A hollow shaft of diameter ratio 3/8 (internal dia to outer dia) is to transmit 375 kW power at 100 rpm. The maximum torque being 20% greater than the mean torque. The shear stress is not to exceed 60 N/mm2 and twist in a length of 4m not to exceed 2o. Calculate its external and internal diameters which would satisfy both the above conditions. (G= 0.85 X 105 N/mm2)

Answer:

The Negative battery cable is used to bolt the vehicle frame to allow the metal structure of the vehicle to serve as a large conductor to carry current.

Explanation:

Battery cable is large automotive cable. Like smaller types of automotive wire, it is available in PVC and cross-linked forms. Click here to read an earlier post explaining the differences between PVC and cross-linked wire insulation.

One type of PVC battery cable is SGT cable. It is rated to 80°C and therefore can be used in starters or battery grounds.  Cross-linked battery cables can also be used in starter and battery ground applications, and they are more resistant to heat, abrasion, and aging than PVC cable.

Two types of cross-linked battery cable are SGX and STX. They are rated to 125°C. Of the three types of battery cable (SGT, SGX, amd STX), STX has the thinnest wall, making it a popular choice for automotive applications with limited space.

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Driving consistently will cause the oil to heat up and break down to create sludge.

And if oil gets too hot, it will cause metal on metal grinding due to the oil thinning out.

Hope this helps you. :)

In the radio communication system, multipath is the propagation phenomenon that causes diffusion or reflection in multiple different directions of a signal.

Multipath is a propagation mechanism that impacts the propagation of signals in radio communication. Multipath results in the transmission of data to the receiving antenna by two or more paths. Diffusion and reflection are the causes that create multiple paths for the signal to be delivered.

Diffraction occurs when a signal bends around sharp corners; while reflection occurs when a signal impinges on a smooth object. When a signal is received through more than one path because of the diffraction or reflection, it creates phase shifting and interference of the signal.

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The water surface elevation must be 110.2631 meters for a flow of 0.03m³ to occur in the cast pipe

For cast iron the chart has 0.0012  from Moody's chart

0.016 for cast iron

\(hf = \frac{flQ^{2}}{12.1d5}\)

\(h1 = h2+\frac{Q^{2} }{12.1} [\frac{0.0012*100}{(12/1000)^5} +\frac{0.0016*150}{(15/100)^5} ]\)

\(h1 = 110m+\frac{0.03^2}{12.1} [\f\frac{0.12}{0.00032} +\frac{0.24}{0.000759} ]\\\\h1 = 110+0.0000744[375+3162.06]\\\\= 110 + 0.2631m\\\\= 110.2631m\)

Therefore the water surface elevation must be 110.2631 meters for a flow of 0.03m³ to occur in the cast pipe

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Answer:

C.

Explanation:

You want to make sure it still works. You don't want to move it periodically though in case of an emergency.

1. Inverse square law states that the intensity of light varies inversely with the square of the distance from the source. It can be represented mathematically as: I = k/d², where I is the intensity of light, d is the distance from the source and k is a constant of proportionality.

This law is illustrated by the fact that as the distance from the source increases, the intensity of light decreases proportionally to the square of the distance.2. Given, diameter of the circular area, d = 200 mRadius of the circular area, r = d/2 = 100 mLamp illuminates a circular area of 200 meters in diameter with the illumination of the disc increasing uniformly from 0.5 meter-candle at the edge to 2 meter-candle at the center. The average illumination can be calculated as follows:Average illumination of the disc, I = (0.5 + 2)/2 = 1.25 meter-candleThe total light received can be calculated as follows:Total light received = (2πr² × I) = (2 × π × 100² × 1.25) = 78,540 lumensAverage candle power of the source can be calculated as follows:Average candle power = Total light received/4π = 78,540/4π = 6250 lumens3. Floodlighting is the use of high-intensity artificial light to illuminate a large area.

The purpose of floodlighting is to provide a bright and uniform light over a large area, typically for outdoor sports fields, stadiums, and other large events. It can be achieved using various types of lighting fixtures, such as floodlights, spotlights, and high-intensity discharge lamps. Suitable diagrams for floodlighting are shown below:

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The list of common malfunctions given in the question belongs to the 'impulse tubing' sensor-related instrument.

Impulse is referred to the sensing process of medium pressure, and tubing that directly connects this sensing process tapping to the transmitters is called impulse tubing. Impulse tubing is used to transmit the pressure signal from a process to a transmitter. Impulse tubing can be affected by several common malfunctions such as leakage, plugged or partially plugged tubing, improperly sized tubing, heat tracing too hot or too cold, excessive vibration, and mechanical damage.

"

Correct question is:

The following list of common malfunctions are for which sensor related instrument?

