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2 years ago

What is the diffrence between a small block and a big block lets see if yall know

Engineering

1 answer:

ivann1987 [24]2 years ago

5 0

Answer:

The difference in weight and size?

Explanation:

It explains itself :P

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Which of the following explains the difference between conservation and preservation?

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3 years ago

• What are some methods currently used to unclog arteries and keep them clear?

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3 years ago

Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compr

mixas84 [53]

Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

$P _1= 15 psia$

$P _1= 100 psia$

We know that work for isothermal process

$W=mRT\ln \dfrac{P_1}{P_2}$

Lets take mass is 1 kg.

So work per unit mass

$W=RT\ln \dfrac{P_1}{P_2}$

We know that for air R=0.287KJ/kg.K

$W=RT\ln \dfrac{P_1}{P_2}$

$W=0.287\times 310.92\ln \dfrac{15}{100}$

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

$\Delta S=-R\ln \dfrac{P_2}{P_1}$

$\Delta S=-0.287\ln \dfrac{100}{15}$

ΔS = -0.544 KJ/Kg.K

3 0

4 years ago

The cantilevered W530 x 150 beam shown is subjected to a 9.8-kN force F applied by means of a welded plate at A. Determine the e

snow_lady [41]

The equivalent force-couple system at O is the force and couple experienced when at point O due to the applied force at point A

The equivalent force couple system at O due to force F are;

Force, F = (8.65·i - 4.6·j) KN

Couple, M₀ ≈ 40.9 k kN·m

The reason the above values are correct is as follows:

The known values for the cantilever are;

The height of the beam = 0.65 m

The magnitude of the applied force, F = 9.8 kN

The length of the beam = 4.9 m

The angle away from the vertical the force is applied = 26°

The required parameter:

The equivalent force-couple system at the centroid of the beam cross-section of the cantilever

Solution:

The equivalent force-couple system is the force-couple system that can replace the given force at centroid of the beam cross-section at the cantilever O ;

The equivalent force $\overset \longrightarrow F$ = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j

Which gives;

The equivalent force $\overset \longrightarrow F$ ≈ (8.65·i - 4.6·j) KN

The couple acting at point O due to the force F is given as follows;

The clockwise moment = 9.8 kN × cos(28°) × 4.9

The anticlockwise moment = 9.8 kN × sin(28°) 0.65/2

The sum of the moments = Anticlockwise moment - Clockwise moments

∴ The sum of the moments, ∑M, gives the moment acting at point O as follows;

M₀ = 9.8 kN × sin(28°) 0.65/2 - 9.8 kN × cos(28°) × 4.9 ≈ 40.9 kN·m

The couple acting at O, due to F, M₀ ≈ 40.9 kN·m

The equivalent force couple system acting at point O due the force, F, is as follows

F = (8.65·i - 4.6·j) KN

M₀ ≈ 40.9 k kN·m

Learn more about equivalent force systems here:

brainly.com/question/12209585

4 0

3 years ago