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3 years ago

Io[iiiiiioooiiijjsjsj

Engineering

2 answers:

-Dominant- [34]3 years ago

8 0

Hungujnggttssyinoiookoiooiioooiiioiiiiioojjjjjkokksjok

Mademuasel [1]3 years ago

6 0

To political, exist is on left down the stair door is on your right.

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WARNING:<br><br> when people put links in the answer it is a virus DO NOT DOWNLOAD IT

kobusy [5.1K]

Answer:

Thank you for this!

Explanation:

I was about to click it on a question I saw.

6 0

3 years ago

Read 2 more answers

A brass alloy rod having a cross sectional area of 100 mm2 and a modulus of 110 GPa is subjected to a tensile load. Plastic defo

nevsk [136]

Answer:

a) the maximum stress that can be applied without plastic deformation is 398.72 N/mm²

b) length of the specimen is 66.97 mm

Explanation:

Given the data in the question;

a) Determine the maximum stress that can be applied without plastic deformation

when know that; maximum stress σ $_{max}$ = F / A

where F is the force in the rod ( 39872 N )

A is the cross-sectional area of the rod ( 100 mm² )

so we substitute;

σ $_{max}$ = 39872 N / 100 mm²

σ $_{max}$ = 398.72 N/mm²

Therefore, the maximum stress that can be applied without plastic deformation is 398.72 N/mm²

strain in the members can be calculated using the expression

ε = σ / E

where σ is the stress in the rod

E is the module of elasticity ( 110 GPa = 110000 N/mm² )

(Sl-L) / L = σ/E

where Sl-L is the change in length of the member

L is the original length of the specimen

so we substitute

(67.21 - L) / L = 398.72 / 110000

110000( 67.21 - L) = 398.72L

7393100 - 110000L = 398.72L

7393100 = 398.72L+ 110000L

7393100 = 110398.72L

L = 7393100 / 110398.72

L = 66.97 mm

Therefore; length of the specimen is 66.97 mm

5 0

3 years ago

Concrete is forced into all parts of its form through a process called?

Leona [35]

Answer:

hydration

Explanation:

home it helps and im too late:D

4 0

3 years ago

Technician A says that in a recirculating-ball steering box when the sector gear and nut teeth

NISA [10]

The second one I believe

3 0

2 years ago

Read 2 more answers

A construction company distributes its products by trucks loaded at its loading station. A backacter in conjunction with trucks

blagie [28]

Answer:

a) $L = 1.5$

$L_q = 0.9$

$W = \dfrac{1 }{8 } \, hour$

$W_q = \dfrac{3}{40 } \, hour$

$P = \dfrac{3}{5 }$

b) The new backacter should be recommended

c) The additional backacter should not be deployed

Explanation:

a) The required parameters are;

L = The number of customers available

$L = \dfrac{\lambda }{\mu -\lambda }$

μ = Service rate

$L_q$ = The number of customers waiting in line

$L_q = p\times L$

W = The time spent waiting including being served

$W = \dfrac{1 }{\mu -\lambda }$

$W_q$ = The time spent waiting in line

$W_q = P \times W$

P = The system utilization

$P = \dfrac{\lambda }{\mu }$

From the information given;

λ = 12 trucks/hour

μ = 3 min/truck = 60/3 truck/hour = 20 truck/hour

Plugging in the above values, we have;

$L = \dfrac{12 }{20 -12 } = \dfrac{12 }{8 } = 1.5$

$P = \dfrac{12 }{20 } = \dfrac{3}{5 }$

$L_q = \dfrac{3}{5 } \times \dfrac{3}{2 } = \dfrac{9}{10 } = 0.9$

$W = \dfrac{1 }{20 -12 } = \dfrac{1 }{8 } \ hour$

$W_q = \dfrac{3}{5 } \times \dfrac{1}{8 } = \dfrac{3}{40 } \, hour$

(b) The service rate with the new backacter = 1.5 minutes/truck which is thus;

μ = 60/1.5 trucks/hour = 40 trucks/hour

$P = \dfrac{12 }{40 } = \dfrac{3}{10}$

$W = \dfrac{1 }{40 -12 } = \dfrac{1 }{38 } \, hour$

$W_q = \dfrac{3}{10 } \times \dfrac{1}{38 } = \dfrac{3}{380 } \, hour$

λ = 12 trucks/hour

Total cost = $mC_s + \lambda WC_w$

m = 1

$C_s$ = GH¢ = 1300

$C_w$ = 400

Total cost with the old backacter is given as follows;

$1 \times 1000 + 12 \times \dfrac{1}{8} \times 400 = \$ 1,600.00$

Total cost with the new backacter is given as follows;

$1 \times 1300 + 12 \times \dfrac{1}{38} \times 400 = \$ 1,426.32$

The new backacter will reduce the total costs, therefore, the new backacter is recommended.

Here μ = 3 min/ 2 trucks = 2×60/3 truck/hour = 40 truck/hour

$\therefore W = \dfrac{1 }{40 -12 } = \dfrac{1 }{38 } \, hour$

Total cost with the one backacter is given as follows;

$1 \times 1000 + 12 \times \dfrac{1}{8} \times 400 = \$ 1,600.00$

Total cost with two backacters is given as follows;

$2 \times 1000 + 12 \times \dfrac{1}{38} \times 400 = \$ 2,126.32$

The additional backacter will increase the total costs, therefore, it should not be deployed.

6 0

3 years ago