List of common malfunctions is as follows:

"

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Answer:

A) F = 0.011 N

B) ΔP = 5.6 N/m²

Explanation:

We are given;

surface tension coefficient; S = 0.07 N/m

Radius; r = 2.5 mm = 0.025 m

A) Formula to find the surface tension force(F) is given by;

F = SL

Where L is effective length = 2πr

F = 0.07 × 2π × 0.025

F = 0.011 N

B) Formula for difference in pressure droplet is;

ΔP = 2S/r

Thus;

ΔP = (2 × 0.07)/0.025

ΔP = 5.6 N/m²

Answer:

In most cases, it will never get that far - the shop will cave and agree to properly diagnose your problem and correctly fix it. Usually, the mechanic that failed to do their job correctly will be back-flagged for the labor and in some cases they'll even be charged for the parts as well

Explanation:

The output of the step-up transformer is 20 V, 0.5 A (Option b).

A step-up transformer increases the voltage while decreasing the current on the output side compared to the input side. The turns ratio of the transformer is given as 10:20, which means that for every 10 turns on the input side, there are 20 turns on the output side.

Since the input voltage is 10 V and the turns ratio is 10:20, the output voltage can be calculated as follows:

Output Voltage = Input Voltage x (Number of Turns on Output Side / Number of Turns on Input Side)

= 10 V x (20 / 10)

= 20 V

Similarly, since the input current is 1.0 A and the turns ratio is 10:20, the output current can be calculated as follows:

Output Current = Input Current x (Number of Turns on Input Side / Number of Turns on Output Side)

= 1.0 A x (10 / 20)

= 0.5 A

Therefore, the output of the step-up transformer is 20 V, 0.5 A (Option b).

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Answer: true

Explanation:

Answer:

Engine oil level.

Brake fluid.

Power steering fluid.

Answer:

(1) 2553 N = 2550 N

(2) 58342 m = 58300 m

(3) 68.534 s = 68.5 s

Explanation:

To round off a number to any significant number start from the last digit, round it off to 1 if the number is up to 5 and to 0 if the last digit is less than 5. Add this 1 or 0 to the preceding digit and continue the process until you are left with three non zero digits, if you are rounding off to three significant figures.

(1) Round off 2553 N to three significant figures.

= 2550 N

(2) Round off 58342 m to three significant figures.

= 58300 m

(3) Round off 68.534 s to three significant figures.

= 68.500 s (zero after decimal point is insignificant)

= 68.5 s

Answer:

Now find the temperature of each surface, we have that the the temperature on the left side of the wall is T∞₁ - Q/h₁A and  the temperature on the right side of the wall is T∞₂ + Q/h₂A.

Note: kindly find an attached diagram to the complete question given below.

Sources: The diagram/image was researched and taken from Slader website.

Explanation:

Solution

Let us consider the rate of heat transfer through the plane wall which can be obtained from the relations given below:

Q = T∞₁ -T₁/1/h₁A = T₁ -T₂/L/kA =T₂ -T∞₂/1/h₂A

= T∞₁  - T∞₂/1/h₁A + L/kA + 1/h₂A

Here

The convective heat transfer coefficient on the left side of the wall is h₁, while the convective heat transfer coefficient on the right side of the wall is h₂. the thickness of the wall is L, the thermal conductivity of the wall material is k, and the heat transfer area on one side of the wall is A. Q is refereed to as  heat transfer.

Thus

Let us consider the convection heat transfer on the left side of the wall which is given below:

Q = T∞₁ -T₁/1/h₁A

T₁ = T∞₁ - Q/h₁A

Therefore the temperature on the left side of the wall is T∞₁ - Q/h₁A

Now

Let us consider the convection heat transfer  on the left side of the wall which is given below:

Q= T₂ -T∞₂/1/h₂A

T₂ = T∞₂ + Q/h₂A

Therefore the temperature on the right side of the wall is T∞₂ + Q/h₂A

Given,Four standard concrete cylinders of 6-inch diameter cross-section were tested under uniformly increasing axial compressive loads 28 days after casting concrete.

The maximum measured loads were 121,000 lbs, 116,700 lbs, 128,800 lbs, and 119,100 lbs.So, the average compressive strength of this concrete in ksi is as follows:The cross-sectional area of the cylinders, A= π r²where r= 6/2= 3 inches= 0.25 feetA = π × (0.25)²= 0.0491 ft²The maximum load on the concrete cylinders can be calculated by using the formula:σ = P/Awhere P = load on the cylinder,

and A = cross-sectional area of the cylinderσ = 121000/0.0491 = 2462631.4 lb/ft²σ = 2462631.4/144 = 17096.5 psi= 17096.5/1000 ksi= 17.1 ksiSimilarly,σ = 116700/0.0491 = 2373864.7 lb/ft²σ = 2373864.7/144 = 16486 psi= 16.5 ksiσ = 128800/0.0491 = 2623481.9 lb/ft²σ = 2623481.9/144 = 18209 psi= 18.2 ksiσ = 119100/0.0491 = 2425757.9 lb/ft²σ = 2425757.9/144 = 16847.7 psi= 16.8 ksiTherefore, the average compressive strength of this concrete in ksi is given by:Average compressive strength = (17.1 + 16.5 + 18.2 + 16.8)/4= 68.6/4= 17.15 ksi (approx)Hence, the average compressive strength of this concrete in ksi is 17.15.

