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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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The dry unit weight of a soil sample is 14.8 kN/m3.
Jlenok [28]

Answer:

See attachment for completed question

Explanation:

Given that; Brainly.com

What is your question?

mkasblog

College Engineering 5+3 pts

The dry unit weight of a soil sample is 14.8 kN/m3.

Given that G_s = 2.72 and w = 17%, determine:

(a) Void ratio

(b) Moist unit weight

(c) Degree of saturation

(d) Unit weight when the sample is fully saturated

See complete solving at attachment

4 0
3 years ago
HELP!
olya-2409 [2.1K]
The thickness is thick
5 0
3 years ago
The device whose operation closely matches the way the clamp-on ammeter works is
Ivanshal [37]

Answer:

The answer is

C. Split phase motor

Explanation:

Clamp meters rely on the principle of magnetic induction to make non contact AC current measurements. Electric current flowing through a wire produces a magnetic field.

Which is similar to basic mode of operation of electric motor and split phase motor is a type of electric motor.

What is a a clamp on meter?

Clamp meters are electrical testers which have wide jaws that are able to clamp around an electrical conductor. Originally designed as a single purpose tool for measuring AC current, clamp meters now include inputs for accepting test leads and other probes that support a wide range of electrical measurements, the jaws of a clamp meter permit work in tight spaces and permits current measurements on live conductors without circuit interruption.

6 0
3 years ago
In highways the far left lane is usually the _____
Ivan
Fastest


(Known as the fast lane)
8 0
3 years ago
Read 2 more answers
A total of 245 kip force is applied to a set of 10 similar bolts. If the spring constant of each bolt is 0.4 Mlb/in and that of
zubka84 [21]

Answer: The net force in every bolt is 44.9 kip

Explanation:

Given that;

External load applied = 245 kip

number of bolts n = 10

External Load shared by each bolt (P_E) = 245/10 = 24.5 kip

spring constant of the bolt Kb = 0.4 Mlb/in

spring constant of members Kc = 1.6 Mlb/in

combined stiffness factor C = Kb / (kb+kc) = 0.4 / ( 0.4 + 1.6)  = 0.4 / 2 = 0.2 Mlb/in

Initial pre load Pi = 40 kip

now for Bolts; both pre load Pi and external load P_E are tensile in nature, therefore we add both of them

External Load on each bolt P_Eb = C × PE = 0.2 × 24.5 = 4.9 kip

So Total net Force on each bolt Fb = P_Eb + Pi

Fb = 4.9 kip + 40 kip

Fb = 44.9 kip

Therefore the net force in every bolt is 44.9 kip

4 0
3 years ago
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