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3 years ago

No question but thx reeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

Engineering

2 answers:

evablogger [386]3 years ago

7 0

Answer:

reeeee

Explanation:

reeeeeeeeeeeeeeeeeeeee

jeka943 years ago

5 0

Answer:

why you doin this

Explanation:

is this so we get free points?

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The sticker inside the door of my car says that the tire pressure should be 32 psig (322 kPa) when the tire is cold. Before a ro

Neko [114]

Answer:

37 psi

Explanation:

For ideal gases this equation applies:

p1*V1/T1 = p2*V2/T2

Since we are assuming volume remains constant:

V2 = V1

p1/T1 = p2/T2

p2 = p1*T2/T1

The temperatures must be in absolute scale.

T1 = 15 + 273 = 288 K

T2 = 60 + 273 = 333 K

Then:

p2 = 32 * 333 / 288 = 37 psi

7 0

3 years ago

What part connect to the tie rod that helps the brake chamber out that deals with slack adjuster

scoray [572]

Answer:

none

Explanation:

7 0

2 years ago

Are considered a form of personal protective equipment (PPE) for eyes.

vampirchik [111]

Answer:

All of the above

Explanation:

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3 years ago

A design that is found to have flaws and must be redesigned and retested is an example of what

Aleksandr [31]

Answer:

A. prototype

Explanation:

7 0

4 years ago

A pin must be inserted into a collar of the same steel using an expansion fit. The coefficient of thermal expansion of the metal

nirvana33 [79]

Answer:

a) the temperature to which the pin must be cooled for assembly is $T_2 = -101.89^ \ ^0}C$

b) the radial pressure at room temperature after assembly is $P_f = 62.8 \ MPa$

c) the safety factor in the resulting assembly = 6.4

Explanation:

Coefficient of thermal expansion $\alpha = 12.3*10^{-6} \ ^0 C$

Yield strength $\sigma_y$ = 400 MPa

Modulus of elasticity (E) = 209 GPa

Room Temperature $T_1$ = 20°C

outer diameter of the collar $D_o = 95 \ mm$

inner diameter of the collar $D_i = 60 \ mm$

pin diameter $D_p$ = $60.03 \ mm$

Clearance c = 0.06 mm

The temperature to which the pin must be cooled for assembly can be calculated by using the formula:

$(D_i - c )-D_p = \alpha * D_p(T_2-T_1)$

$(60-0.06)-60.03=12.3*10^{-6}*60.03(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}$ $(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}T_2$ $\ \ - \ \ 0.01476738$

$-0.09 + 0.01476738$ = $7.38369*10^{-4}T_2$

−0.07523262 = $7.38369*10^{-4}T_2$

$T_2 = \frac{-0.07523262}{7.38369*10^{-4}}$

$T_2 = -101.89^ \ ^0}C$

To determine the radial pressure at room temperature after assembly ;we have:

$P_f = \frac{E * (D_p-D_i)(D_o^2-D_1^2)}{D_i*D_o} \\ \\ \\ P_f = \frac{209*10^9* 0.03(95^2-60^2)}{60*95^2} \\ \\ P_f = 62815789.47 \ Pa \\ \\ P_f = 62.8 \ MPa$

c) the safety factor of the resulting assembly is calculated as:

safety factor = $\frac{Yield \ strength }{walking \ stress}$

safety factor = $\frac{400}{62.8}$

safety factor = 6.4

Thus, the safety factor in the resulting assembly = 6.4

4 0

3 years ago