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IgorC [24]
3 years ago
11

Crank OA rotates with uniform angular velocity 0  4 rad/s along counterclockwise. Take OA= r= 0.5

Engineering
1 answer:
mafiozo [28]3 years ago
3 0
CRABK DAT SOUIJA BOI LIKE OOOOO WATCH ME WATCH ME OOOO LASAGNA Can you lend me 700$ because I used my toaster as a bath heater and now my legs are gone plz I need money for bandaids
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A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 50.0 MPa-m1/2 . If, during service
Studentka2010 [4]

Answer:

The answer is "0.0199044586 \ m\\\\".

Explanation:

The crucial stress essential for activating the spreading of the crack is \sigma, the hardness of the strain break is K, as well as the area long of a break is a, for dimensionless Y. Its equation of the length of its surface of the fracture is 50.1 MPa \sqrt{m} on K, 200MPa on \sigma, and 1 on Y.

a = \frac{1}{\pi} (\frac{K}{Y \ \sigma}  )^2 \\\\

  = \frac{1}{\pi} ( \frac{50.0}{1 \times 200} )^2 \\\\= \frac{1}{3.14} ( \frac{1}{4})^2 \\\\= \frac{1}{3.14} ( \frac{1}{16}) \\\\=\frac{1}{50.24}\\\\=0.0199044586 \ m\\\\

4 0
3 years ago
Some easy points points you ​
Mila [183]

Answer:

thanks -_-

Explanation:

be cuz it nice

5 0
3 years ago
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What are some homophones​
ryzh [129]

Answer:

accessary, accessory.

ad, add.

ail, ale.

air, heir.

aisle, I'll, isle.

all, awl.

allowed, aloud.

alms, arms.

5 0
3 years ago
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What is 12 cm theodolite​
Klio2033 [76]

Answer:

Answer below :)

Explanation:

<u>SIZE OF THEODOLITE:</u> A theodolite is designated by diameter of the graduated circle on the lower plate. The common sizes are 8 cm to 12 cm while<em> 14 cm</em> to <em>25 cm</em> instrument are used for triangulation work.

4 0
3 years ago
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A model of a submarine, 1:15 scale, is to be tested at 180 ft/s in a wind tunnel with standard sea-level air, while theprototype
ella [17]

Answer:

Explanation:

Given

scale i.e. L_r=1:15

Using Reynolds number similarity

(Re)_m=(Re)_p

(\frac{Vl}{\nu })_m=(\frac{Vl}{\nu })_p

Properties of air

\nu _{air}=1.57\times 10^{-4} ft/s

Properties of sea water

\nu _{sea}=1.26\times 10^{-5} ft/s

\left ( \frac{V_ml_M}{\nu _m}\right )=\left ( \frac{V_pl_p}{\nu _p}\right )

V_p=V_m\left ( \frac{l_m}{l_p}\right )\left ( \frac{\nu _p}{\nu _m}\right )

V_p=180\times \frac{1}{15}\times \frac{1.26\times 10^{-5}}{1.57\times 10^{-4}}

V_p=0.963\ ft/s

8 0
3 years ago
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