Answer:
First we must know the efficiency of the gearbox in which energy is lost by friction between its components, we divide this efficiency by the nominal power and find the real power.
With this value knowing the speed of rotation of the gearbox we select an engine that has a power just above.
Also keep in mind that the motor torque must be greater than the resistance offered by the gearbox and the 0.5Kw machine
If a steel cable is rated to take 800-lb and the steel has a yield strength of 90,000psi, what is the diameter of the cable?
Answer:
d = 2.69 mm
Explanation:
Assuming the cable is rated with a factor of safety of 1.
The stress on the cable is:
σ = P/A
Where
σ = normal stress
P: load
A: cross section
The section area of a circle is:
A = π/4 * d^2
Then:
σ = 4*P / (π*d^2)
Rearranging:
d^2 = 4*P / (π*σ)
[tex]d = \sqrt{4*P / (\pi*\sigma)}[/tex]
Replacing:
[tex]d = \sqrt{4*800 / (\pi*\90000)} = 0.106 inches[/tex]
0.106 inches = 2.69 mm
You are riding in an elevator that is going up at 10 ft/s. You are holding your cell phone 5 ft above the floor when it suddenly slips out of your hand and falls to the floor. Will it hit the elevator floor in more time, equal time, or less time than it would take if the elevator were standing still? Show work justifying your answer.
Answer:
It falls at the same speed in both cases.
Explanation:
If I were standing still the phone would be in free fall after slipping out of my hand.
I set a frame of reference with origin on the ground and the positive Y axis pointing up.
It would slip at t0 = 0, from a position Y0 = 5 ft, with a speed of Vy0 = 0.
It would be subject to an gravitational acceleration of -32.2 ft/s^2.
Since acceleration is constant:
Y(t) = Y0 + Vy0 * t + 1/2 * 4 * t^2
When it hits the floor at t1 it will be at Y(t1) = 0
0 = 5 + 0 * t1 - 16.1 * t1^2
16.1 * t1^2 = 5
t1^2 = 5 / 16.1
[tex]t1 = \sqrt{0.31} = 0.55 s[/tex]
If the elevator is standing still it would take 0.55 s to hit the ground.
Now, if the elevator is moving up at 10 ft/s.
The frame of reference will have its origin at the place the floor of the elevator is at t = 0, and stay there as the elevetor moves. The floor of trhe elevator will have a position of Ye = 10 * t
Vy0 = 10 ft/s because it will be moving initially at the same speed as the elevator.
And it will hit the floor of the elevator not at 0, but at
Ye = 10 * t2
So:
10 * t2 = 5 + 10 * t2 - 16.1 * t2^2
0 = 5 - 16.1 * t2^2
16.1 * t1^2 = 5
t1^2 = 5 / 16.1
[tex]t1 = \sqrt{0.31} = 0.55 s[/tex]
It falls at the same speed in both cases.
A carpenter uses a hammer to strike a nail. Approximate the hammer's weight of 1.8lbs, as being concentrated at the head, and assume that at impact the head is traveling in the -j direction. If the hammer contacts the nail at 50mph and the impact occurs over 0.023 seconds, what is the magnitude of the average force exerted by the nail of the hammer?
Answer:
The average force F exherted by the nail over the hammer is 178.4 lbf.
Explanation:
The force F exherted by the nail over the hammer is defined as:
F = |I|/Δt
Where I and Δt are the magnitude of the impact and the period of time respectively. We know that the impact can be calculated as the difference in momentum:
I = ΔP = Pf - Pi
Where Pf and Pi are the momentum after and before the impact. Recalling for the definition for momentum:
P = m.v
Where m and v are the mass and the velocity of the body respectively. Notice that final hummer's momentum is zero due to the hammers de-acelerate to zero velocity. Then the momentum variation will be expressed as:
ΔP = - Pi = -m.vi
The initial velocity is given as 50 mph and we will expressed in ft/s:
vi = 50 mph * 1.47 ft/s/mph = 73.3 ft/s
By multiplyng by the mass of 1.8 lbs, we obtain the impulse I:
|I|= |ΔP|= |-m.vi| = 1.8 lb * 73.3 ft/s = 132 lb.ft/s
Dividing the impulse by a duration of 0.023 seconds, we finally find the force F:
F = 132 lb.ft/s / 0.023 s = 5740 lb.ft/s^2
Expressing in lbf:
F = 5740 lb.ft/s^2 * 0.031 lbf/lb.ft/s^2 = 178.4 lbf
What is a "gob" as described in the glass making process?
Explanation:
Step1
Gob is the primary element in the glass production. Gob is the hot molten glass which can shape according to the shape of container.
Step2
Gob is filled in the mold in glass production. The shape of gob is different for different glass product in glass production process. So, the shape of gob is very important parameter for glass production. This gob is filled in the mold by guided method or the gravity method.
The manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 mA for 40 continuous hours. During that time, the voltage will drop from 1.5 V to 1.0 V. Assume the drop in voltage is linear with time. (2 points) How many seconds is 40 hrs? (5 points) Plot the battery voltage as a function of time. Each axis needs a label (what is being plotted), scale (the values along the axis), and units. (7 points) Plot the battery power as a function of time. Write an equation for the power from 0 hours to 40 hours. (6 points) Remember that power is the derivative of energy with respect to time so energy is the integral of power over a given time period. There are two ways to find the energy. One is to calculate the area under the power curve from 0 hours to 40 hours. The second is to perform the integration of the power function from 0 to 40 hours. Find how much energy does the battery delivers in this 40 hour interval using both methods. The numerical answer is 1620 J. You must show the correct method to get credit.
Answer:
a) 144.000 s
b) and c)Battery voltage and power plots in attached image.
[tex]V=-\frac{0.5}{144000} t + 1.5 V[tex]
[tex]P(t)=-(31.25X10^{-9}) t+0.0135[/tex] where D:{0<t<40} h
d) 1620 J
Explanation:
a) The first answer is a rule of three
[tex]s=\frac{3600s * 40h}{1h} = 144000s[/tex]
b) Using the line equation with initial point (0 seconds, 1.5 V)
[tex]m=\frac{1-1.5}{144000-0} = \frac{-0.5}{144000}[/tex]
where m is the slope.
[tex]V-V_{1}=m(x-x_{1})[/tex]
where V is voltage in V, and t is time in seconds
[tex]V=m(t-t_{1}) + V_{1}[/tex] and using P and m.
[tex]V=-\frac{0.5}{144000} t + 1.5 V[tex]
c) Using the equation V
POWER IS DEFINED AS:
[tex] P(t) = v(t) * i(t) [tex]
so.
[tex] P(t) = 9mA * (-\frac{0.5}{144000} t + 1.5) [tex]
[tex]P(t) = - (31.25X10^{-9}) t + 0.0135[/tex]
d) Having a count that.
[tex]E = \int\limits^{144000}_{0} {P(t)} \, dt = \int\limits^{144000}_{0} {v(t)*i(t)} \, dt[/tex]
[tex]E = \int\limits^{144000}_{0} {-\frac{0.5}{144000} t + 1.5*0.009} \, dt = 1620 J[/tex]
A 1200-kg car moving at 20 km/h is accelerated
ata constant rate of 4 m/s2 up to a speed of 75km/h.
What are the force and total time required?
Answer:
Force on the car will be 4800 N and time required to cover this distance 13.75 sec
Explanation:
We have given mass of the car = 1200 kg
Initial velocity u = 20 km/h
Final velocity v = 75 km/h
Acceleration [tex]a=4m/sec^2[/tex]
From the first equation of motion we know that
v = u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time
So [tex]75=20+4\times t[/tex]
t = 13.75 sec
From second law of motion we know that [tex]F=ma[/tex]
So force [tex]F=1200\times 4=4800N[/tex]
An unknown immiscible liquid seeps into the bottom of an openoil
tank. Some measurements indicate that the depth of theunkown liquid
is 1.5m and the depth of the oil ( specific weight =8.5
kN/m3) floating on top is 5m. A pressure gageconnected
to the bottom of the tank reads 65 kPa. What is thespecific gravity
of the unkown liquid?
Answer:
1.53
Explanation:
Given:
depth of the unknown liquid = 1.5m
the depth of the oil floating on top = 5m
specific weight of oil, γ = 8.5 kN/m³
Total pressure at the bottom = 65 kPa = 65 kN/m²
let the specific weight of the unknown liquid be " γ' "
Now,
The total pressure
= Pressure due to the unknown liquid + Pressure due to floating liquid
or
65 = γ' × 1.5 + γ × 5
or
65 = γ' × 1.5 + 8.5 × 5
or
22.5 = γ' × 1.5
or
γ' = 15 kN/m³
Also,
Specific gravity = [tex]\frac{\textup{Specific weight of unknown liquid}}{\textup{Specific weight of water}}[/tex]
specific weight of water = 9.81 kN/m³
or
Specific gravity = [tex]\frac{15}{\9.81}[/tex] = 1.53
To calculate the specific gravity of the unknown liquid, you use the pressure reading from the tank, along with the height and specific weight of the oil, and then compare it to the specific weight of water.
Explanation:The question revolves around finding the specific gravity of an unknown liquid using a pressure gauge reading at the bottom of a tank containing both oil and the unknown liquid. To find the specific gravity, we use the equation for pressure caused by a static fluid column, which is P = h⋅γ, where P is pressure, h is the height of the fluid column, and γ is the specific weight of the fluid. Based on the given pressure reading and the specific weight of the oil, we can determine the specific weight of the unknown liquid. Once we have that, the specific gravity is found by dividing the specific weight of the unknown liquid by the specific weight of water (9.81 kN/m3, since the specific weight of water is its density (1000 kg/m3) times the acceleration due to gravity (9.81 m/s2)).
