Answer:
Explanation:
[tex]496=\frac{480\times 60}{demand}[/tex]
demand per day = 58 pans
Due to availability of two workers we can have parallel we can have deep drawing and trimming operations simultaneously.
Hence the cycle time would be the greater time of the two operations.
cycle time = 450 seconds
[tex]\text{capacity of work cell}=\frac{\text{available working time}}{\text{cycle time}}[/tex]
[tex]\text{capacity of work cell}=\frac{480\times 60}{450}[/tex]
[tex]\text{capacity of work cell}=64 ~pans[/tex] (which is greater than the demand of 58 pans)
Therefore the work cell has sufficient capacity and time (496 sec.>cycle time 450 sec) to meet the demand.
b)
Required daily production is 58 pans
Define chart name the different types of charts explain any three types of charts
Answer:
There are several different types of charts and graphs. The four most common are probably line graphs, bar graphs and histograms, pie charts, and Cartesian graphs. They are generally used for, and are best for, quite different things. ... Pie charts to show you how a whole is divided into different parts.
#carryonlearning
What are the initial questions that a systems analyst must answer to build an initial prototype of the system output.
Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.
Answer:
The answer is "15 N".
Explanation:
Please find the complete question in the attached file.
In frame B:
For just slipping:
[tex]\to \frac{P}{2} \cos \theta =mg \sin \theta\\\\\to P=2 mg \tan \theta \\\\[/tex]
[tex]=2 \times 1 \times g \times \tan 37^{\circ}\\\\ =2 \times 10 \times \frac{3}{4}\\\\ =15 \ N[/tex]
A 1m3 tank containing air at 25℃ and 500kPa is connected through a valve to
another tank containing 5kg of air at 35℃ and 200kPa. Now the valve is opened,
and the entire system is allowed to reach thermal equilibrium, which is at 20℃
(Take: Ru = 8.314 kJ / kg.K).
Answer:
The right answer is "2.2099 m³".
Explanation:
Given:
Mass,
m = 5 kg
Temperature,
T = 35℃
or,
= 35 + 273
Pressure,
P = 200 kPa
Gas constant,
R = 0.2870 kj/kgK
By using the ideal gas equation,
The volume will be:
⇒ [tex]PV=mRT[/tex]
or,
⇒ [tex]V=\frac{mRT}{P}[/tex]
By substituting the values, we get
[tex]=\frac{5(0.2870)(35+273)}{200}[/tex]
[tex]=\frac{441.98}{200}[/tex]
[tex]=2.2099 \ m^3[/tex]
What must you do to become ASE certified as an automotive technician?
Answer:
To become ASE certified, you must pass an ASE test and have relevant hands-on work experience. The amount of work experience required can vary by test, and is specified in detail here. ASE recommends submitting the form after you've registered to take an ASE certification test.
Good luck!
Explanation:
Answer: One theme in White Fang is adapting in order to survive. White Fang finally submits to Gray Beaver. He also copes with fighting other dogs. White Fang changes his behaviors so that he can live.
Explanation: its the sample response
what is the best glide speed for your training airplane
1.5 nautical miles per 1,000 feet
Your shifts productivity is Slow because one person is not pulling his share. The rest of the team is Getting upset.
Answer:
you are right but then you ddnt ask a question
In a true Brayton cycle, the pressure ratio is 9. Air input temperature to the cycle 300 K pressure is 100 kPa. The maximum temperature in the cycle is 1300 K. Compressor and turbine their yields are equal to each other. Net work obtained from the cycle is 225 kJ / kg. Accordingly, the cycle find the overall yield. The specific temperatures are variable.
