The system is stimulated, via the voltage source, with a pulse of height 2 and width 4 s. Determine the voltage across the resistor.

Answer:

Hello the table which is part of the question is missing and below are the table values

For a 5V belt the available diameters are : 5.5, 5.8, 5.9, 6.2, 6.3, 6.6, 12.5, 13.9, 15.5, 16.1, 18.5, 20.1

Answers:

belt size = 140 in with diameter of 20.1n

actual speed of belt = 288.49 in/s

actual center distance = 49.345 in

Explanation:

Given data :

Electric motor (driver sheave) speed (w1) = 950 rpm

Driven sheave speed (w2) = 250 rpm

pick D1 ( diameter of driver sheave)  = 5.8 in  ( from table )

To select an appropriate belt size we apply the equation for the velocity ratio to get the diameter first

VR = [tex]\frac{w1}{w2}[/tex] = 950 / 250

also since the speed of  belt would be constant then ;

Vb = w1r1 = w2r2 ------- equation 1

r = d/2

substituting the value of r into equation 1

equation 2 becomes : [tex]\frac{w1}{w2} = \frac{d2}{d1}[/tex]    = VR

Appropriate belt size ( d2) can be calculated as

d2 = [tex]\frac{w1d1}{w2}[/tex] = [tex]\frac{950 * 5.8}{250}[/tex] = 22.04

From the given table the appropriate belt size would be : 20.1 because it is the closest to the calculated value

next we have to determine the belt length /size

[tex]L = 2C + \frac{\pi }{2} ( d1+d2) + \frac{(d2-d1)^2}{4C}[/tex]

inputting  all the values into the above equation including the value of C as calculated below

L ≈ 140 in

Calculating the center distance

we use this equation to get the ideal center distance

[tex]d2< C_{ideal} < 3( d1 +d2)[/tex]

22.04 < c < 3 ( 5.8 + 20.1 )

22.04 < c < 77.7

the center distance is between 22.04 and 77.7  but taking an average value

ideal center distance would be ≈ 48 in

To calculate the actual center distance we use

[tex]C = \frac{B+\sqrt{B^2 - 32(d2-d1)^2} }{16}[/tex] -------- equation 3

B = [tex]4L -2\pi (d2 + d1 )[/tex]

inputting all the values into (B)

B = 140(4) - 2[tex]\pi[/tex]( 20.01 + 5.8 )

B ≈ 399.15 in

inputting all the values gotten Back to equation 3 to get the actual center distance

C = 49.345 in ( actual center distance )

Calculating the actual belt speed

w1 = 950 rpm = 99.48 rad/s

belt speed ( Vb) = w1r1 = w1 * [tex]\frac{d1}{2}[/tex]

                           = 99.48 * 5.8 / 2 = 288.49 in/s

Answer: the thermal conductivity of the sample is 22.4 W/m . °C

Explanation:

We already know that the thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.

ASSUMPTIONS

1. Steady operating conditions exist since the temperature readings do not change with time.

2. Heat losses through lateral surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples.

3. The apparatus possess thermal symmetry

ANALYSIS

The electrical power consumed by resistance heater and converted to heat is:

Wₐ = VI = ( 110 V ) ( 0.4 A ) = 44 W

Q = 1/2Wₐ = 1/2 ( 44 A )

Now since only half of the heat generated flows through each samples because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possess thermal symmetry. The heat transfer area is the area normal to the direction of heat transfer. which is the cross- sectional area of the cylinder in this case; so

A = 1/4πD² = 1/3 × π × ( 0.05 m )² = 0.001963 m²

Now Note that, the temperature drops by 15 degree Celsius within 3 cm in the direction of heat flow, the thermal conductivity of the sample will be

Q = kA ( ΔT/L ) → k = QL / AΔT

k = ( 22 W × 0.03 m ) / (0.001963 m² × 15°C )

k = 22.4 W/m . °C

Answer:

a) cross sectional area of the core = 0.0187 m²

b) The secondary voltage on no-load = 413 V

c) The primary currency and power factor on no load is 1.21 A and 0.168 lagging respectively.

Explanation:

See attached solution.

Answer:

a. Tm = 0.3192min.

b. MRR = 396.91mm^{3}/s.

Explanation:

Given the following data;

Drill diameter, D = 12.7mm

Depth, L = 60mm

Cutting speed, V = 25m/min = 25,000m

Feed, F = 0.30mm/rev

To find the cutting time;

Cutting time, Tm =?