The equation to find the specific gravity (SG) of the unknown liquid is SG = γunknown / γwater, where γunknown is the specific weight of the unknown liquid calculated from the pressure reading at the bottom of the tank, and γwater is the specific weight of water.
A 30-seat turboprop airliner whose mass is 14,000 kg takes off from an airport and eventually achieves its cruising speed of 620 km/h at an altitude of 10,000 m. For g = 9.78 m/s^2, determine the change in kinetic energy and the change in gravitational potential energy of the airliner, each in kJ.
Answer:
Ek=207.569MJ
Ep=136.92MJ
Explanation:
kinetic energy is that energy that bodies with mass have when they have movement, while the potential energy is due to the height and mass that bodies have.
taking into account that the plane starts from rest at a height of zero, the following equations are taken to calculate the kinetic energy (Ek) and the potential energy (Ep)
[tex]Ek=0.5mv^{2} \\Ep=mgh[/tex]
where
m=mass=14000kg
v=speed=620km/h=172.2m/s
h=altitude=1000m
g=gravity=9.78 m/s^2
solving
Ek=(0.5)(14000)(172.2)^2=207569880J=207.569MJ
Ep=(14000)(9.78)(1000)=136920000=136.92MJ
A walker’s cassette tape player uses four AA batteries in series to provide 6V to the player circuit. The four alkaline battery cells store a total of 200 watt-seconds of energy. If the cassette player is drawing a constant 10mA from the battery pack, how long will the cassette operate at normal power?
Answer:
The cassette player will operate at normal power for 3333.33 seconds.
Explanation:
The first step is to identify the operating voltage and the operating current with the purpose to determine the power that the cassette player consumes. Remember that power equals the voltage multiplied by the current [tex]P=V\times I[/tex], where P is in Watts (W), V is in Volts (V) and I is in Amperes (A).
The problem says that four batteries are connected in series to provide a voltage of 6V to the player circuit. So, the operating voltage is 6V, [tex]V=6V[/tex]
Then, the problem says that cassette player draws a constant current of 10mA from the battery pack. So, the operating current is 10mA or 0.01A, [tex]I=0.01A[/tex]
From previous, it could be said that the cassette player consumes 0.06W.
[tex]P=V\times I=(6V)\times (0.01A)=0.06W[/tex]
Now, the idea is to calculate how long the cassette will operate at 0.06W.
The problem says that the battery pack stores [tex]200\, W\cdot s[/tex], it means that the battery pack could provide 200W in a second; after a second, the battery pack will not work properly. However, the battery pack just have to provide 0.06W so, it will last more time. For calculating that, you must divide the total power per time the cell can provide by the power that the cassette player needs to work.
[tex]t=\frac{200W\cdot s}{0.06W}=3333.33s[/tex]
As you can see, the W units are canceled and second remains.
Thus, the cassette player will operate at normal power for 3333.33 seconds.
A spacecraft component occupies a volume of 8ft^3 and weighs 25 lb at a location where the acceleration of gravity is 31.0 ft/s^2. Determine its weight, in pounds, and its average density, in lbm/ft^3, on the moon, where g=5.57 ft/s^2.
Answer:
The weight is 4.492 lb
The density is [tex]0.5614 lbm/ft^{3}[/tex]
Solution:
As per the question:
Volume of spacecraft component, [tex]V_{s} = 8ft^{3}[/tex]
Mass of the component of spacecraft, [tex]m_{s} = 25 lb[/tex]
Acceleration of gravity at a point on Earth, [tex]g_{E} = 31.0 ft/s^{2}[/tex]
Acceleration of gravity on Moon, [tex]g_{M} = 5.57 ft/s^{2}[/tex]
Now,
The weight of the component, [tex]w_{c} = m_{c}\times \frac{g_{M}}{g_{E}}[/tex]
[tex]w_{c} = 25\times \frac{5.57}{31.0}[/tex]
[tex]w_{c} = 4.492 lb[/tex]
Now,
Average density, [tex]\rho_{avg}[/tex]
[tex]\rho_{avg} = \farc{w_{s}}{V_{s}}[/tex]
[tex]\rho_{avg} = \farc{4.492}{8} = 0.5614 lbm/ft^{3}[/tex]
Answer:
density=3.125pounds/ft^3
weight=4.35lbf
Explanation:
Density is a property of matter that indicates how much mass a body has in a given volume.
It is given by the following equation.