Answer:
i did not known answer but anobody help you
In a CNC machining operation, the has to be moved from point (5, 4) to point(7, 2)along a circular path with center at (7,2). Before starting operation, the tool is at (5, 4).The correct G and M code for this motion is
Answer: hello your question is incomplete below is the complete question
answer:
N010 GO2 X7.0 Y2.0 15.0 J2.0 ( option 1 )
Explanation:
Given that the NC machining has to be moved from point ( 5,4 ) to point ( 7,2 ) along a circular path
GO2 = circular interpolation in a clockwise path
G91 = incremental dimension
hence the correct option is :
N010 GO2 X7.0 Y2.0 15.0 J2.0
how does load transfer of space needle
Answer:
The Space Needle is a cut away with minimal residual deflection due to load transfer.
A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape until the tank pressure drops to 200 kPa. Assuming that the argon remaining in the tank experiences a reversible adiabatic process, find the final mass of argon in the tank. Since you don't have argon gas tables, assume cp, cv, k as needed at some appropriate temperature(s).
Answer:
Final mass of Argon= 2.46 kg
Explanation:
Initial mass of Argon gas ( M1 ) = 4 kg
P1 = 450 kPa
T1 = 30°C = 303 K
P2 = 200 kPa
k ( specific heat ratio of Argon ) = 1.667
assuming a reversible adiabatic process
Calculate the value of the M2
Applying ideal gas equation ( PV = mRT )
P₁V / P₂V = m₁ RT₁ / m₂ RT₂
hence : m2 = P₂T₁ / P₁T₂ * m₁
= (200 * 303 ) / (450 * 219 ) * 4
= 2.46 kg
Note: Calculation for T2 is attached below
where are the field poles mounted on an alternator
Answer:
The magnetic field for this type of alternator is established by a set of stationary field poles mounted on the periphery of the alternator frame. The field flux created by these poles is cut by conductors inserted in slots on the surface of the rotating armature.
Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 24 C and exits at 1 bar. If the refrigerant undergoes a throttling process, what is the quality of the refrigerant exiting the expansion valve.
Answer:
[tex]h_{1} = h_2} = 293.45 KJ/kg[/tex].
The quality of the refrigerant exiting the expansion valve is
[tex]x_{2}=0.193596[/tex].
Explanation:
Fluid given Ammonia.
Inlet 1:-
Temperature [tex]T_{1}[/tex] = [tex]24^{o} C[/tex].
Pressure [tex]P_{1}[/tex] = 10 bar.
Exit 2:-
Pressure [tex]P_{2}[/tex] = 1 bar.
Solution:-
Set the leak rate to zero and choose a non-zero value for the proportional feedback gain.Restart the simulation and turn on the outflow valve.What happens to the liquid level in the tank?Repeat this process with higher and lower values for the proportional feedback gain.What happens when the proportional feedback gain is increased?What happens when it is decreased?Find the proportional gain that will reach steady state the quickest without oscillationin the state of the valve and restart the simulation.What is the system time constant, as determined from the tank level versus time plot.
Answer:
Explanation:
The proportional gain K is usually a fixed property of the controller . If proportional gain is increased , The sensitivity of the controller to error is increased but the stability is impaired. The system approaches the behaviour of on off controlled system and it response become oscillatory
A 20-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 15°, what must be the length of the bar?
Answer:
1.887 m
Explanation:
(15 *pi)/180
= 0.2618 rad
Polar moment
= Pi*d⁴/32
= (22/7*20⁴)/32
= 15707.96
Torque on shaft
= ((22/7)*20³*110)/16
= 172857.14
= 172.8nm
Shear modulus
G = 79.3
L = Gjθ/T
= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8
= 1.887 m
The length of the bar is therefore 1.887 meters
what are some quality assurance systems
A micromechanical resonator is to be designed to have a Q factor of 1000 and a natural frequency of 2 kHz. Determine the system-damping factor and the system bandwidth.
Answer:
Explanation:
Given:
Q factor, =1000
natural frequency, [tex]f_n=2000~Hz[/tex]
Damping factor, [tex]\zeta=?[/tex]
Bandwidth, BW=?
