Answer: Application.
Explanation:
The question on wether to contine the use of cadavers in the lab for test is being centered around its application. Cadaver which is same as a corpse or dead body is used in crash site during automobil test in lab, some of this cadavers are been disrespected with their applications in the automobile industries because many didn’t consent to be used in those experiments or test.
Solve the compound inequality. 3x − 4 > 5 or 1 − 2x ≥ 7
Two wooden members of uniform rectangular cross section are joined using a simple glued scarf splice. The maximum allowable shearing stress and maximum allowable normal stress in the glued splice is 50 MPa and 100 MPa, respectively. The cross-section area of the glued member is 400 mm2. (a) What should the value of the angle be to achieve maximum load Fmax? (b) What is the magnitude of the maximum load Fmax?
Answer:
a). α = 26.57
b). Maximum load is 50 .kN
Explanation:
a).
The normal force is given by
N = σ A cosec β
where, σ is the normal stress
A is the cross sectional area
Similarly, shear force is given by
S= τ A cosec β
where, τ is the shearing stress
Now from the figure,
tan β = S/N
= τ/σ
Therefore, [tex]$\beta = \tan^{-1}(2)$[/tex] = 63.43
α = 90 - β = 26.57
b).
The normal force is given by
[tex]$N=(100\times 10^6)(400\times 10^{-6}) \text{ cosec}\ 63.43$[/tex]
[tex]$N=44.78\times 10^3$[/tex] N
We have
[tex]$\Sigma F_y=0$[/tex]
∴ N - F sin β = 0
⇒ F = N / sin β
= [tex]$\frac{44.72\times 10^3}{\sin(63.43)} = 50\times 10^3 N$[/tex]
Similarly,
The shear force is given by
S = τ A cosec β
= [tex]$(50\times 10^6)(400\times 10^{-6}) \text{ cosec}\ 63.43 = 22.36\times 10^3 N$[/tex]
[tex]$\Sigma F_x=0$[/tex]
∴ S - F cos β = 0
⇒ F = S / cos β
[tex]$\frac{22.36\times 10^3}{\cos(63.43)} = 49.99\times 10^3 N$[/tex]
Therefore, force is 50 kN.
How can you safely lift and support a vehicle
Answer:
Place the jack under the part of the vehicle that it should contact when raised. If you're using jack stands, place them near the jack. If you place your jack incorrectly, you can injure your car. To find the proper place to position the jack for your particular vehicle, check your owner's manual.
To create an effective study schedule, a student must
guess the amount of time needed for studying.
O write down all activities and commitments.
O allow the same amount of time each day for studying.
O budget periods of time to study daily, even if it is interrupted.
Answer:
B) write down all activities and commitments.
Explanation:
also i found the answer on quizlet hope this works:))
Answer:
write down all activities and commitments.
Explanation:
i had got the same question on my unit test and it was the answer it was right
You are designing an airplane to carry liquid cargo that will slosh and move side to side in its container. This could make the plane unstable. What type of airfoil would you use for the wing? Why. Answer in a full sentence or more
What allows negative feedback to control a system
Answer:
Negative feedback control of the amplifier is achieved by applying a small part of the output voltage signal at Vout back to the inverting ( – ) input terminal via the feedback resistor
Answer:
The system has parts that sense the amount of output
Explanation:
Apex cirtified
A company that produces footballs uses a proprietary mixture of ideal gases to inflate their footballs. If the temperature of 230 grams [g] of gas mixture in a 15-liter [L] tank is maintained at 465 degrees Rankine [°R] and the tank is pressurized to 135 pound-force per square inch [psi], what is the molecular weight of the gas mixture in units of grams per mole
Answer:
The molecular weight of the gas mixture is 35.38 g/mol.
Explanation:
The molecular weight of the gas can be found using the following equation:
[tex] M = \frac{m}{n} [/tex]
Where:
m: is the mass = 230 g
n: is the number of moles
First, we need to find the number of moles using Ideal Gas Law:
[tex] PV = nRT [/tex]
Where:
P: is the pressure = 135 psi
V: is the volume = 15 L
R: is the gas constant = 0.082 L*atm/(K*mol)
T: is the temperature = 465 °R (K = R*5/9)
[tex]n = \frac{PV}{RT} = \frac{135 psi*\frac{1 atm}{14.6959 psi}*15 L}{0.082 L*atm/(K*mol)*465*(5/9) K} = 6.50 moles[/tex]
Finally, the molecular weight of the gas is:
[tex] M = \frac{m}{n} = \frac{230 g}{6.50 moles} = 35.38 g/mol [/tex]
Therefore, the molecular weight of the gas mixture is 35.38 g/mol.
I hope it helps you!