[tex]Tm = \frac{L}{Fr}[/tex] .......eqn 1

We would first solve for the feed rate (F);

[tex]Fr = NF[/tex] .......eqn 2

But we need to find the rotational speed (N);

[tex]N= \frac{V}{\pi *D}[/tex]

[tex]N= \frac{25000}{3.142*12.7}[/tex]

[tex]N= \frac{25000}{39.90}[/tex]

N = 626.57rev/min.

Substiting N into eqn 2;

[tex]Fr = NF[/tex]

Fr = 626.57 * 0.30

Fr = 187.97mm/min.

Substiting F into eqn 1;

[tex]Tm = \frac{L}{Fr}[/tex]

[tex]Tm = \frac{60}{187.97}[/tex]

Tm = 0.3192min.

Therefore, the cutting time is 0.3192 minutes.

For the material removal rate (MRR);

[tex]MRR = \frac{\pi *D^{2}Fr}{4}[/tex]

[tex]MRR = \frac{3.142*12.7^{2}*187.97}{4}[/tex]

[tex]MRR = \frac{3.142*161.29*187.97}{4}[/tex]

[tex]MRR = \frac{95258.16}{4}[/tex]

[tex]MRR = 23814.54mm^{3}/min[/tex]

Time in seconds, we divide by 60;

MRR = 23814.54/60 =396.91mm^{3}/s.

Therefore, the material removal rate (MRR) is 396.91mm^{3}/s.

Answer:

true

Explanation:

true

Answer:

I believe this is true

Explanation:

If your looking at something and you look at something else everything is still in perfect view and clear, in focus.

hope this helps :)

Answer:

Engineering drawing abbreviations and symbols are used to communicate and detail the characteristics of an engineering drawing.

There are many abbreviations common to the vocabulary of people who work with engineering drawings in the manufacture and inspection of parts and assemblies.

Technical standards exist to provide glossaries of abbreviations, acronyms, and symbols that may be found on engineering drawings. Many corporations have such standards, which define some terms and symbols specific to them; on the national and international level, like BS8110 or Eurocode 2 as an example.

Explanation:

Answer:

[tex]\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]

Explanation:

Given that

Shear modulus= G

Sectional area = A

Torsional load,

[tex]t(x) = p sin( \frac{2\pi}{ L} x)[/tex]

For the maximum value of internal torque

[tex]\dfrac{dt(x)}{dx}=0[/tex]

Therefore

[tex]\dfrac{dt(x)}{dx} = p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}\\ p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}=0\\cos( \frac{2\pi}{ L} x)=0\\ \dfrac{2\pi}{ L} x=\dfrac{\pi}{2}\\\\x=\dfrac{L}{4}[/tex]

Thus the maximum internal torque will be at x= 0.25 L

[tex]t(x)_{max} = \int_{0}^{0.25L}p sin( \frac{2\pi}{ L} x)dx\\t(x)_{max} =\left [p\times \dfrac{-cos( \frac{2\pi}{ L} x)}{\frac{2\pi}{ L}} \right ]_0^{0.25L}\\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]

Answer:

59°C

Explanation:

Given that, Cc = McCp,c = 5000 /3600 × 4178 = 5803.2(W/K)

and Ch = MhCp,h = 10000 / 3600 × 4188 = 11634.3(W/K)

Therefore the minimum and maximum heat capacities are:

Cmin = Cc = 5803.2(W/K)

Cmax = Ch = 11634.3(W/K)

The capacity ratio is:

Cr = Cmin / Cmax = 0.499 = 0.5

The maximum possible heat transfer rate is:

Qmax = Cmin (Th,i - Tc,i) = 5803.2 (80 - 20) = 348192(W)

And the number of transfer units is: NTU = UA / Cmin = 11600 / 5803.2 = 1.99

Given that from the appropriate graph in the handouts we can read  = 0.7. So the actual heat transfer rate is: Qact = Qmax = 0.7 × 348192 = 243734.4(W)

Hence, the outlet hot temperature is: Th,o = Th,i - Qact / Ch = 59°C

Answer and Explanation:

The coefficient of determination also called "goodness of fit" or R-squared(R²) is used in statistical measurements to understand the relationship between two variables such that changes in one variable affects the other. The level of relationship or the degree to which one affects the other is measured by 0 to 1 whereby 0 means no relationship at all and 1 means one totally affects the other while figures in between such 0.40 would mean one variable affects 40% of the other variable.