We have the relation:
[tex]Q=\frac{1}{2\zeta}[/tex]
[tex]\zeta=\frac{1}{2Q}[/tex]
[tex]\zeta=\frac{1}{2\times 1000}[/tex]
[tex]\zeta=5\times 10^{-4}[/tex]
Bandwidth:
[tex]BW=\frac{f_n}{Q}[/tex]
[tex]BW=\frac{2000}{1000}[/tex]
[tex]BW=2~Hz[/tex]
Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging water at the rate of 12.5litres per second. Take the frictional coefficient ,f, as being 0.008
Answer: [tex]10.631\times 10^3\ N/m^2[/tex]
Explanation:
Given
Discharge is [tex]Q=12.5\ L[/tex]
Diameter of pipe [tex]d=150\ mm[/tex]
Distance between two ends of pipe [tex]L=800\ m[/tex]
friction factor [tex]f=0.008[/tex]
Average velocity is given by
[tex]\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s[/tex]
Pressure difference is given by
[tex]\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa[/tex]
Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW of energy to the water and the head loss of the flow is 10 m. Determine the velocity of the water leaving the pump and discharging into tank B.
Complete Question
Complete Question is attached below.
Answer:
[tex]V'=5m/s[/tex]
Explanation:
From the question we are told that:
Diameter [tex]d=0.10m[/tex]
Power [tex]P=4.0kW[/tex]
Head loss [tex]\mu=10m[/tex]
[tex]\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu[/tex]
[tex]\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10[/tex]
[tex]H_m=(\frac{200*10^3}{1000*9.8}-10)[/tex]
[tex]H_m=10.39m[/tex]
Generally the equation for Power is mathematically given by
[tex]P=\rho gQH_m[/tex]
Therefore
[tex]Q=\frac{P}{\rho g H_m}[/tex]
[tex]Q=\frac{4*10^4}{1000*9.81*10.9}[/tex]
[tex]Q=0.03935m^3/sec[/tex]
Since
[tex]Q=AV'[/tex]
Where
[tex]A=\pi r^2\\A=3.142 (0.05)^2[/tex]
[tex]A=7.85*10^{-3}[/tex]
Therefore
[tex]V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}[/tex]
[tex]V'=5m/s[/tex]
A 0.82-in-diameter aluminum rod is 5.5 ft long and carries a load of 3000 lbf. Find the tensile stress, the total deformation, the unit strains, and the change in the rod diameter.
Answer:
Tensile stress = 0.1855Kpsi
Total deformation = 0.0012243 in
Unit strain = 1.855 *10^-5 or 18.55μ
Change in the rod diameter = 5.02 * 10^ -6 in
Explanation:
Data given: D= 0.82 in
L = 5.5 ft * 12 = 66 in
load (p) = 3000 (Ibf) /32.174 = 93.243 Ibm
Area = (π/4) D² = (π/4) 0.82² = 0.502655 in²
∴ Tensile stress Rt = P/A = 93.243/0.502655 = 185.50099 pound/in²
Rt = 0.1855 Kpsi
∴ Total deformation = PL / AE = Rt * L/ Eal
= 0.1855 * 10³ * 66 / 10000 * 10³
= 0.0012243 in
∴the unit strains = total deformation / L = 0.0012243/ 66
=0.00001855 = 1.855 *10^-5
= 18.55μ
∴ Change in rod Δd/ d = μ ΔL/L
= (0.33) 1.855 *10^-5 * 0.82
= 5.02 * 10^ -6 in
Using 1.5 V batteries, a switch, and three lamps, devise a circuit to apply 4.5 V across eitherone lamp, two lamps in series, or three lamps in series with a single-control switch. Draw theschematic.
Answer: the attached picture is the answer.
