graph plot of line distance versus the transmission speed for a bit length of 1000 bits is; plotted below with Lν = 2 × 10¹¹ m/s
What is the number of bits on a transmission line?
We are given;
Bit length; B = 1000 bits
Propagation velocity; V = 2 × 10⁸ m/s
Formula for bit length of a link is expressed as;
B = ν × L/V
where;
V is propagation velocity
L is the distance
ν is the transmission speed
B is bit length
Thus, plugging in the relevant values gives;
1000 = ν × L/(2 × 10⁸)
Thus;
Lν = 1000 × 2 × 10⁸
Lν = 2 × 10¹¹ m/s
Thus, find the attached image of a graph showing the line distance versus the transmission speed for a bit length of 1000 bits.
Read more about Bit length at; https://brainly.com/question/16612919
The rate of heat transfer between a certain electric motor and its surroundings varies with time as , Q with dot on top equals negative 0.2 open square brackets 1 minus e to the power of negative (0.05 t )end exponent close square brackets , where t is in seconds and Q with dot on top is in kW. The shaft of the motor rotates at a constant speed of omega equals 100 space r a d divided by s (about 955 revolutions per minute, or RPM) and applies a constant torque of tau equals 18 space N m to an external load. The motor draws a constant electric power input equal to 2.0 kW. For the motor, provide an expression to show the change in total energy, increment E, as a function of time.
Answer:
the required expression is [tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]
Explanation:
Given the data in the question;
Q = -0.2[ 1 - [tex]e^(-0.05t)[/tex] ]
ω = 100 rad/s
Torque T = 18 N-m
Electric power input = 2.0 kW
now, form the first law of thermodynamics;
dE/dt = dQ/dt + dw/dt = Q' + w'
dE/dt = Q' + w' ------ let this be equation 1
w' is the net power on the system
w' = [tex]w_{elect[/tex] - [tex]w_{shaft[/tex]
[tex]w_{shaft[/tex] = T × ω
we substitute
[tex]w_{shaft[/tex]= 18 × 100
[tex]w_{shaft[/tex] = 1800 W
[tex]w_{shaft[/tex] = 1.8 kW
so
w' = [tex]w_{elect[/tex] - [tex]w_{shaft[/tex]
w' = 2.0 kW - 1.8 kW
w' = 0.2 kW
hence, from equation 1, dE/dt = Q' + w'
we substitute
dE/dt = -0.2[ 1 - [tex]e^{(-0.05t)[/tex] ] + 0.2
dE/dt = -0.2 + 0.2[tex]e^{(-0.05t)[/tex] ] + 0.2
dE/dt = 0.2[tex]e^{(-0.05t)[/tex]
Now, the change in total energy, increment E, as a function of time;
ΔE = [tex]\int\limits^t_0}\frac{dE}{dt} . dt[/tex]
ΔE = [tex]\int\limits^t_0} 0.2e^{(-0.05t)} dt[/tex]
ΔE = [tex]\int\limits^t_0} \frac{0.2}{-0.05} [e^{(-0.05t)}]^t_0[/tex]
[tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]
Therefore, the required expression is [tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]
Iman wanted to find out at what temperature butter melts. She placed a cube of solid butter in a beaker. She put the beaker on a hot plate. She turned up the temperature dial one degree every two minutes. Then she recorded the temperature at which the butter started to melt.
Iman wanted to determine if her results were reliable. She decided to repeat the experiment. Which statement describes the main purpose of repeating the experiment?
to test a different substance
to see if the results are similar
to practice completing experiments
to determine if butter will melt twice
Answer:
Ans: To see if the results are similar
Sasha goes back through her architectural design books for inspiration and designs the new country club to have a Romanesque feel, using rounded arches for the doorways and windows and constructing the walls with extra thickness to give a feeling of imposing massiveness to the building. The client is mystified by Sasha’s approach. He says he can’t visualize it working at all and demands she rethink the entire design. What would be best for Sasha to do in response to this criticism?
She should create a computer animated view of the design to walk the client through it.
She should get a better idea for what the client wants and expects.
She should try a different architectural style for her design.
She should build a model of her design to show the client how it will look when finished.
Una cuerda fuertemente estirada tiene sus puntos extremos fijos en x=0 y x=L. Si se le da un desplazamiento inicial F(x)=ax(L-x) desde la posición de equilibrio. Donde a es una constante y luego se suelta, encuentre el desplazamiento en cualquier tiempo t>0. Donde c=1/π.