You have a piece of film paper that is 3 in x 5 in. You fix it inside the back of a pinhole camera with a focal length of 5.5 in. You want to use it to take a picture of your team’s mascot – a giant guinea pig that just barely fits in a 4 ft. tall cube. The picture will be taken directly in front, from a stool that places the aperture 2 ft. above the ground. You have to determine how far away the camera must be from your mascot to get a good portrait that fills up the whole film paper, without cutting any part of him off. How far apart should the camera and the mascot be to take the portrait? Show your work.
Answer:
In order to take a portrait, the distance of the mascot from the camera should be approximately 7.33 feet
Explanation:
The size of the film paper = 3 in. × 5 in.
The focal length of the camera = 5.5 in.
The height and width of the guinea pig = 4 ft.
The height of the aperture above the ground = 2 ft.
Therefore, we have;
Magnification = Height of image/(Height of object)
Withe the 3 in. wide film, we have;
Magnification = 3 in./(4 ft.) = 3 in./(48 in.) = 0.0625
Magnification = Length of camera/(Distance of object from pin hole)
∴ Length of camera/(Distance of object from pin hole) = 0.0625
Length of camera = Focal length of the camera = 5.5 in.
Therefore;
5.5 in./(Distance of object from pin hole) = 0.0625
Distance of object from pin hole = 5.5/0.0625 = 88 inches = 7.33 ft
Therefore, the camera should be approximately 7.33 ft. from the mascot to take a portrait.
In this exercise we have to use the magnification knowledge to calculate the distance that the photograph should be taken, thus we have to:
Distance of the mascot from the camera should be approximately 7.33 feet
To calculate the best distance to take the photo, we have that some information must be taken into account such as:
Size of the film paper: [tex](3)*(5) in[/tex] Focal length of the camera: [tex]5.5 in[/tex] Height and width of the guinea pig: [tex]4 ft[/tex] Height of the aperture above the ground: [tex]2 ft[/tex]
Therefore, we have that the formula of magnification is:
[tex]Magnification = Height \ of \ image/(Height \ of \ object)[/tex]
With the 3 in wide film, we have;
[tex]Magnification = 3 in/(4 ft) \\= 3 in/(48 in) = 0.0625 in[/tex]
Rewriting the magnification formula as:
[tex]Magnification = Length \ of \ camera/(Distance \ of \ object \ from \ pin \ hole)[/tex]
Substituting the values already known we have the equation will be matched as:
[tex]Length\ of \ camera/(Distance\ of\ object \ from \ pin \ hole) = 0.0625\\Length \ of \ camera = Focal \ length \ of \ the \ camera = 5.5 in.[/tex]
Therefore;
[tex]5.5 /(Distance \ of \ object\ from \ pin\ hole) = 0.0625 in\\Distance \ of \ object\ from \ pin \ hole = 5.5/0.0625\\ = 88 inches = 7.33 ft[/tex]
See more about distance at brainly.com/question/989117
Explain how engineering and science-related?
Answer:
While scientists study how nature works, engineers create new things, such as products, websites, environments, and experiences. Because engineers and scientists have different objectives, they follow different processes in their work.
Explanation:
An analog baseband audio signal with a bandwidth of 4kHz is transmitted through a transmission channel with additive white noise. The channel is assumed to be distortionless, and the power spectral density of white noise, No/2 is 10 WHz. An RC low-pass filter with a 3-dB bandwidth of 8 kHz is used at the receiver to limit the output noise power. Calculate the output noise power.
Answer:
2k20
Explanation:
4k ✈
A standard carbon resistor has a gold band to indicate + 5% tolerance. If its resistance is 3,500 , what are the upper and lower limits for its resistance? OA . 3495 - 3505 2 OB. 3300 Q - 3600 0 OC. 3325 N - 3675 OD 3450 - 35500
Answer:
C. 3325 Ω - 3675 Ω
Explanation:
5% of 3500 Ω is ...
0.05 × 3500 = 175
The lower limit is this amount less than the nominal value:
3500 -175 = 3325
The upper limit is the nominal value plus the tolerance:
3500 +175 = 3675
The lower and upper limits are 3325 Ω and 3675 Ω, respectively.
for high-volume production runs, machining parts from solid material might not be the best choice of manufacturing operations because
Answer:
There are actually multiple types of processes a manufacturer uses, and those can be grouped into four main categories: casting and molding, machining, joining, and shearing and forming.
Explanation:
The displacement volume of an internal combustion engine is 3 liters. The processes within each cylinder of the engine are modeled as an air-standard Diesel cycle with a cutoff ratio of 2.5. The state of the air at the beginning of compression is fixed by P1=95kPa, T1=22oC, and V1 = 3.17 liters.