Explanation:
Assuming:
the switch position connect to 1, hence 4.5V exist at across lamp1
the switch position connects to 2 hence 4.5 V exist across lamp 1 and lamp 2
the switch position connects to 3, hence, 4.5 V exist across lamp 1, lamp 2 and lamp 3.
bending stress distribution is a.rectangle b.parabolic c.curve d.i section
CO2 enters an adiabatic nozzle, operating at steady state, at 200 kPa, 1500 K, 5 m/s and exits at 100 kPa, 1400 K. The exit area of the nozzle is 10 cm2. Using the PG model, determine the exit velocity
Answer:
[tex]v_2=549.2 m/s\\[/tex]
Explanation:
Given:
[tex]P_1=2500kPa\\T_1=1500 k\\V_1=5 m/s\\P_2=100 kPa\\T_2=1400 k\\A_2=10 cm^2[/tex]
Solution:
For [tex]Co_2[/tex] y=1.4
Since Nozzle is adiababic
So,
[tex]h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}\\\frac{v_2^2}{2}=(h_2-h_2)+\frac{r^2}{2}\\v_2^2=2(h_1-h_2)+v_1^2\\v_2=\sqrt{2(h_1-h_2)+v_1^2}[/tex]
Now,
[tex]h_1-h_2=Cp_1T_1-CP_2T_2\\h_1-h_2=(1989-1838.2)*10^3\\ =150.8 * 10^3\\Cp for co_2\\C_{p1}=1.326 kj/kg\\C_{p2}=1.313 kj/kg\\v_2=\sqrt{301600+25}\\ =549.2 m/s[/tex]
The propeller shaft of the submarine experiences both torsional and axial loads. Draw Mohr's Circle for a stress element on the outside surface of the solid shaft. Determine the principal stresses, the maximum in-plane shear stress and average normal stress using Mohr's Circle.
Answer: Attached below is the missing detail and Mohr's circle.
i) б1 = 9.6 Ksi
б2 = -10.7 ksi
ii) 10.2 Ksi
iii) -0.51Ksi
Explanation:
First step :
direct compressive stress on shaft
бd = P / π/4 * d^2
= -20 / 0.785 * 5^2 = -1.09 Ksi
shear stress at the outer surface due to torsion
ζ = 16*T / πd^3
= (16 * 250 ) / π * 5^3 = 010.19 Ksi
Calculate the Principal stress, maximum in-plane shear stress and average normal stress
Using Mohr's circle ( attached below )
i) principal stresses:
б1 = 4.8 cm * 2 = 9.6 Ksi
б2 = -5.35 cm * 2 = -10.7 ksi
ii) maximum in-plane shear stress
ζ = radius of Mohr's circle
= 5.1 cm = 10.2 Ksi ( Given that ; 1 cm = 2Ksi )
iii) average normal stress
= 9.6 + ( - 10.7 ) / 2
= -0.51Ksi
James the Pilot James is a pilot. He is wearing a flight suit. He flies to Paris. He loves flying. 1. James is a a) teacher b) doctor c) pilot. whatisthe 2. He is wearing a a) shirt b) t-shirt c) flight suit. 3. Where does he fly to? a) Italy b) Luxembourg c) Paris http https://whatistheurl.com Please visit our site for worksheets and charts
Answer:
1.c
2.c
3.c
Explanation:
James is a pilot, whistle. He is wearing a flight suit. Paris is the palace where does he fly to. Hence, option C, C, and C are correct.
What is the point of a flight suit?When flying an aircraft, such as a military aircraft, a glider, or a helicopter, one must wear a full-body suit called a flight suit. These outfits are typically meant to keep the user warm and are also functional (they have many of pockets) (including fire ). In most cases, it looks like a jumpsuit.
The G suit, sometimes known as a "anti-G suit," is a one-piece jumpsuit that shields a pilot from the pressure of G forces pressing down on him and causing discomfort or unconsciousness.
The traditional attire for pilots of military and commercial aircraft, helicopters, and even gliders is flight suits or flyers coveralls. In areas where there is a risk of fire, ground personnel—including aircrews—often wear flight suits as well.
