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Vedmedyk [2.9K]

2 years ago

10

In a vehicle with front disc brakes, the vehicle pulls to the left when braking. What causes this?

2 answers:

IgorC [24]2 years ago

8 0

The answer is b !!!

IrinaVladis [17]2 years ago

3 0

The answer to your question is A

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Which of these is the BEST description of

strojnjashka [21]

Answer:

i would say C but i may be wrong have a great day

Explanation:

3 0

3 years ago

For a column that is pinned at both ends, the critical buckling load can be calculated as, Pcr = π2 E I /L^2 where E is Young's

gulaghasi [49]

When a slender member is subjected to an axial compressive load, it may fail by a ... Consider a column of length, L, cross-sectional Moment of Inertia, I, having Young's Modulus, E. Both ends are pinned, meaning they can freely rotate ... p2EI L2 ... scr, is the Euler Buckling Load divided by the columns cross-sectional area

6 0

3 years ago

A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima

zvonat [6]

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

$\sigma_c = \frac{P}{A}$

$\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}$

$\sigma_c = 81.5MPa$

Through Goodman's equations the combined effort by fatigue and compression is expressed as:

$\frac{\sigma_a}{S_e}+\frac{\sigma_c}{\sigma_u}=\frac{1}{Fs}$

Where,

$\sigma_a=$ Fatigue limit for comined alternating and mean stress

$S_e =$ Fatigue Limit

$\sigma_c=$ Mean stress (due to static load)

$\sigma_u =$ Ultimate tensile stress

$Fs =$ Security Factor

We can replace the values and assume a security factor of 1, then

$\frac{\sigma_a}{320}+\frac{81.5}{400}=\frac{1}{1}$

Re-arrenge for $\sigma_a$

$\sigma_a = 254.8Mpa$

We know that the stress is representing as,

$\sigma_a = \frac{M_c}{I}$

Then,

Where $M_c$ =Max Moment

I= Intertia

The inertia for this object is

$I=\frac{\pi d^4}{64}$

Then replacing and re-arrenge for $M_c$

$M_c = \frac{\sigma_a*\pi*d^3}{32}$

$M_c = \frac{260.9*10^6*\pi*0.05^3}{32}$

$M_c = 3201.7N.m$

Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm

5 0

3 years ago

What color is an orange? its for my 8th grade

Ksenya-84 [330]

Answer:

hmmmm i think orange I may be wrong....

Explanation:

5 0

2 years ago

Air enters a diffuser operating at steady state at 540°R, 15 lbf/in.2, with a velocity of 600 ft/s, and exits with a velocity of

yKpoI14uk [10]

Answer: Hello the question is incomplete below is the missing part

Question: determine the temperature, in °R, at the exit

answer:

T2= 569.62°R

Explanation:

T1 = 540°R

V2 = 600 ft/s

V1 = 60 ft/s

h1 = 129.0613 ( value gotten from Ideal gas property-air table )

<em>first step : calculate the value of h2 using the equation below </em>

assuming no work is done ( potential energy is ignored )

h2 = [ h1 + ( V2^2 - V1^2 ) / 2 ] * 1 / 32.2 * 1 / 778

∴ h2 = 136.17 Btu/Ibm

From Table A-17

we will apply interpolation

attached below is the remaining part of the solution

8 0

2 years ago