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3 years ago

HIIIIIIIIII htrdcvbnjuytresdxcvbnjhytrdefghjkmnbvcdertyhjmn i'm soo bored

Engineering

2 answers:

Aloiza [94]3 years ago

8 0

Answer:

My name is Yoshikage Kira

Explanation:

My name is Yoshikage Kira. I'm 33 years old. My house is in the northeast section of Morioh, where all the villas are, and I am not married. I work as an employee for the Kame Yu department stores, and I get home every day by 8 PM at the latest. I don't smoke, but I occasionally drink. I'm in bed by 11 PM, and make sure I get eight hours of sleep, no matter what. After having a glass of warm milk and doing about twenty minutes of stretches before going to bed, I usually have no problems sleeping until morning. Just like a baby, I wake up without any fatigue or stress in the morning. I was told there were no issues at my last check-up. I'm trying to explain that I'm a person who wishes to live a very quiet life. I take care not to trouble myself with any enemies, like winning and losing, that would cause me to lose sleep at night. That is how I deal with society, and I know that is what brings me happiness. Although, if I were to fight I wouldn't lose to anyone.

sweet [91]3 years ago

3 0

I’m bored too hejdhejfnfndndnd

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Replace the black box portion of the circuit with the Norton's equivalent circuit. ______ for the load can now be calculated usi

photoshop1234 [79]

Answer:

Portion

Explanation:

8 0

3 years ago

Determine the required dimensions of a column with a square cross section to carry an axial compressive load of 6500 lb if its l

exis [7]

Answer: 0.95 inches

Explanation:

A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.

The length= 64 inches

Ends are fixed Le= 64/2 = 32 inches

Factor Of Safety (FOS) = 3. 0

E= 10.6× 10^6 ps

σy= 4000ps

The square cross-section= ia^4/12

PE= π^2EI/Le^2

6500= 3.142^2 × 10^6 × a^4/12×32^2

a^4= 0.81 => a=0.81 inches => a=0.95 inches

Given σy= 4000ps

σallowable= σy/3= 40000/3= 13333. 33psi

Load acting= 6500

Area= a^2= 0.95 ×0.95= 0.9025

σactual=6500/0.9025

σ actual < σallowable

The dimension a= 0.95 inches

5 0

3 years ago

Read 2 more answers

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

4 years ago

How do you measure 1/8 teaspoon if you<br> only have a 1/4 teaspoon?

abruzzese [7]

Take half of it sir

8 0

3 years ago

Read 2 more answers

A batch of parts is produced on a semi-automated production machine in a sequential batch production operation. Batch quantity i

vredina [299]

Answer:

a) 4.21 min

b) 21.9666 hrs

c) 1.3657 Pc/hr

Explanation:

Given that;

Batch quantity = 300 units

time setup = 55min

unload/loading time = 0.75

processing time = 3.46

a) Average cycle time;

Average Cycle Time TC = loading time + processing time

TC = 0.75 + 3.46

Tc = 4.21 min

b) Time to complete the batch

Time to complete the batch Tb = setup time + process time + non operation time

Tb = (55min * 1hr/60min) + (300 * 3.46 * 1hr/60min) + (300 * 0.75 * 1hr/60min)

Tb = 0.9166 + 17.3 + 3.75

Tb = 21.9666 hrs

c) Average production rate

Average production rate Rp = 1 / ( Tb / batch size)

we substitute

Rp = 1 / ( 21.9666 / 300 )

Rp = 1 / 0.7322

Rp = 1.3657 Pc/hr

4 0

3 years ago