ρ=m/v (1)
where
ρ=density
m=mass
v=volume
The weight on the other hand is the force which the earth (or the moon) attracts to a body with mass, this force is given by the following equation
W=mg (2)
W=weight
m=mass
g=gravity
to solve this problem we have to calculate the mass of the component using the ecuation number 2
W=mg
m=w/g
w=25lbf=804.35pound .ft/s^2
g=32.2ft/s^2
m=804.35/32.2=25 pounds
density
ρ=m/v (1)
ρ=25pounds/8ft^3
ρ=3.125pounds/ft^3 =density
weight in the moon
W=mg
W=(25pounds)(5.57ft/s^2.)=139.25pound .ft/s^2=4.35lbf
What is the range of a 32-bit unsigned integer?
Answer:
0 to 4294967295
Explanation:
Unsigned integers have only positive numbers and zero. Their range goes from zero to (2^n)-1.
In the case of a 32-bit unsigned integer this would be.
(2^32) - 1 = 4294967296 - 1 = 4294967295
So the range goes from 0 to 4294967295 or form zero to about 4.3 billion.
The minus one term is because the zero takes one of the values.
A piston-cylinder device contains 0.1 m3 of liquid water and 0.9 m² of water vapor in equilibrium at 800 kPa. Heat is transferred at constant pressure until the temperature reaches 350°C. Determine the initial temperature, total mass and final volume of the water. Show the process on a P-v diagram with respect to saturation lines.
Answer:
Initial temperature = 170. 414 °C
Total mass = 94.478 Kg
Final volumen = 33.1181 m^3
Diagram = see picture.
Explanation:
We can consider this system as a close system, because there is not information about any output or input of water, so the mass in the system is constant.
The information tells us that the system is in equilibrium with two phases: liquid and steam. When a system is a two phases region (equilibrium) the temperature and pressure keep constant until the change is completed (either condensation or evaporation). Since we know that we are in a two-phase region and we know the pressure of the system, we can check the thermodynamics tables to know the temperature, because there is a unique temperature in which with this pressure (800 kPa) the system can be in two-phases region (reach the equilibrium condition).
For water in equilibrium at 800 kPa the temperature of saturation is 170.414 °C which is the initial temperature of the system.
to calculate the total mass of the system, we need to estimate the mass of steam and liquid water and add them. To get these values we use the specific volume for both, liquid and steam for the initial condition. We can get them from the thermodynamics tables.
For the condition of 800 kPa and 170.414 °C using the thermodynamics tables we get:
Vg (Specific Volume of Saturated Steam) = 0.240328 m^3/kg
Vf (Specific Volume of Saturated Liquid) = 0.00111479 m^3/kg
if you divide the volume of liquid and steam provided in the statement by the specific volume of saturated liquid and steam, we can obtain the value of mass of vapor and liquid in the system.
Steam mass = *0.9 m^3 / 0.240328 m^3/kg = 3.74488 Kg
Liquid mass = 0.1 m^3 /0.00111479 m^3/kg = 89.70299 Kg
Total mass of the system = 3.74488 Kg + 89.70299 Kg = 93,4478 Kg
If we keep the pressure constant increasing the temperature the system will experience a phase-change (see the diagram) going from two-phase region to superheated steam. When we check for properties for the condition of P= 800 kPa and T= 350°C we see that is in the region of superheated steam, so we don’t have liquid water in this condition.
If we want to get the final volume of the water (steam) in the system, we need to get the specific volume for this condition from the thermodynamics tables.
Specific Volume of Superheated Steam at 800 kPa and 350°C = 0.354411 m^3/kg
We already know that this a close system so the mass in it keeps constant during the process.
If we multiply the mass of the system by the specific volume in the final condition, we can get the final volume for the system.
Final volume = 93.4478 Kg * 0.354411 m^3/kg = 33.1189 m^3
You can the P-v diagram for this system in the picture.
For the initial condition you can calculate the quality of the steam (measure of the proportion of steam on the mixture) to see how far the point is from for the condition on all the mix is steam. Is a value between 0 and 1, where 0 is saturated liquid and 1 is saturated steam.
Quality of steam = mass of steam / total mass of the system
Quality of steam = 3.74488 Kg /93.4478 Kg = 0,040 this value is usually present as a percentage so is 4%.
Since this a low value we can say that we are very close the saturated liquid point in the diagram.
The initial temperature of the water is 170.41 °C.
The initial volume of the water is 1 cubic meter.
The mass of the water is 4.620 kilograms.
The final volume of the water is 1.62828 cubic meters.
The process is described in the [tex]P-\nu[/tex] diagram attached below.