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Answer:

Forces like wind and water break down rocks through the processes of weathering and erosion. ... Forces like wind and water move the rock pieces. They mix with matter like sand to become sediment. Weathering and erosion help shape Earth's surface.

Answer:

As an agent of erosion, the wind will quickly break the rock into different rock types.

Explanation:

Answer:

when car vroom it go electric

Explanation:

i think then it goes to some spinny thing that spins and make the electric and the electic spins the spinny thing and the smoke comes outside the back

Answer:

Induced draft is a mechanical draft created by air pulled through the boiler firebox by a blower located in the breaching after the boiler.

Explanation:

Explanation:

1 meter ≈ 3.281 ft.

so,

1 m² = 1m×1m ≈ 3.281 × 3.281 = 10.764961 ft²

now we have 22.3 ft² of skin.

that would be then

22.3/10.764961 = 2.071535605 ≈ 2.1 m² or even more rounded ≈ 2 m²

When a roof slope is 2/12 or less, the rafters are often referred to as flat-roof rafters or low-slope roof rafters.

In traditional residential construction, low-slope roofs are often constructed using a different method than steep-slope roofs with angled rafters. Instead of using rafters, low-slope roofs typically employ a system of joists or trusses.

Here are a few common methods used for constructing low-slope roofs:

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It is to be noted that all UHRS platforms are optimized for the popular kinds of internet browser applications.

The Universal Human Relevance System (UHRS) is a crowdsourcing platform that allows for data labeling for a variety of AI application situations.

Vendor partners link people referred to as "judges" to offer data labeling at scale for us. All UHRS judges are bound by an NDA, ensuring that data is kept protected.

A browser is a software tool that allows you to see and interact with all of the knowledgeon the World Wide Web. Web sites, movies, and photos are all examples of this.

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Answer:

Fast Fourier ( FFT ) algorithms speed up computation of Fourier coefficients by simply reducing the the computing time of a traditional direct form  Fourier series. it  achieves this by breaking complex DFTS into smaller DFTS to reduce its complexity and in turn reduce its computing time

Explanation:

Fast Fourier ( FFT ) algorithms speed up computation of Fourier coefficients by simply reducing the the computing time of a traditional direct form  Fourier series. it  achieves this by breaking complex DFTS into smaller DFTS to reduce its complexity and in turn reduce its computing time. an example of such FFT is Cooley-Tukey algorithm

Answer:

A.

Explanation:

The voltage change across a resistor is negative when you traverse in the opposite direction of the current.

Explanation:

mathematics education grade 7

The two physical resources that are abstracted by the hypervisor and presented to virtual machines are:

B. CPU (Central Processing Unit)

E. Memory

Explanation:

1. CPU (Central Processing Unit): The hypervisor abstracts the physical CPU(s) of the host machine and presents them to virtual machines as virtual CPUs (vCPUs). Each virtual machine is allocated a portion of the CPU resources, allowing them to run their own operating systems and execute their own processes independently. The hypervisor manages the scheduling and allocation of CPU resources among the virtual machines.

2. Memory: The hypervisor also abstracts the physical memory (RAM) of the host machine and presents it to virtual machines as virtual memory. Each virtual machine is assigned a specific amount of memory that it can utilize for running its applications and processes. The hypervisor handles the management and allocation of memory resources, ensuring that each virtual machine receives its allocated memory and preventing conflicts or overlaps.

It's important to note that the other options mentioned are not directly abstracted by the hypervisor:

A. Libraries: Libraries are software components that contain precompiled functions and routines. They are not physical resources that can be abstracted by the hypervisor.

C. Application: Applications are software programs that run on top of the operating system. The hypervisor virtualizes the underlying hardware resources for virtual machines, but it does not directly abstract or manage specific applications.

D. Desktop: The desktop refers to the user interface and graphical environment provided by the operating system. The hypervisor focuses on abstracting and managing hardware resources rather than the desktop interface itself.

In summary, the hypervisor abstracts and presents the physical CPU and memory resources of the host machine to virtual machines as virtual CPUs and virtual memory, respectively. This allows virtual machines to have their own isolated computing environments while sharing the underlying hardware resources efficiently.

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Answer:

somewhere around 34.2223 meters thick but that's what I am estimating.

Real power absorbed by a load is important for engineers as it helps them to ensure that the load is supplied with the correct power and that the power system is operating efficiently.

Power is the ability to act or make decisions in a way that has an influence on others and their environment. It is the ability to control and command, to motivate others, and to shape the future. It is a form of strength and authority, and can be seen in many different forms from physical strength to money, status, and knowledge. Power can be used to create positive change and empower others, or it can be used to oppress and control.

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Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length  = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod \(\overline T = 548 \ K\)

\(\rho = 7900 \ kg/m^3\)

\(K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K\)

\(\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657\)

Here, we can't apply the lumped capacitance method, since Bi > 0.1

\(\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\\)

\(0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81\)

\(t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins\)

However, on a single rod, the energy extracted is:

\(\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J\)

Hence, for centerline temperature at 50 °C;

The surface temperature is:

\(T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}\)