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3 years ago

BRO FNAF ANIMATRONICS ALL OF THEM INCLUDING THE BOOKS AND THE OTHER ONES IN FNAF SPECIAL DELIVERY!

Engineering

3 answers:

Nezavi [6.7K]3 years ago

9 0

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MrRissso [65]3 years ago

8 0

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YNwastaken2 years ago

0 0

heh heh heh what.

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A wooden pallet carrying 540kg rests on a wooden floor. (a) a forklift driver decides to push it without lifting it.what force m

kicyunya [14]

Answer:

The appropriate solution is "1481.76 N".

Explanation:

According to the question,

Mass,

m = 540 kg

Coefficient of static friction,

$\mu_s$ = 0.28

Now,

The applied force will be:

⇒ $F=\mu_s mg$

By substituting the values, we get

$=0.28\times 540\times 9.8$

$=1481.76 \ N$

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2 years ago

A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion

kati45 [8]

Answer:

$X_B = 1.8 \times 10^{-3} m = 1.8 mm$

Explanation:

Given data:

Diffusion constant for nitrogen is $= 1.85\times 10^{-10} m^2/s$

Diffusion flux $= 1.0\times 10^{-7} kg/m^2-s$

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen concentration is 0.5 kg/m^3 ......?

from fick's first law

$J = D \frac{C_A C_B}{X_A X_B}$

Take C_A as point on which nitrogen concentration is 2 kg/m^3

$x_B = X_A + D\frac{C_A -C_B}{J}$

Assume X_A is zero at the surface

$X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}$

$X_B = 1.8 \times 10^{-3} m = 1.8 mm$

4 0

3 years ago

4. Water vapor enters a turbine operating at steady state at 1000oF, 220 lbf/in2 , with a volumetric flow rate of 25 ft3/s, and

hodyreva [135]

Yes i is the time of the day you get to frost the moon and back and then you can come over and then go to hang out with me me and then go to hang out

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3 years ago

A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at 2308C by rejectin

lord [1]

Answer:

hello your question is incomplete attached below is the missing part and also attached is the solution

answer: a) 0.4801

b) 5.398 kw

c) 2.14

d) 12.72

Explanation:

The quality of the refrigerant at the evaporator inlet

h4 = hf4 + x4(hfx4)

Refrigeration load

Ql = m(h1-h4)

COP of the refrigerator

Ql / m(h2-h1) - Qm

Theoretical maximum refrigeration load

( Ql )max = COPr.rev * [m(h2-h1) - Qin]

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