How to determine the water properties in a piston-cylinder device
By steam tables we find the missing properties of water:
State 1[tex]p = 800\,kPa[/tex], [tex]T = 170.41\,^{\circ}C[/tex], [tex]\nu = 0.21643\,\frac{m^{3}}{kg}[/tex], [tex]h = 2563.62\,\frac{kJ}{kg}[/tex] (Liquid-Vapor Mix) ([tex]x = 90\,\%[/tex])
State 2[tex]p = 800\,kPa[/tex], [tex]T = 350\,^{\circ}C[/tex], [tex]\nu = 0.35442\,\frac{m^{3}}{kg}[/tex], [tex]h = 3162.2\,\frac{kJ}{kg}[/tex] (Superheated Vapor)
Now we proceed to calculate the initial temperature, initial volume, mass and the final volume of the water:
Initial temperature[tex]T_{1} = 170.41\,^{\circ}C[/tex]
The initial temperature of the water is 170.41 °C. [tex]\blacksquare[/tex]
Initial volume[tex]V_{1} = 0.9\,m^{3}+0.1\,m^{3}[/tex]
[tex]V_{1} = 1\,m^{3}[/tex]
The initial volume of the water is 1 cubic meter. [tex]\blacksquare[/tex]
Mass[tex]m = \frac{V_{1}}{\nu_{1}}[/tex] (1)
[tex]m = 4.620\,kg[/tex]
The mass of the water is 4.620 kilograms. [tex]\blacksquare[/tex]
Final volume[tex]V_{2} = m\cdot \nu_{2}[/tex] (2)
[tex]V_{2} = 1.62828\,m^{3}[/tex]
The final volume of the water is 1.62828 cubic meters. [tex]\blacksquare[/tex]
The process is described in the [tex]P-\nu[/tex] diagram attached below. [tex]\blacksquare[/tex]
To learn more on piston-cylinder devices, we kindly invite to check this verified question: https://brainly.com/question/6334891
A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs are inboard channels set on a diameter of 18m. for a flow of 12900m^3/d, calculate the overflow rate, detention time,and weir loading.
Answer:
overflow rate 20.53 m^3/d/m^2
Detention time 2.34 hr
weir loading 114.06 m^3/d/m
Explanation:
calculation for single clarifier
[tex]sewag\ flow Q = \frac{12900}{2} = 6450 m^2/d[/tex]
[tex]surface\ area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2[/tex]
[tex]surface area = 314.16 m^2[/tex]
volume of tank[tex] V = A\times side\ water\ depth[/tex]
[tex]=314.16\times 2 = 628.32m^3[/tex]
[tex]Length\ of\ weir = \pi \times diameter of weir[/tex]
[tex] = \pi \times 18 = 56.549 m[/tex]
overflow rate =[tex] v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2[/tex]
Detention time[tex] t_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr[/tex]
weir loading[tex]= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m[/tex]
calculate the viscosity(dynamic) and kinematic viscosity of airwhen
the temperature is 288.15K and the density is 1.23kg/m3.
Answer:
(a) dynamic viscosity = [tex]1.812\times 10^{-5}Pa-sec[/tex]
(b) kinematic viscosity = [tex]1.4732\times 10^{-5}m^2/sec[/tex]
Explanation:
We have given temperature T = 288.15 K
Density [tex]d=1.23kg/m^3[/tex]
According to Sutherland's Formula dynamic viscosity is given by
[tex]{\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}[/tex], here
μ = dynamic viscosity in (Pa·s) at input temperature T,
[tex]\mu _0[/tex]= reference viscosity in(Pa·s) at reference temperature T0,
T = input temperature in kelvin,
[tex]T_0[/tex] = reference temperature in kelvin,
C = Sutherland's constant for the gaseous material in question here C =120
[tex]\mu _0=4\pi \times 10^{-7}[/tex]
[tex]T_0[/tex] = 291.15
[tex]\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-s[/tex]when T = 288.15 K
For kinematic viscosity :
[tex]\nu = \frac {\mu} {\rho}[/tex]
[tex]kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec[/tex]
Gas is kept in a 0.1 m diameter cylinder under the weight of a 100 kg piston that is held down by a spring with a stiffness k = 5 kN / m. If the gauge pressure of the gas is 300 kPa, how much is the spring compressed?
Answer:
The spring is compressed by 0.275 meters.
Explanation:
For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of weight of the piston and the force the spring exerts on the piston
Mathematically we can write
[tex]Force_{pressure}=Force_{spring}+Weight_{piston}[/tex]
we know that
[tex]Force_{pressure}=Pressure\times Area=300\times 10^{3}\times \frac{\pi \times 0.1^2}{4}=750\pi Newtons[/tex]
[tex]Weight_{piston}=mass\times g=100\times 9.81=981Newtons[/tex]
Now the force exerted by an spring compressed by a distance 'x' is given by [tex]Force_{spring}=k\cdot x=5\times 10^{3}\times x[/tex]
Using the above quatities in the above relation we get
[tex]5\times 10^{3}\times x+981=750\pi \\\\\therefore x=\frac{750\pi -981}{5\times 10^{3}}=0.275meters[/tex]
Calculate the rate at which body heat is conducted through the clothing of a skier in a steady- state process, given the following data: the body surface area is 1.80 m and the clothing is 1.00 cm thick; the skin surface temperature is 33.0 C and the outer surface of the clothing is at 1.00 C the thermal conductivity of the clothing is 0.040 W/m K
The rate of heat transfer through the skier's clothing is approximately 230.4 watts.
