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2 years ago

Please help me with this. Picture

Engineering

1 answer:

qwelly [4]2 years ago

5 0

Answer:

^547.'":8765456\⇆⊇⊇ys p)(ay^547.'":8765456\⇆⊇⊇ys p)(*

iuytrT&*7654567iol;';";lk↓ωω*&65∴∀Hgtre

6и876&*b n™Ο65656&^cxCv876и876&*b n™‰‰ay^547.'":87765^&* kIKUYtb

Explanation:

=yx^z3

ÜÜÜ

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An exit sign must be:Colored in a way that doesn’t attract attentionIlluminated by a reliable light sourceAt least 3 inches tall

Ksenya-84 [330]

Answer:

Red

Explanation:

5 0

2 years ago

An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a rate of 1.5 ft3/s th

photoshop1234 [79]

Answer:

mechanical power used to overcome frictional effects in piping is 2.37 hp

Explanation:

given data

efficient pump = 80%

power input = 20 hp

rate = 1.5 ft³/s

free surface = 80 ft

solution

we use mechanical pumping power delivered to water is

${W_{u}}= \eta {W_{pump}}$ .............1

put here value

${W_{u}}$ = (0.80)(20)

${W_{u}}$ = 16 hp

and

now we get change in the total mechanical energy of water is equal to the change in its potential energy

$\Delta{E_{mech}} = {m} \Delta pe$ ..............2

$\Delta {E_{mech}} = {m} g \Delta z$

and that can be express as

$\Delta {E_{mech}} = \rho Q g \Delta z$ ..................3

$\Delta {E_{mech}} = (62.4lbm/ft^3)(1.5ft^3/s)(32.2ft/s^2)(80ft)[\frac{1lbf}{32.2lbm\cdot ft/s^2}][\frac{1hp}{550lbf \cdot ft/s}]$ ......4

solve it we get

$\Delta {E_{mech}} = 13.614$ hp

so here

due to frictional effects, mechanical power lost in piping

we get here

${W_{frict}} = {W_{u}}-\Delta {E_{mech}}$

put here value

${W_{frict}}$ = 16 -13.614

${W_{frict}}$ = 2.37 hp

so mechanical power used to overcome frictional effects in piping is 2.37 hp

4 0

2 years ago

Select four items that an industrial engineer must obtain in order to practice in the field.

alex41 [277]

Answer:

Professional engineering license

Bachelor's degree

Computer science classes

job recommendations

3 0

1 year ago

Read 2 more answers

Pls help me it’s due today

hichkok12 [17]

Answer:

C. 14.55

Explanation:

12 x 10 = 120

120 divded by 10 is 12

so now we do the left side

7 x 3 = 21 divded by 10 is 2

so now we have 14

and the remaning area is 0.55

so 14.55

6 0

2 years ago

A ball bearing has been selected with the bore size specified in the catolog as 35.000 mm to 35.020 mm. Specify appropriate mini

Fofino [41]

Answer:

the minimum shaft diameter is 35.026 mm

the maximum shaft diameter is 35.042mm

Explanation:

Given data;

D-maximum = 35.020mm and d-minimum = 35.000mm

we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6

so From table, Selection of International Trade Grades metric series

the grade tolerance are;

ΔD = IT7(0.025 mm)

Δd = IT6(0.016 mm)

Also from Table "Fundamental Deviations for Shafts" metric series

Sf = 0.026

D-maximum

Dmax = d + Sf + Δd

we substitute

Dmax = 35 + 0.026 + 0.016

Dmax = 35.042 mm

therefore the maximum diameter of shaft is 35.042mm

d-minimum

Dmin = d + Sf

Dmin = 35 + 0.026

Dmin = 35.026 mm

therefore the minimum diameter of shaft is 35.026 mm

8 0

2 years ago