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4 years ago

How do I put a picture on here for my graph equation??

Engineering

2 answers:

Pavel [41]4 years ago

8 0

Answer:

Take a screenshot of it and when you go to question you will see a paperclip looking thing

Explanation:

Ipatiy [6.2K]4 years ago

5 0

While writing your question you should have a button at the bottom to add a picture. You should also have a button to take a picture, once you hit the button you pick/take a photo and it should add it for people to look at to help answer your question

You might be interested in

Air is compressed by a 40-kW compressor from P1 to P2. The air temperature is maintained constant at 25Â°C during this process a

AlexFokin [52]

Answer:

the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Explanation:

Given the data in the question;

From the first law of thermodynamics;

dQ = dU + dW ------ let this be equation 1

where dQ is the heat transfer, dU is internal energy and dW is the work done.

from the question, the process is isothermal ( internally reversible process )

Thus, the change in internal energy is 0

dU = 0

given that; Air is compressed by a 40-kW compressor from P1 to P2

since it is compressed, dW = -40 kW

we substitute into equation 1

dQ = 0 + ( -40 kW )

dQ = -40 kW

Now, change in entropy of air is;

ΔS $_{air$ = dQ / T

given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K

so we substitute

ΔS $_{air$ = -40 kW / 298.15 K

ΔS $_{air$ = -0.13416 ≈ -0.1342 kW/K

Therefore, the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

7 0

3 years ago

A sprinter reaches his maximum speed in 2.5sec from rest with constant acceleration. He then maintains that speed and finishes t

liubo4ka [24]

The maximum Speed of the Sprinter from the velocity time graph of his motion is; 11.98 m/s

<h3>How to find the maximum speed?</h3>

We are given;

Initial Speed; u = 2.5 s

Total distance; d = 100 m

Total time; T = 9.6 s

The total distance is;

d = ¹/₂(9.6 + (9.6 - 2.5) * v

where v is maximum speed.

Thus;

¹/₂(9.6 + (9.6 - 2.5) * v = 100

16.7v = 200

v = 200/16.7

v = 11.98 m/s

Read more about Maximum Speed at; brainly.com/question/4931057

#SPJ1

3 0

2 years ago

Is a street the same as a avenue

-BARSIC- [3]

they're essentially the same thing so i'd say yes

5 0

3 years ago

Read 2 more answers

Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 35 MPa m . It has bee

GaryK [48]

Answer: Fracture will not occur since Kc (32.2 MPa√m) ∠ KIc (35 MPa√m).

Explanation:

in this question we are asked to determine if an aircraft will fracture for a given fracture toughness.

let us begin,

from the question we have that;

stress = 325 MPa

fracture toughness (KIc) = 35 MPa√m

the max internal crack length = 1.0 m

using the formula;

Y = KIc/σ√(πα) ---------------(1)

solving for Y we have;

Y = 35 (MPa√m) / 250 (MPa) √(π × 2×10⁻3/2m)

Y = 2.50

so to calculate the fracture roughness;

Kc = Y × σ√(πα) = 2.5 × 3.25√(π × 1×10⁻³/2) = 32.2 MPa√m

Kc = 32.2 MPa√m

From our results we can say that fracture will not occur since Kc (32.2 MPa√m) is less than KIc (35 MPa√m) of the material.

cheers i hope this helps!!!!

8 0

3 years ago

Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the

Law Incorporation [45]

Answer:

B A and C

Explanation:

Given:

Specimen σ $_{max}$ σ $_{min}$

A +450 -150

B +300 -300

C +500 -200

Solution:

Compute the mean stress

σ $_{m}$ = (σ $_{max}$ + σ $_{min}$ )/2

σ $_{mA}$ = (450 + (-150)) / 2

= (450 - 150) / 2

= 300/2

σ $_{mA}$ = 150 MPa

σ $_{mB}$ = (300 + (-300))/2

= (300 - 300) / 2

= 0/2

σ $_{mB}$ = 0 MPa

σ $_{mC}$ = (500 + (-200))/2

= (500 - 200) / 2

= 300/2

σ $_{mC}$ = 150 MPa

Compute stress amplitude:

σ $_{a}$ = (σ $_{max}$ - σ $_{min}$ )/2

σ $_{aA}$ = (450 - (-150)) / 2

= (450 + 150) / 2

= 600/2

σ $_{aA}$ = 300 MPa

σ $_{aB}$ = (300- (-300)) / 2

= (300 + 300) / 2

= 600/2

σ $_{aB}$ = 300 MPa

σ $_{aC}$ = (500 - (-200))/2

= (500 + 200) / 2

= 700 / 2

σ $_{aC}$ = 350 MPa

From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.

Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.

In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.

7 0

4 years ago