Given the information you provided, we can calculate the rate of heat transfer through the skier's clothing using the following formula for heat conduction:
Q = k * A * ΔT / L
where:
Q is the heat transfer rate (W)
k is the thermal conductivity of the clothing (W/m K)
A is the body surface area (m²)
ΔT is the temperature difference between the skin and the outer surface of the clothing (K)
L is the thickness of the clothing (m)
Let's plug in the values:
k = 0.040 W/m K
A = 1.80 m²
ΔT = 33.0°C - 1.0°C = 32.0 K (convert temperature difference from Celsius to Kelvin)
L = 0.01 m (convert centimeter to meter)
Q = 0.040 W/m K * 1.80 m² * 32.0 K / 0.01 m
Q = 230.40 W
A Carnot heat engine operates between 1000 deg F and 50 deg F, producing 120 BTU of work. What is the heat input to the engine?
Answer:
184.6 BTU
Explanation:
The thermal efficiency for a Carnot cycle follows this equation:
η = 1 - T2/T1
Where
η: thermal efficiency
T1: temperature of the heat source
T2: temperature of the heat sink
These temperatures must be in absolute scale:
1000 F = 1460 R
50 F = 510 R
Then
η = 1 - 510/1460 = 0.65
We also know that for any heat engine:
η = L / Q1
Where
L: useful work
Q1: heat taken from the source
Rearranging:
Q1 = L / η
Q1 = 120 / 0.65 = 184.6 BTU
In this exercise we will deal with the Carnot cycle and calculate how much heat was initially placed, so we have that:
The heat input to the engine was 184.6 BTU
How does the Carnot Cycle work?The theoretical Carnot cycle is formed by two isothermal transformations and two adiabatic transformations. One of the isothermal transformations is used for the temperature of the hot source, where the expansion process takes place, and the other for the cold source, where the compression process takes place.
The thermal efficiency for a Carnot cycle follows this equation:
[tex]\eta= 1 - T_2/T_1[/tex]
Where:
η: thermal efficiencyT1: temperature of the heat sourceT2: temperature of the heat sinkThese temperatures must be in absolute scale:
[tex]\eta = 1 - 510/1460 = 0.65[/tex]
Rearranging:
[tex]Q_1 = L / \eta\\Q_1 = 120 / 0.65 = 184.6 BTU[/tex]
See more about carnot cycle at brainly.com/question/21126569
What is an isochoric process? b) Can heat be exchanged in an isochoric process? c) A 100L container holding an ideal gas at an initial pressure of 10MPa is raised to a pressure of 15MPa. How much work is done?
Answer:
a)A constant volume process is called isochoric process.
b)Yes
c)Work =0
Explanation:
Isochoric process:
A constant volume process is called isochoric process.
In constant volume process work done on the system or work done by the system will remain zero .Because we know that work done give as
work = PΔV
Where P is pressure and ΔV is the change in volume.
For constant volume process ΔV = 0⇒ Work =0
Yes heat transfer can be take place in isochoric process.Because we know that temperature difference leads to transfer of heat.
Given that
Initial P=10 MPa
Final pressure =15 MPa
Volume = 100 L
Here volume of gas is constant so the work work done will be zero.
An automobile tire with a volume of 0.8 m3 is inflated to a gage pressure of 200 kPa. Calculate the mass of air in the tire if the temperature is 23 °C.
The mass of the air in the tire is 1040.192 grams.
We must know the concept of the Ideal Gas Equation to solve this question.
What is Ideal Gas Equation?The Ideal Gas Equation shows the empirical relationship between the Volume, Pressure, Temperature, and the number of moles in a given substance.
Using Ideal Gas Equation:
PV = nRT
From the given information:
The volume of the air O₂ = 0.8 m³ = 0.8 × 1000 L = 800 LThe pressure = 200 KPaThe temperature = 23°C = (273 + 23) K = 296 KThe rate constant = 8.31446 L.kPa.K⁻¹.mol⁻¹[tex]\mathbf{200 kPa \times 800 \ L = n \times 8.31446 L.kPa.K^{-1}.mol^{-1} \times 296 \ K }[/tex]
[tex]\mathbf{n = \dfrac{200 kPa \times 800 \ L}{8.31446 \ L.kPa.K^{-1}.mol^{-1} \times 296 \ K}}[/tex]
[tex]\mathbf{n =65.012 \ mol}[/tex]
From the relation;
Number of moles = mass/molar mass.