Thus, option C, C, and C are correct.
For more information about point of a flight suit, click here:
https://brainly.com/question/12302183
#SPJ2
How much energy does it take to boil water for pasta? For a one-pound box of pasta
you would need four quarts of water, which requires 15.8 kJ of energy for every degree
Celsius (°C) of temperature increase. Your thermometer measures the starting
temperature as 48°F. Water boils at 212°F.
a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?
b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?
c. [2 pts] How much energy is required to heat the four quarts of water from
48°F to 212°F (boiling)?
Answer:
a. 164 °F b. 91.11 °C c. 1439.54 kJ
Explanation:
a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?
Since the starting temperature is 48°F and the final temperature which water boils is 212°F, the number of degrees Fahrenheit we would need to raise the temperature is the difference between the final temperature and the initial temperature.
So, Δ°F = 212 °F - 48 °F = 164 °F
b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?
To find the degree change in Celsius, we convert the initial and final temperature to Celsius.
°C = 5(°F - 32)/9
So, 48 °F in Celsius is
°C₁ = 5(48 - 32)/9
°C₁ = 5(16)/9
°C₁ = 80/9
°C₁ = 8.89 °C
Also, 212 °F in Celsius is
°C₂ = 5(212 - 32)/9
°C₂ = 5(180)/9
°C₂ = 5(20)
°C₂ = 100 °C
So, the number of degrees in Celsius you must raise the temperature is the temperature difference between the final and initial temperatures in Celsius.
So, Δ°C = °C₂ - °C₁ = 100 °C - 8.89 °C = 91.11 °C
c. [2 pts] How much energy is required to heat the four quarts of water from
48°F to 212°F (boiling)?
Since we require 15.8 kJ for every degree Celsius of temperature increase of the four quarts of water, that is 15.8 kJ/°C and it rises by 91.11 °C, then the amount of energy Q required is Q = amount of heat per temperature rise × temperature rise = 15.8 kJ/°C × 91.11 °C = 1439.54 kJ
The system is initially moving with the cable taut, the 15-kg block moving down the rough incline with a speed of 0.080 m/s, and the spring stretched 39 mm. By the method of this article, (a) determine the velocity v of the block after it has traveled 99 mm, and (b) calculate the distance d traveled by the block before it comes to rest.
Solution :
The spring is expanded by 2 times of the block when it moves down an inclined by x times.
Here, [tex]$x_1$[/tex] = 39 mm
[tex]x_2[/tex] = 225 mm
a). From the work energy principal,
Work forces = kinetic energy
[tex]$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$[/tex]
[tex]$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$[/tex]
[tex]$9.75= 7.5(V^2_2-0.08^2)$[/tex]
[tex]$1.3= V^2_2-0.08^2$[/tex]
[tex]$V_2=1.14\ m/s$[/tex]
b). calculating the distance travelled by the block before it comes to rest.
Substitute the value of [tex]V_2[/tex] in (1),
[tex]$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$[/tex]
[tex]$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$[/tex]
[tex]$98.43x - 100(4x^2+0.156x)+0.048=0$[/tex]
[tex]$98.43x - 400x^2-15.6x+0.048=0$[/tex]
[tex]$82.83x - 400x^2+0.048=0$[/tex]
[tex]$ 400x^2- 82.83x-0.048=0$[/tex]
x = 0.20 m
Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manometer fluid of mercury (density: 13,600 kg/m3) to achieve uncertainty of 5% (i.e., 2.5 m/s) and 1 % (0.5 m/s).