In making a decision as an engineer while using the coefficient of determination, one would try to understand the relationship between variables under consideration and make decisions based on figures obtained from calculating coefficient of determination. In other words when there is a 0 coefficient then there is no relationship between variables and an engineer would make his decisions with this in mind and vice versa.

Answer:

a) 42 mm

b) 144.4 MPa

Explanation:

Load P = 200 kN = 200 x 10^3 N

Torque T = 1.5 kN-m = 1.5 x 10^3 N-m

maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa

diameter of shaft d = ?

From T = τ * [tex]\frac{\pi }{16}[/tex] * [tex]d^{3}[/tex]

substituting values, we have

1.5 x 10^3 = 100 x 10^6 x [tex]\frac{3.142 }{16}[/tex] x [tex]d^{3}[/tex]

[tex]d^{3}[/tex] = 7.638 x 10^-5

d = [tex]\sqrt[3]{7.638 * 10^-5}[/tex] = 0.042 m = 42 mm

b) Normal stress = P/A

where A is the area

A = [tex]\frac{\pi d^{2} }{4}[/tex] = [tex]\frac{3.142*0.042^{2} }{4}[/tex] = 1.385 x 10^-3

Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = 144.4 MPa

Answer:

E = 2940 J

Explanation:

It is given that,

Mass, m = 12 kg

Position at which the object is placed, h = 25 m

We need to find the potential energy of the mass. It is given by the formula as follows :

E = mgh

g is acceleration due to gravity

[tex]E=12\times 9.8\times 25\\\\E=2940\ J[/tex]

So, the potential energy of the mass is 2940 J.

Given:

We have given two statements.

Statement 1: Proper footwear may include both leather and steel-toed shoes.

Statement 2:  Leather-soled shoes provide slip resistance.

Find:

Which statement is true.

Solution:

A slip-resistant outsole is smoother and more slip-resistant than other outsole formulations when exposed to water and oil. A smoother outsole in rubber ensures a slip-resistant shoe can handle a slippery floor more effectively.

Slip resistant shoes have an interlocked tread pattern that does not close the water in, enabling the slip resistant sole to touch the floor to provide better slip resistance.

Leather-soled shoes don't provide slop resistance.

Therefore, both the Technicians are wrong.

From the statements made by both technician A and Technician B, we can say that; both technicians are wrong.

We are given the statements made by both technicians;

Technician A: Proper footwear may include both leather and steel-toed shoes.

Technician B: Leather-soled shoes provide slip resistance.

Now, they are talking about safety shoes to be worn in workshops.

A shoe that is Slip resistant will have rubber soles and tread patterns that can help to have better grip of wet or greasy floors.

This is the type of shoe that should be worn by technicians in the workshop.

Thus, Technician A is wrong because proper footwear does not include leather shoes.

Similarly, technician B is also wrong because leather shoes are not safety shoes.

Read more about slip resistant shoes at; https://brainly.com/question/17411739

Answer: 255

255 turns are required to create 25 ohms of  secondary impedance.

Explanation:

Given that,

Number of turns in primary wire N₁ = 900

impedance on Primary wire Z₁ = 400 ohms

Number of turns in Secondary wire N₂ = ?

impedance on Secondary wire Z₂ = 25 ohms

we know that, the relationship between turn and impedance is

Zp / Zs = ( Np / Ns )²

(Primary impedance / secondary impedance) = Number of turns in primary wire / Number of turns in secondary wire)²

there fore

Z₁ / Z₂ = ( N₁ / N₂ )²

Now we substitute

( 400 / 25 ) = ( 900 / N₂ )²

400 / 25 = 900² / N₂²

we cross multiple to get our N₂

400 × N₂² =  900² × 25

N₂² = ( 900² × 25 ) / 400

N₂² = ( 810000 × 25 ) / 400

N₂² = 20250000 / 400

N₂² = 50625

N₂ = √50625

N₂ = 225

Therefore 255 turns are required to create 25 ohms of  secondary impedance.

Answer:

at  t = 45 s :  

To = 61.7⁰c,  T1 = 55.6⁰c, T2 = 49.5⁰c, T3 = 34.8⁰C

Explanation:

Wall thickness = 0.12 m

thermal diffusivity = 1.5 * 10^-6 m^2/s

Δt ( time increment ) = 300 s

Δ x   = 0.03 m ( dividing wall thickness into 4 parts assuming the system to be one dimensional )

using the explicit finite-difference technique

Detailed solution is attached below