Mass of air (O₂) = number of moles × molar mass
Mass of air (O₂) = 65.012 mol × 16 g/mol
Mass of air (O₂) = 1040.192 grams
Learn more about ideal gas equation here:
https://brainly.com/question/25290815
A town has a 1-million-gallon storage capacity water tower. If the density of water is 62.41b/ft and local acceleration due to gravity is 32.1 ft/s, what is the force in lbf the structural base must provide to support the water in the tower?
Answer:
[tex]w = 8.316\times 10^6 lb[/tex]
Explanation:
force that structural base must be provided with is equal to the weight of water
we know that
[tex]\rho =\frac{mass}{V}[/tex]
[tex]m =\rho V[/tex]
We know that
w = mg
so we have
[tex]w = \rho Vg[/tex]
density of water is 62.4 lb/ft^3
V = 1 milllion = 100,000 gallons
[tex]g = 32.1 ft/s^2[/tex]
[tex]w = 62.4[\times10^6 gallons \frac{0.134}{1 gallon}] [32.1 ft/s^2 \frac{0.3048 m/s^2}{ft/s^2} \frac{m/s^2}{9.8 m/s^2}][/tex]
[tex]w = 8.316\times 10^6 lb[/tex]
A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the tube if the stress is 120 MPa?
Answer:
119.35 mm
Explanation:
Given:
Inside diameter, d = 100 mm
Tensile load, P = 400 kN
Stress = 120 MPa
let the outside diameter be 'D'
Now,
Stress is given as:
stress = Load × Area
also,
Area of hollow pipe = [tex]\frac{\pi}{4}(D^2-d^2)[/tex]
or
Area of hollow pipe = [tex]\frac{\pi}{4}(D^2-100^2)[/tex]
thus,
400 × 10³ N = 120 × [tex]\frac{\pi}{4}(D^2-100^2)[/tex]
or
D² = tex]\frac{400\times10^3+30\pi\times10^4}{30\pi}[/tex]
or
D = 119.35 mm
Answer:
D =119.35 mm
Explanation:
given data:
inside diameter = 100 mm
load = 400 kN
stress = 120MPa
we know that load is given as
[tex]P = \sigma A[/tex]
where:
P=400kN = 400000N
[tex]\sigma = 120MPa[/tex]
[tex]A =(\frac{1}{4} \pi D^2 - \frac{1}{4}\pi (100^2)[/tex]
[tex]A=\frac{1}{4} \pi (D^2 - 10000)[/tex]
putting all value in the above equation to get the required diameter value
[tex]400 = 120*\frac{1}{4} \pi (D^2 - 10000)[/tex]
solving for
D =119.35 mm
Metallic implants: a) Would you use pure iron as an implant material in the body? Please justify your answer. b) Suggest, giving also an example, how this material can be improved (in case you think it can be improved), in order to be used in the body.
Answer:
a)No, I wouldn't.
b) It can be improved through alloys.
Explanation:
Iron is a metal with lower resistance to corrosion, this means it will show visible signs of oxidation in living tissue, this corrosion happens so rapidly that supply and migration of oxygen can't pace with the consumption of the oxidant so that the contact tissue becomes starving of oxygen, this is the reason why I wouldn't use pure iron as an implant material.
Iron can be used in alloys to improve its properties, reducing the corrosion, but keeping the iron's resistance for implants in places with higher mechanical stress, one example is Iron-Chromium-Nickel alloys.
I hope you find this information useful! Good luck!
There are two questions about SolidWorks.
1. Why is it good practice to fully define
sketchgeometry?
2. Describe what happens if you do not select a planeprior to
clicking on the Sketch icon when creating a sketch.
Answer:
1. It is a good practice to fully define a sketch to avoid having erroneous dimensions on the faces of a solid, this avoids that when it is required to make an assembly with the drawn part appear assembly errors.
2. The 2D sketch should always be done on a plane, so solidworks would ask you to select a plane on which you want to make the sketch, on the other hand, if it is a 3D sketch, solidworks allows you to do it without the need for Select any plane.
Discuss the differences between conduction and convective heat transfer.
Answer:
Basically there are two principal differences between the convection and conduction heat transfer
Explanation:
The conduction heat transfer is referred to the transfer between two solids due a temperature difference, while for, the convective heat transfer is referred to the transfer between a fluid (liquid or gas) and a solid. Also, they used different coefficients for its calculation.
We can include on the explanation that conduction thermal transfer is due to temperature difference, while convection thermal transfer is due to density difference.
A piston-cylinder apparatus has a piston of mass 2kg and diameterof
10cm resting on a body of water 30 cm high atmospheric pressureis
101 kpa, and the temperature of water 50 degrees Celsius. What is
the mass of water in the container.
