steep safety ramps are built beside mountain highway to enable vehichles with fedective brakes to stop safely. a truck enters a 1000 ft ramps at a high speed vo and travels 600ft in 7 s at constant deceleration before its speed is reduced to uo/2. Assuming the same constant deceleration.

Determine:
a. The additional time required for the truck to stop.
b. The additional distance traveled by the truck.

In this exercise we have to calculate the formula that will be able to determine the length of the cantilevered, like this:

[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]

So to determinated the maximum tensile and compreensive stress due to bending we can describe the formula as:

[tex]\sigma_b = \frac{MC}{I}[/tex]

Where,

So making this formula for the max, we have:

[tex]\sigma_c=\frac{MC}{I} \\\sigma_T=-\sigma_c=-\frac{MC}{I}\\\sigma_{max}=M_{max}\\[/tex]

With all this information we can put the formula as:

[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]

See more about stress in the beam at brainly.com/question/23637191

Answer: 164.2253 MPa

Explanation:

First we find the half internal crack which is  = length of surface crack /2

so α = 8.6 /2 = 4.3mm ( 4.3×10⁻³m )

Now we find the dimensionless parameter using the critical stress crack propagation equation

∝ = K / Y√πα

stress level ∝ = 112Mpa

fracture toughness K = 26Mpa

dimensionless parameter Y = ?

SO working the formula

Y = K / ∝√πα

Y = 26 / 112 (√π × 4.3× 10⁻³)

Y = 1.9973

We are asked to find stress level for internal crack length of  4m

so half internal crack is  = length of surface crack /2

4/2 = 2mm ( 2 × 10⁻³)

from the previous formula critical stress crack propagation equation

∝ = K / Y√πα

∝ = 26 / 1.9973 √(π × 2 × 10⁻³)

∝ = 164.2253 Mpa

Answer:

From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.

Explanation:

Heat generated Q = 200 kW

power input W = 75 kW

Temperature of heated region [tex]T_{h}[/tex] = 293 K

Temperature of heat source [tex]T_{c}[/tex] = 273 K

For this engine,

coefficient of performance COP = Q/W  = 200/75 = 2.67

The maximum theoretical COP obtainable for a heat pump is given as

COP = [tex]\frac{T_{h} }{T_{h} - T_{c} }[/tex] =  [tex]\frac{293 }{293 - 273 }[/tex] = 14.65

From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.

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:

A to C = 6.4 km

Explanation:

A to B = 4 km

B to C = 5 km

A to C =  using pythagorean theorem

a² + b² = c²

a = A to B = 4

b = B to C = 5

c = A to C

c² = 4² + 5²

c = 6.4 km (A to C)

According to the scenario, the distance between town C from town A is found to be 6.40 Km.

The background that this question depends on is known as the direction-based question. These types of questions completely depend on the distance of moving bodies like cars, persons, or any other objects as well with respect to the initial position.

According to the question,

The distance between town A to town B = 4 km.

The distance between town B to town C = 5 km.

Now, according to the Pythagoras theorem, the distance between town C to town A is as follows:

[tex]AC^2[/tex] = [tex]AB^2 +BC^2[/tex].

[tex]AC^2[/tex] = [tex]4^2+5^2[/tex]

[tex]AC^2[/tex] = 16 + 25 = 41.

AC = √41 = 6.40 km.

Therefore, the distance between town C from town A is found to be 6.40 Km.

To learn more about Pythagoras' theorem, refer to the link:

https://brainly.com/question/343682

#SPJ5

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:

the required stress level  at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa

Explanation:

From the given information; the objective is to compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.

The Critical Stress for a maximum internal crack can be expressed by the formula:

[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]

[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]

where;

[tex]\sigma_c[/tex] = critical stress required for initiating crack propagation

[tex]K_{lc}[/tex] = plain stress fracture toughness = 26 Mpa

Y = dimensionless parameter

a = length of the internal crack

given that ;

the maximum internal crack length is 8.6 mm

half length of the internal  crack will be 8.6 mm/2 = 4.3mm

half length of the internal  crack a = 4.3 × 10⁻³ m

From :

[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]

[tex]Y= \dfrac{26}{112 \times \sqrt{\pi \times 4.3 \times 10 ^{-3}}}[/tex]

[tex]Y= \dfrac{26}{112 \times0.1162275716}[/tex]

[tex]Y= \dfrac{26}{13.01748802}[/tex]

[tex]Y=1.99731315[/tex]

[tex]Y \approx 1.997[/tex]

For this same component and alloy, we are to also compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.

when the length of the internal crack a = 3mm

half  length of the internal  crack will be 3.0 mm / 2 = 1.5 mm

half length of the internal  crack a =1.5 × 10⁻³ m

From;

[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]

[tex]\sigma_c = \dfrac{26}{1.997 \sqrt{\pi \times 1.5 \times 10^{-3}}}[/tex]

[tex]\sigma_c = \dfrac{26}{0.1370877444}[/tex]

[tex]\sigma_c =189.6595506[/tex]

[tex]\sigma_c =[/tex] 189.66 MPa

Thus; the required stress level  at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa

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.

Hey,

Who plays a role in the financial activities of a company?

O D. Everyone at the company, including managers and employees

Answer:

Everyone at the company, including managers and employees

Explanation:

Explanation:

1. increase This due to increase in the pore volume.

2.increase . This is due to the fact that more liquid phase will be present at the firing.

3. Increase. This increase is because of the fact that clay on cooling forms glass.Thus, gaining more strength as the liquid phase formed fills in pore volume.

4. Increase, Rate of corrosion depends upon the surface area exposed.Since, upon vitrification surface area would increase, therefore corrosion increases.

5. Increase , glass has higher thermal conductivity than the pores it fills.

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:

[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:

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.

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