Determine the net work per cycle, in kJ, the power developed by the engine, in kW, and the thermal efficiency, if the cycle is executed 1000 times per min.
Answer:
1) The power developed by the engine is 14705.7739 kW
2) The thermal efficiency is approximately 61.5%
Explanation:
The given parameters are;
P₁ = 95 kPa
T₁ = 22°C
V₁ = 3.17 liters
The cutoff ratio = 2.5
Displacement volume = 3 liters
The number of times the cycle is executed per minute = 1000 times per minute
We have;
The displacement volume = V₁ - V₂ = 3 l
V₁ = 3.17 l
V₂ = 3 - 3.17 = 0.17 l
Compression ratio = V₁/V₂ = 3.17/0.17 ≈ 18.65
P₂/P₁ = P₂/(95 kPa) = (V₁/V₂)^(k) = 18.65^1.4
P₂ = (95×18.65^(1.4)) ≈ 5710.5 kPa
T₂/T₁ = (V₁/V₂)^(k - 1)
T₂/(295 K)= (18.65)^(1.4 - 1)
T₂ = 295 * (18.65)^(1.4 - 1) = 950.81 K
The cutoff ratio = V₃/V₂ = 2.5
T₃ = T₂ × V₃/V₂ = 2.5 * 950.81 K = 2377.025 K
[tex]Q_{in}[/tex] = [tex]C_p[/tex]×(T₃ - T₂) = 1.006 × (2377.025 - 950.81) = 1,434.77 kJ/kg
T₄ = T₃ × (V₃/V₄)^(k-1) =
Therefore,
[tex]T_4 = T_3 \times \left (\dfrac{r_c}{r} \right )^{k - 1} = 2377.025 \times \left( \dfrac{2.5}{18.65} \right )^{1.4 - 1} \approx 1064 \ K[/tex]
T₄ ≈ 1064 K
[tex]Q_{out}[/tex] = [tex]-C_v \times (T_4 - T_1)[/tex]
[tex]C_v = C_p/k = 1.006/1.4 \approx 0.7186 \ kJ/kg[/tex]
∴ [tex]Q_{out}[/tex] = 0.7186×(1064 - 295) = 552.6034 kJ/kg
1) The net work = [tex]Q_{in}[/tex] - [tex]Q_{out}[/tex] = 1,434.77 kJ/kg - 552.6034 kJ/kg ≈ 882.17 kJ/kg
The number of cycle per minute = 1000 rpm
The number of cycle per minute = 1000 rpm/60 = 16.67 cycles per second
The power developed by the engine = The number of cycles per second × The net work of the engine
Therefore;
The power developed by the engine = 16.67 cycles/second × 882.17 kJ/kg
The power developed by the engine = 14705.7739 kW
2) Efficiency, [tex]\eta _{th}[/tex], is given as follows;
[tex]\eta _{th} = \dfrac{Q_{in}-Q_{out}}{Q_{in}} \times 100 = 1 - \dfrac{Q_{out}}{Q_{in}} \times 100= 1 - \dfrac{552.6034}{1434.77}\times 100 \approx 61.5\%[/tex]
Therefore, the thermal efficiency ≈ 61.5%.
- The four leading causes of death in the
construction industry include electrical
incidents, struck-by incidents, caught-in or
caught-between incidents, and
a. vehicular incidents
b. falls
C. radiation exposure
d. chemical burns
Use the diagram to determine total resistance. (Round the FINAL answer to one decimal place.) Note: Use the rated voltage and wattage to determine resistance. Note: Ra and Rb represent the resistance of the wire. The 120 volts of the motor and lamps is their rated voltage and not the actual voltage. I know the answer is 22.7 Ohms. I get very close, but I am not 100%.. I need to be 100%. If you can also: with the same diagram I need the total current as well! Thank you very much.
Answer:
22.6622 Ω ≈ 22.7 Ω5.29516 A ≈ 5.3 AExplanation:
For an electric motor, 1 hp = 746 W, so 1/2 hp = 373 W. Then the total wattage of the load is ...
Pl = 373 +100 +100 +75 = 648 W
The equivalent resistance is ...
R = V^2/P = (120 V)^2/(648 W) = 22 2/9 Ω
Adding the wiring resistance, we get a circuit resistance of ...
22.2222 +0.22 +0.22 = 22.6622 Ω . . . . total circuit resistance
Then the circuit current is ...
120 V/(22.6622 Ω) ≈ 5.29516 A
Circuit resistance is about 22.7 Ω; circuit current is about 5.3 A.
_____
Exact values are 5099/225 ohms and 27000/5099 amps. Of course, these assume the motor and lamps are purely resistive, which they are not.