Answer:
M =2.33 kg
Explanation:
given data:
mass of piston - 2kg
diameter of piston is 10 cm
height of water 30 cm
atmospheric pressure 101 kPa
water temperature = 50°C
Density of water at 50 degree celcius is 988kg/m^3
volume of cylinder is [tex]V = A \times h[/tex]
[tex]= \pi r^2 \times h[/tex]
[tex]= \pi 0.05^2\times 0.3[/tex]
mass of available in the given container is
[tex]M = V\times d[/tex]
[tex] = volume \times density[/tex]
[tex]= \pi 0.05^2\times 0.3 \times 988[/tex]
M =2.33 kg
A loan for $85,000 is to be paid in 10 yearly payments. Each payment is larger by $100.00. Compute the first, the fourth, and the last payment, if the time value of money is 10% per year.
Answer:
1st payment = $9350
4th payment = $9650
last 10th payment = $10250
Explanation:
given data
loan = $85000
time = 10 year
each payment larger = $100
time value of money = 10% per year
to find out
first, the fourth, and the last payment
solution
we find here actual value at end of 1st year that is
actual value = loan amount + 10% of loan
actual value N = 85000 + (10% ×85000 )
actual value N = $93500
so
1st payment is = [tex]\frac{actual value}{total time}[/tex]
1st payment is = [tex]\frac{93500}{10}[/tex]
1st payment = $9350
and
4th payment = 1st payment + 3× each payment larger
4th payment = 9350 + 3×100
4th payment = $9650
and
last 10th payment = 4th payment + 6× each payment larger
last 10th payment = 9650 + 6× 100
last 10th payment = $10250
If 3.7 grams of a gas contains 3.7 × 10^22 molecules, what is the molar mass of this gas in units of g/mol?
Answer:
Molar mass of the gas will be 60.65 gram/mole
Explanation:
We have given mass of gas = 3.7 gram
Gas contains [tex]3.7\times 10^{22}[/tex]
We know that any gas contain [tex]6.022\times 10^{23}[/tex] molecules in 1 mole
So number of moles [tex]=\frac{3.7\times 10^{22}}{6.022\times 10^{23}}=0.061[/tex]
We know that number of moles [tex]n=\frac{mass\ in\ gram}{molar\ mass}[/tex]
So [tex]0.061=\frac{3.7}{molar\ mass}[/tex]
Molar mass = 60.65 gram/mole
What is the origin of the "horsepower"? Why would anyone wish to express power in the unit of horsepower? How many watts are in one horsepower?
Answer:
Horsepower unit was first time used by James Watt in 1782. The story refers that James Watt uses worked with pony to charge coal from the mines. According to that story, he have the need of a unit to measure the force from one of this animals. He founds that they can move 22.000 lbs per minute, so he (arbitrarily) increase this measure in 50% been the unit Horsepower in 33.000 lb/feet per minute.
Explanation:
This measure unit can measure "work" or "force". In the SI correspond to move up 75 Kg, to 1 meter high, in one second.
1 HP = (330 lb) x (100 feet)/1min = 33000 lb x feet/min
Why would anyone wish to express power in HP?
Is a practical unit, because reduce the amount of digits in a specific value. Also, his use is very spread specially in mechanical applications.
1 HP= 746 W (0,746 kW).
Yield and tensile strengths and modulus of elasticity . with increasing temperature. (increase/decrease/independent)
Answer:
Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.
Explanation:
The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.
Mathematically we can write,
[tex]E(t)=E_o[1-a\frac{T}{T_m}][/tex]
where,
E(t) is the modulus of elasticity at any temperature 'T'
[tex]E_o[/tex] is the modulus of elasticity at absolute zero.
[tex]T_{m}[/tex] is the mean melting point of the material
Hence we can see that with increasing temperature modulus of elasticity decreases.
In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.
A rectangular tank is filled with water to a depth of 1.5 m. Its longest side measures 2.5 m. What is the moment of the force about the base on this side? Is it. a) 5.5 kN.m b) 9.2 kN.m c) 13.8 kN.m d) 27.6 kN.m e) 41.4 kN.m
Answer:
The correct answer is option 'c': 13.8 kNm
Explanation:
We know that moment of a force equals
[tex]Moment=Force\times Arm[/tex]
The hydro static force is given by [tex]Force=Pressure\times Area[/tex]
We know that the hydrostatic pressure on a rectangular surface in vertical position is given by [tex]Pressure=\rho \times g\times h_{c.g}[/tex]
For the given rectangular surface we have [tex]h_{c.g}=\frac{h}{2}=\frac{1.5}{2}=0.75m[/tex]
Thus applying the values we get force as
[tex]Force=1000\times 9.81\times 0.75\times 1.5\times 2.5=27.59kN[/tex]
This pressure will act at center of pressure of the rectangular plate whose co-ordinates is given by h/3 from base
Thus applying the calculated values we get
[tex]Moment=27.59\times \frac{1.5}{3}=13.8kN.m[/tex]