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3 years ago

15

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1 answer:

kifflom [539]3 years ago

5 0

Answer: C. Exchanging ideas with the group will help build his knowledge and reputation

Explanation:

Professional networking refers to the building of professional relationships. This helps in establishing connections that are mutually-beneficial with the people that are in ones profession.

Based on the scenario given, the reason for Vijay to join a networking group is that exchanging ideas with the group will help build his knowledge and reputation.

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Doubling the diameter of a solid, cylindrical wire doubles its strength in tension.

julsineya [31]

Answer:

True ❤️

-Solid by solid can make Cylindrical wire doubles Strengths in tension

4 0

3 years ago

Need help fast I been stuck in this for the longest

gavmur [86]

Answer:

3rd and 4rth

Explanation:

a geologist studies the earth and both of these have something to do with the earth

6 0

3 years ago

A circular column is fixed at the base and not supported at the top. If the column needs to be 15ft and hold 10kips, what is the

muminat

Answer:

The required size of column is length = 15 ft and diameter = 4.04 inches

Explanation:

Given;

Length of the column, L = 15 ft

Applied load, P = 10 kips = 10 × 10³ Psi

End condition as fixed at the base and free at the top

thus,

Effective length of the column, $\L_e$ = 2L = 30 ft = 360 inches

now, for aluminium

Elastic modulus, E = 1.0 × 10⁷ Psi

Now, from the Euler's critical load, we have

$P =\frac{\pi^2EI}{L_e^2}$

where, I is the moment of inertia

on substituting the respective values, we get

$10\times10^3 =\frac{\pi^2\times1.0\times10^7\times I}{360^2}$

or

I = 13.13 in⁴

also for circular cross-section

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

thus,

13.13 = $\frac{\pi}{64}\times d^4$

or

d = 4.04 inches

The required size of column is length = 15 ft and diameter = 4.04 inches

3 0

3 years ago

First person to awnser gets brainlyest!!

Andrews [41]

Answer:

ananswer my question please

7 0

3 years ago

Read 2 more answers

a ten station assembly machine has ideal cycle time of 6 sec. the fraction defect rate at each station 0.005 and defect always j

MAXImum [283]

Answer:

$T_{P}=(2.6667)(10^{-3})h$

Explanation:

Let's write the equation of the production rate for the assembly machine :

$T_{P}=T_{C}+(n).(m).(p).(T_{D})$

Where $T_{P}$ is the production rate for the assembly machine.

Where $T_{C}$ is the ideal cycle time

Where n is the number of stations.

Where m is the number stations that get jam when the defect occurs.

Where p is the defect rate at each station.

And where $T_{D}$ is the average downtime per breakdown

We are looking for the hourly production rate ⇒

$1h=60min\\1min=60s$ ⇒

$1h=3600s$ ⇒

$6s=\frac{(6s)(1h)}{(3600s)}= \frac{1}{600}h$

$60min=1h$ ⇒

$1.2min=\frac{(1.2min)(1h)}{(60min)}=0.02h$

$T_{P}=\frac{1}{600}h+(10)(1.0)(0.005)(0.02h)=\frac{1}{375}h=(2.6667)(10^{-3})h$

m = 1.0 in the equation.

3 0

3 years ago