1. Looking at the case study provided under the Companion Material section, what is the main problem that is addressed in this case study? Maintenance workers painted over the pipes. Engineers lost all of their work because of the flood. The water pipes broke and flooded the space center. There was no paperwork for the maintenance work done on the pipes.
Incomplete question. However, I provided information that could assist you in identifying the main problem or issue addressed in any case study.
Explanation:
First, note that a case study is simply a learning aid that allows one to learn from a real-life scenario.
To determine the main problems of a case study one needs to:
Read the case as many times as possible to become familiar with the message been expressed. For example, by highlighting or underlining the most important facts it can help you to discover the main problem or issue. Check for any facts provided in the case study, by so doing you can identify the most important problems.Thus, by taking these few steps you may be able to determine the main problem in that case study.
A picture of a ____ is the label pictogram on a chemical that warms of a product's acute toxicity
Answer:skull and crossbones
Explanation:
12. You need to be at the lift controls whenever the lift is in motion A) True B)False
the answer to that would be a) true
Wheel grinders need be equipped with an
Answer:
wheel guard
Explanation:
to protect our hands and reduce spark
While out on the International Space Station, an engineer was able to gather a sample of a new type of unidentified rock. What knowledge will the engineer use to predict the potential of this new material?
Answer:
The engineer will conduct a variety of tests, including chemical, mechanical, electrical, and physical examinations, to determine the potential of the new material.
Explanation:
They will need to test the material, this will also help to determine its malleability.
Hope this helps!
An eyewash station needs to be located no more than 10 seconds or less than 55 feet away from a work area.
Answer: True is correct
A third-degree burn may look charred and black or dry and white.
Explanation:
This ball is technology too! It can be
rolled, kicked, or thrown. Is that a need
or a want?
That is an example of a want because you don't need the ball.
Which statement demonstrates the most scientific observation?
Which items are NOT found on a
door?*
5 points
Cladding
Moulding
Weatherstrip
Check Strap
Striker
All of the above
None of the above
Answer:
None of the above cause thats what i put
__________ use cleaning solutions that eventually become spent and must be disposed of properly
a sump pits
b car washers
c part washers
d tire cleaners
Answer:Part washers
Explanation:
I did it..
The only option that uses cleaning solutions that become spent and must be disposed properly is;
Option C; Parts Washers
To answer this question, let us examine each option;
Option A; Sump Pits; These are basins at the lowest part of specific areas designed to allow all water collected in that basin area to drain into it. This has no business with cleaning solution.
Option B; Car washers; These are like machine dryers used to clean the exterior and interior parts of motor vehicles.
Option C; Part Washers; These are mechanical devices that are designed for cleaning to remove debris, grime, oil dirt, paint, and other substances that could potentially cause contamination from parts in order to get them ready for assembling, packaging, or even coating. They are basically used to clean parts to get them ready for functional use.
Option D; Tire cleaner is just used to clean tire but is not something that becomes spent and should be disposed properly as it is more like a liquid.
The only correct option is Option C.
Read more at; https://brainly.com/question/25024807
parallel circuits???
PLEASE GIVE BRAINLIST
A parallel circuit has two or more paths for current to flow through. Voltage is the same across each component of the parallel circuit. The sum of the currents through each path is equal to the total current that flows from the source.
HOPE THIS HELPED
Help me for this question
A sign that has a white background with a
green panel with white lettering is a
a. general information sign
b. safety instruction sign
c. caution sign
d. danger sign
Answer:
A
Explanation:
general information sign
safety is orange
caution is yellow
danger is red
The heat transfer rate per unit area in a thermal circuit is equivalent to what quantity in an electric circuit? A. voltage B. current C. resistance D. inductance
Answer:
A. voltage
Explanation:
The rate of heat transfer per unit area in a thermal circuit is equivalent to the voltage in an electrical circuit.
This is because, within an electrical circuit the voltage is supplied by a generator, which can be batteries, batteries, or others. The moment the electrical voltage acts within a circuit, it will consist of the gradient of electrical potential between two points, which will cause the electrons to move from the point with the highest concentration to the point with the lowest concentration (similar to the transfer of heat from the thermal circuit) generating the electric current.
What is the period if the clock frequency is 3.5 GHz?
Answer:
Period = 0.2857 nanoseconds
Explanation:
We are told that frequency = 3.5 GHz
This is simply 3.5 × 10^(9) Hz
Now, from wave equations, Period is given by the formula;
Period = 1/frequency
Thus;
Period = 1/(3.5 × 10^(9))
Period = 0.2857 × 10^(-9) seconds
From conversions, we can simplify the answer.
1 second = 10^(-9) nanoseconds
Thus, 0.2857 × 10^(-9) seconds = 0.2857 × 10^(-9) × 10^(-9) nanoseconds = 0.2857 nanoseconds