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Zinaida [17]
3 years ago
6

I really need help on this!

Engineering
2 answers:
Dafna1 [17]3 years ago
8 0
C: benchmark because I have done this before
den301095 [7]3 years ago
3 0
The answer is D!
Plz mark brainliest
You might be interested in
A rope is wrapped three and a half times around a cylinder. Determine the range of force T exerted on
Mademuasel [1]

The range of force exerted at the end of the rope is; 285.7 N to 1,000 N.

<h3>What is the Net horizontal force?</h3>

The net horizontal force of the cylinder when it is at equilibrium position can be found by applying Newton's second law of motion. Thus;

∑F = 0

F - μF_n = 0

We are given;

F_n = 5 kN = 5000 N

μ = 0.2

Thus;

F - 0.2(5,000) = 0

F - 1,000 = 0

F = 1,000 N

The strength of the applied force will be increasing as the number of turns of the rope increases. Thus;

Minimum force = Total force/number of turns of rope

Since rope is wrapped three and half times, then;

number of turns = 3.5

Thus;

minimum force = 1,000/3.5

minimum force = 285.7 N

Thus, the range of force exerted at the end of the rope is 285.7 N to 1,000 N.

Read more about net horizontal force at; brainly.com/question/26957287

7 0
2 years ago
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3.5 × 10
babunello [35]

Answer:

The magnitude of the maximum stress that exists at the tip of the internal crack is 3900.183MPa (565703.38 psi).

Explanation:

The magnitude of the maximum stress that exists at the tip of an internal crack can be estimated using fracture mechanics. If a material that have internal crack is loaded with tensile stress, the crack tends to widen due to the tensile force created by the stress, and this leads to a concentration of stress near the tip of the crack, thereby expanding the crack length and finally leading to failure of the material.

Fracture mechanics involves the study of propagation of cracks in a material. It is used to predict the conditions at which failure of a material will likely occur.

A crack may appear in the surface of the material (surface crack) or at the interior of the material (internal crack), but the same formula can be used to estimate the magnitude of the maximum stress at the tip of the crack.

The formula is given as Πm=2Πo√(a/r)

Where,

Πm is the maximum stress

Πo is the tensile stress

r is the radius of curvature

a is the crack length.

The major difference between the surface crack and the internal crack is the value for their crack length.

Crack length can be understood as the length of a crack at which the crack becomes unstable under certain applied stress.

For surface crack, crack length=a

For internal crack, crack length=2a because internal cracks forms two surfaces of a complete sphere, while the surface crack forms one surface of a sphere.

Given in the question,

Radius of curvature, r=3.5×10^(-4) mm=0.00035mm

Crack length, a=5.5×10^(-2) mm

For internal crack, a=[(crack length)/2]={[5.5×10^(-2)]/2}=0.0275mm

Tensile stress, Πo=220Mpa

Maximum stress=Πm

Using,

Πm=2Πo√(a/r)

=(2×220)√(0.0275/0.00035)

=440√78.571

=3900.183MPa (565703.38 psi)

4 0
3 years ago
Refrigerant 134a is the working fluid in a vaporcompression heat pump that provides 35 kW to heat a dwelling on a day when the o
Dennis_Churaev [7]

Answer:

Hello, dear.

For the answer please see the explanation below.

Explanation:

The compressor power is:

2.212kW

(b) The refrigeration capacity is:

3.62 tons

(c) The coefficient of performance is:

5.75

5 0
4 years ago
(10 points) A single crystal in the titanium cable is oriented so that the [001] direction is parallel to the applied load. If t
Ulleksa [173]

Answer:yep

Explanation:

4 0
3 years ago
A surveyor knows an elevation at Catch Basin to be elev=2156.77 ft. The surveyor takes a BS=2.67 ft on a rod at BM Catch Basin a
fenix001 [56]

Answer:

the elevation at point X is 2152.72 ft

Explanation:

given data

elev = 2156.77 ft

BS = 2.67 ft

FS = 6.72 ft

solution

first we get here height of instrument that is

H.I = elev + BS   ..............1

put here value

H.I =  2156.77 ft + 2.67 ft  

H.I = 2159.44 ft

and

Elevation at point (x) will be

point (x)  = H.I - FS   .............2

point (x)  = 2159.44 ft  - 6.72 ft

point (x)  = 2152.72 ft

3 0
4 years ago
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