Answer is in a photo. I couldn't attach it here, but I uploaded it to a file hosting. link below! Good Luck!
bit.[tex]^{}[/tex]ly/3a8Nt8n
Some wire of radius is 1.262mm has a resistance of 20Ω. Determine the resistance of a wire of the same length and material if the diameter is 1.6mm
Answer:
12.44Ω
Explanation:
For a question such as this, the assumption must be that the resistance is inversely proportional to the square of the diameter. The resistance will be ...
20Ω × (1.262/1.6)² ≈ 12.44Ω
Answer:
12.44[tex]ohms[/tex]
the resistance of a wire of the same length and material if the diameter is 1.6 mm of the given measuresExplanation:
hope it helps
Calculate the power required for heating a 1.5 kg sample of water for 10 minutes in a thermal system. What is the amount of energy required to raise the temperature of the water from 25°C to 60°C ? How much minimum amount of power does the heater have to supply per unit time? Why does the actual power rating of the heater need to be higher than this minimum amount? (The specific heat of water is 4.18 J/g°C .)
*60 points to anyone who can help*
Answer:
Power= Heat energy\ time
heat energy= mc∆T
specific heat of water = 4180
1.5*4180*(60-25)
=219450
time = 10× 60= 600 secs
power = 219450/600
Power= 365.75 Watt
=0.37 KW
Explanation:
the actual has to be bigger because the heater might be required to handle more ...also to accommodate extra energy.
The boost converter of Fig. 6-8 has parameter Vs 20 V, D 0.6, R 12.5 , L 10 H, C 40 F, and the switching frequency is 200 kHz. (a) Determine the output voltage. (b) Determine the average, maximum, and minimum inductor currents. (c) Determine the output voltage ripple. (d) Determine the average current in the diode. Assume ideal components.
Answer:
a) the output voltage is 50 V
b)
- the average inductor current is 10 A
- the maximum inductor current is 13 A
- the maximum inductor current is 7 A
c) the output voltage ripple is 0.006 or 0.6%V₀
d) the average current in the diode under ideal components is 4 A
Explanation:
Given the data in the question;
a) the output voltage
V₀ = V[tex]_s[/tex]/( 1 - D )
given that; V[tex]_s[/tex] = 20 V, D = 0.6
we substitute
V₀ = 20 / ( 1 - 0.6 )
V₀ = 20 / 0.4
V₀ = 50 V
Therefore, the output voltage is 50 V
b)
- the average inductor current
[tex]I_L[/tex] = V[tex]_s[/tex] / ( 1 - D )²R
given that R = 12.5 Ω, V[tex]_s[/tex] = 20 V, D = 0.6
we substitute
[tex]I_L[/tex] = 20 / (( 1 - 0.6 )² × 12.5)
[tex]I_L[/tex] = 20 / (( 0.4)² × 12.5)
[tex]I_L[/tex] = 20 / ( 0.16 × 12.5 )
[tex]I_L[/tex] = 20 / 2
[tex]I_L[/tex] = 10 A
Therefore, the average inductor current is 10 A
- the maximum inductor current
[tex]I_{Lmax[/tex] = [V[tex]_s[/tex] / ( 1 - D )²R] + [ V
given that, R = 12.5 Ω, V[tex]_s[/tex] = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz
we substitute
[tex]I_{Lmax[/tex] = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]
[tex]I_{Lmax[/tex] = [20 / 2 ] + [ 60 / 20 ]
[tex]I_{Lmax[/tex] = 10 + 3
[tex]I_{Lmax[/tex] = 13 A
Therefore, the maximum inductor current is 13 A
- The minimum inductor current
[tex]I_{Lmax[/tex] = [V[tex]_s[/tex] / ( 1 - D )²R] - [ V
given that, R = 12.5 Ω, V[tex]_s[/tex] = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz
we substitute
[tex]I_{Lmin[/tex] = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]
[tex]I_{Lmin[/tex] = [20 / 2 ] -[ 60 / 20 ]
[tex]I_{Lmin[/tex] = 10 - 3
[tex]I_{Lmin[/tex] = 7 A
Therefore, the maximum inductor current is 7 A
c) the output voltage ripple
ΔV₀/V₀ = D/RCf
given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz
we substitute
ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )
ΔV₀/V₀ = 0.6 / 100
ΔV₀/V₀ = 0.006 or 0.6%V₀
Therefore, the output voltage ripple is 0.006 or 0.6%V₀
d) the average current in the diode under ideal components;
under ideal components; diode current = output current
hence the diode current will be;
[tex]I_D[/tex] = V₀/R
as V₀ = 50 V and R = 12.5 Ω
we substitute
[tex]I_D[/tex] = 50 / 12.5
[tex]I_D[/tex] = 4 A
Therefore, the average current in the diode under ideal components is 4 A
(TCO 3) Linear-phase filters are desirable in many applications because Group of answer choices they have a sharper frequency response. they do not cause phase-distortion of an input signal. they have better stop-band performance by eliminating side-lobes. they have the smallest number of filter coefficients compared to nonlinear filters.
