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Y_Kistochka [10]
3 years ago
14

HELPthe overall width of a part is dimensioned as 3.00 ± 0.02. what is the tolerance

Engineering
2 answers:
Svetllana [295]3 years ago
7 0

poopoopoopoopoopoopoopoopoopoopoopoopoopoopoopoopoopoo

Explanation:

MariettaO [177]3 years ago
3 0

Answer:

Not knowing the units the tolerance is 0.02.  I would presume mm but hopefully your question has more detail.  

Explanation:

The tolerance is the portion after the main dimension (+/- 0.02).  In our case we have bilateral tolerance since there is tolerance in both directions (positive and negative).  If you were building a part the acceptable range would be 2.98 to 3.02 based on the tolerance provided.  

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An iron-carbon alloy initially containing 0.286 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1200°
Fantom [35]

Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

carbon concentration cx = 0.215 wt%

time = 7 hr

D =  7.5 \times 10^{-11} m^2/s

for 0.225% carbon concentration following formula is used

\frac{cx -co}{cs -co} = 1 - erf(\frac{x}{2\sqrt{DT}})

where, erf stand for error function

\frac{cx -co}{cs -co} = \frac{0.215 -0.286}{0 -0.286} =0.248

0.248 = 1 - erf(\frac{x}{2\sqrt{DT}})

erf(\frac{x}{2\sqrt{DT}}) = 1 - 0.248

erf(\frac{x}{2\sqrt{DT}}) = 0.751

from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815

from given table

\frac{x}{2\sqrt{DT}} = 0.815

x = 2\times 0.815 \times \sqrt{7.5 \times 10^{-11}\times (7\times 3600)

x = 2.39\times 10^{-3} m

x = 0.002395 mm

8 0
3 years ago
A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower
PIT_PIT [208]

Answer:

γ_{xy} =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²;    1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ_{xy} = displacement / total height

     = 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also,  τ = Gγ_{xy}

By comapring both equations, we get

P/A = Gγ_{xy}   ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

6 0
3 years ago
The author states that chemical engineering is one of the most difficult and complex aspects of engineering. Why do you think th
marishachu [46]

Answer:

hope this helps

Explanation:

answers:

1. Chemical engineering is most difficult because it's a mix of physics, chemistry and math

2. Stoichiometry is so important because it shows how materials react, interact and play off each other

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4. i have no idea sorry :(

5. This is explaining how there are rules and regulations to make the workplace safe. it can be accomplished by following those rules and regulations

8 0
3 years ago
Read 2 more answers
Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve
Dmitry [639]

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature T_1 = -16^0\ C

Quality x_1 = 0.2

Outlet:

Temperature T_2 = -16^0 C

Quality  x_2 = 1

The following data were obtained at saturation properties of R134a at the temperature of -16° C

v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\  v_g = 0.1247 \ m^3 /kg

v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\  v_1 = 0.0255 \ m^3/kg \\ \\ \\  v_2 = v_g = 0.1247 \ m^3/kg

m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\  m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6  \\ \\ \mathbf{m = 0.0534 \ kg/sec}

\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2  \\ \\  \rho_1v_1 = \rho_2v_2   \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}

3 0
3 years ago
Don't break or crush mercury-containing lamps because mercury powder may be released.
alexandr402 [8]
A.

It would be released without a doubt so be careful!

Hope this helps :)
5 0
3 years ago
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