Answer:
a) Δh = 2 cm, b) Δh = 0.4 cm
Explanation:
Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1
y₁ = y₂
subtitute
P₁ = P₂ + ½ ρ v₂²
P₁ -P₂ = ½ ρ v²
where ρ is the density of fluid
now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface
ΔP =ρ_{Hg} g h - ρ g h
ΔP = (ρ_{Hg} - ρ) g h
as the static pressure we can equalize the equations
ΔP = P₁ - P₂
(ρ_{Hg} - ρ) g h = ½ ρ v²
v = [tex]\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}[/tex]
in this expression the densities are constant
v = A √h
A =[tex]\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }[/tex]
They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³
we look for the constant
A = [tex]\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }[/tex]
A = 454.55
we substitute
v = 454.55 √h
to calculate the uncertainty or error of the velocity
h = [tex]\frac{1}{454.55^2} \ v^2[/tex]
Δh = [tex]\frac{dh}{dv}[/tex] Δv
[tex]\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}[/tex]
Suppose we have a height reading of h = 20 cm = 0.20 m
a) uncertainty 2.5 m / s ( 0.05)
[tex]\frac{\delta v}{v} = 0.05[/tex]
[tex]\frac{\Delta h}{h}[/tex] = 2 0.05
Δh = 0.1 h
Δh = 0.1 20 cm
Δh = 2 cm
b) uncertainty 0.5 m / s ( Δv/v= 0.01)
[tex]\frac{\Delta h}{h}[/tex] = 2 0.01
Δh = 0.02 h
Δh = 0.02 20
Δh = 0.1 20 cm
Δh = 0.4 cm = 4 mm
Hans, a scheduler at Mangel-Wurzel Transport, gets a call from a regular customer needing to move of rock and soil, which Hans knows from previous experience has an average density of . Hans has available a dump truck with a capacity of and a maximum safe load of .Calculate the number of trips the dump truck will have to make to haul the customer's load away.
Complete Question:
Hans, a scheduler at Mangel-Wurzel Transport, gets a call from a regular customer needing to move 19.8m³ of rock and soil, which Hans knows from previous experience has an average density of 650/kgm³. Hans has available a dump truck with a capacity of 4m³ and a maximum safe load of 3700kg. Required: calculate the number of trips the dump truck will have to make to haul the customer's load away.
Answer:
Mangel-Wurzel Transport
The number of trips that the dump truck will have to make to haul the customer's load away is:
= 5 trips.
Explanation:
a) Data and Calculations:
Volume of customer's load (rock and soil) = 19.8m³
Density of load = 650 kg/m³
Mass of load = Volume of load * Density of load
= 19.8m³ × 650 kg/m³
= 12,870 kg
The maximum safe load (mass) of the dump truck = 3,700 kg
Volume of the dump truck = 4m³
Assuming the truck is to carry 4m³ of the load.
The mass of load that the 4m³ capacity truck can carry = 4m³ × 650kg/m³
= 2,600kg
Quick Check:
Mass = 2,600kg < 3,700 kg, satisfying required conditions.
The number of trips that the truck would make to haul the customer's load away is, therefore, calculated as follows:
Number of trips = N
N = total volume of load/ volume per trip
N = 19.8/4
N = 4.95
N = 5 trips approx.
Atmospheric pressure is 101 kPa. Pressure inside a tire is measured using a typical tire pressure gage to be 900 kPa. Find gage pressure and absolute pressure in the tire. ___________________________________________________________________
Answer:
The gage and absolute pressures are 900 and 1001 kilopascals, respectively.
Explanation:
The gage pressure ([tex]P_{g}[/tex]), in kilopascals, is the difference between absolute ([tex]P_{abs}[/tex]) and atmospheric pressures ([tex]P_{atm}[/tex]), measured in kilopascals. If we know that [tex]P_{g} = 900\,kPa[/tex] and [tex]P_{atm} = 101\,kPa[/tex], then the gage and absolute pressures are, respectively:
[tex]P_{g} = 900\,kPa[/tex]
[tex]P_{abs} = P_{atm} + P_{g}[/tex]
[tex]P_{abs} = 101\,kPa + 900\,kPa[/tex]
[tex]P_{abs} = 1001\,kPa[/tex]
The gage and absolute pressures are 900 and 1001 kilopascals, respectively.