Answer:
They do not cause phase-distortion of an input signal. ( option 2 )
Explanation:
Linear-phase filters are desirable in many applications simply because they do not cause phase-distortion of an input signal.
The signals inputted into a Linear-phase filter ( FIR type ) are delayed equally as frequency is delayed in the pass band but they are are distorted.
The main special function of the Linear-phase filter is to reduce signal distortion.
180 kg/h of a liquid feed enters an extraction column with 98% paraffin oil and 2% ethyl alcohol at room temperature and pressure. Liquid absorbent is 100% water. Required recovery (extraction) of ethyl alcohol from paraffin oil is 97%. Given that paraffin oil and water are immiscible with each other and the extraction column operates in a counter current fashion. Assume isothermal and isobaric conditions and that only ethyl alcohol is absorbed. Equilibrium relation between mass fractions of ethyl alcohol in water and paraffin oil is given by Xwater = 1.7544 * Xparaffin oil (a) Find minimum amount of water required (Answer – 97.5 kg/hr) (b) If you have 1.5 times minimum water amount, how many equilibrium stages are required? (Answer ~ 6)
The most common type of pressure gauge is the 1. Piston 2. Linkage 3. Bourdon Tube 4. 5. Temperature
Answer:
Bourdon Tube gauge
Explanation:
The most popular type of pressure gauge in several countries is the Bourdon pressure tube gauge, that is used to determine medium and high loads. Bourdon tube will measure pressures ranging between 600 mbar - 4,000 bar. While the inner pressure is greater than the exterior pressure, the tube pushes forward, and vise versa.
flhsmv lesson 6 circle the road way user who must yield the right of way
Answer:
entering drivers
Explanation:
Drivers entering a traffic circle must yield the right-of-way to those already in the traffic circle. (Among other things, this rule helps avoid gridlock.)
Answer:
entering driversthe road way user who must yield the right of wayAir enters the combustor of a jet engine at Pressure=10 atm, T=1000R and M=0.2. Fuel is injected and burned with fuel-air ratio (by mass) of 0.06. The heat released during the combustion is 4.50000000 ft-lb per slug of fuel. Assuming one-dimensional frictionless flow with r=1.4 for the fuel-air mixture, calculate M2, p2 and T2 at the exit of combustor?
(HVAC) PLEASE HELP neeed helpp
If you are designing a space heater to warm a room and are required by safety codes to keep its surface temperature below 350 K (170 F or 77 C), the only variable you have to work with is the surface area, and you want to minimize that to make it convenient and inexpensive. Neglecting convection entirely, and assuming all the heat is transferred as infrared light, suppose it is to radiate 1200 W. How much area would it have
Answer:
A = 1.41 m²
Explanation:
The ara can be found out by using Stefan Boltzman Law:
[tex]P = \sigma AT^4[/tex]
where,
A = Area = ?
P = Radiation Power = 1200 W
σ = Stefan Boltzman Constant = 5.6703 x 10⁻⁸ W/m².k⁴
T = surface temperature = 350 k
Therefore:
[tex]1200\ W = (5.6703\ x\ 10^{-8}W/m^2.k^4)(A)(350\ k)^4\\\\A = \frac{1200\ W}{850.9\ W/m^2}[/tex]
A = 1.41 m²
The two major forces opposing the motion of a vehicle moving on a level road are the rolling resistance of the tires, Fr, and the aerodynamic drag force of the air flowing around the vehicle, Fd, given respectively by Fr, = fW, Fd= CdA1/2 rhoV2 where f and Cd are constants known as the rolling resistance coefficient and drag coefficient, respectively, W and A are the vehicle weight and projected frontal area, respectively, V is the vehicle velocity, and rho is the air density. For a passenger car with W = 3,550 lbf, A = 23.3 ft^2, and Cd = 0.34, and where f = 0.02 and rho = 0.08 lbm/ft^3.
Required:
Determine the power required, in HP, to overcome rolling resistance and aerodynamic drag when V is 55 mph.
Answer:
The power required to overcome rolling resistance and aerodynamic drag is 19.623 h.p.
Explanation:
Let suppose that vehicle is moving at constant velocity. By Newton's Law of Motion, the force given by engine must be equal to the sum of the rolling resistance and the aerodynamic drag force of the air. And by definition of power, we have the following formula:
[tex]\dot W = \left(f\cdot W +\frac{\rho\cdot C_{D}\cdot A\cdot v^{2}}{2\cdot g_{c}} \right)\cdot v[/tex] (1)
Where:
[tex]\dot W[/tex]- Power, in pounds-force-feet per second.
[tex]f[/tex] - Rolling resistance coefficient, no unit.
[tex]W[/tex] - Weight of the passanger car, in pounds-force.
[tex]\rho[/tex] - Density of air, in pounds-mass per cubic feet.
[tex]C_{D}[/tex] - Drag coefficient, no unit.
[tex]A[/tex] - Projected frontal area, in square feet.
[tex]v[/tex] - Vehicle speed, in feet per second.
[tex]g_{c}[/tex] - Pound-mass to pound-force ratio, in pounds-mass to pound-force.
If we know that [tex]f = 0.02[/tex], [tex]W = 3,550\,lbf[/tex], [tex]\rho = 0.08\,\frac{lbm}{ft^{3}}[/tex], [tex]C_{D} = 0.34[/tex], [tex]A = 23.3\,ft^{2}[/tex], [tex]v = 80.685\,\frac{ft}{s}[/tex] and [tex]g_{c} = 32.174\,\frac{lbm}{lbf}[/tex], then the power required by the car is:
[tex]\dot W = \left(f\cdot W +\frac{\rho\cdot C_{D}\cdot A\cdot v^{2}}{2\cdot g_{c}} \right)\cdot v[/tex]
[tex]\dot W = 10901.941\,\frac{lbf\cdot ft}{s}[/tex]
[tex]\dot W = 19.623\,h.p.[/tex]
The power required to overcome rolling resistance and aerodynamic drag is 19.623 h.p.
Technician A says that collapsible steering columns may use a pyrotechnic charge. Technician B says that collapsible brake pedals assemblies reduce the risk of trapping the drivers feet. Who is right
Answer:
I think A is correct
The two technicians are right in their given assessment about the use of collapsible steering columns and collapsible brake pedals to reduce the risk of trapping the driver's feet.
What is Safe Driving?This refers to the vehicular movement from one point to another, obeying traffic rules, and respect for other drivers and pedestrians.
Hence, we can see that with regards to safe driving, the opinions of both Technicians A and B are both correct as they both want the safety of the car and the driver.
Read more about safe driving here:
https://brainly.com/question/26084890
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Air at 1.00 atm is passed through a horizontal 1.5-in. Schedule 40 (steam-jacketed) steel pipe at a velocity of 2.10 ft/s and an inlet and outlet temperature of 25.0 °C and 95.5 °C, respectively. The pipe wall temperature is maintained at 121.1 °C. (HINT: You may neglect the viscosity correction in your calculations.) a) Assuming only heat transfer by forced convection, how long must
Answer:
missing question; How long must the heated section be
answer : 5.898 meters
Explanation:
Given parameters:
Diameter of steel pipe = 1.5 inches = 0.0381 m
velocity of air = 2.10 ft/s = 0.64 m/s
Inlet temperature = 25°C
Outlet temperature = 95.5°C
pipe wall temperature = 121.1°C
Average temperature = ( 25 + 95.5 ) / 2 = 60.25°
determine Reynolds number of air at 60.25°
= VD / v = ( 0.64 * 0.0381 ) / ( 1.83 * 10^-5 )
= 1332.46 this means the flow is laminar
hence mass flow rate ( m ) = eAv = 1.08 * πd^2 * velocity
= 1.08 * 0.0046 * 0.64 = 0.0032 g/s
attached below is the remaining part of the solution
A stainless steel ball (=8055 kg/m3, Cp= 480 J/kgK) of diameter D =15 cm is removed from theoven at a uniform temperature of 350oC. The ball is then subjectedto the flow of air at 1 atm pressure and 30oC with a velocityof 6 m/s. The surface temperature of the ball eventuallydrops to 250oC. Determine the average convection heat transfercoefficient during this cooling process and estimate howlong this process has taken.
Answer:
i) 25.04 W/m^2 .k
ii) 23.82 minutes = 1429.2 secs
Explanation:
Given data:
Diameter of steel ball = 15 cm
uniform temperature = 350°C
p = 8055 kg/m^3
Cp = 480 J/kg.k
surface temp of ball drops to 250°C
average surface temperature = ( 350 + 250 ) / 2 = 300°C
i) Determine the average convection heat transfer coefficient during the cooling process
Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table " properties of air " contained in your textbook
average convection heat transfer coefficient = 25.04 W/m^2 .k
ii) Determine how long this process has taken
Time taken by the process = 23.82 minutes = 1429.2 seconds
Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs
attached below is the detailed solution of the given question