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3 years ago

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During an office party, an office worker claims that a can of cold beer on his table warmed up to 20oC by picking up energy from

the surrounding air, which is 25oC. Is there any truth to his claim? Explain.

1 answer:

Contact [7]3 years ago

4 0

Answer:

Yes

Explanation:

As we know that heat transfer take place from high temperature body to low temperature body.

In the given problem ,the temperature of the air is high as compare to the temperature of can of bear ,so the heat transfer will take place from air to can of bear and at the last stage when temperature of can of bear will become to the temperature of air then heat transfer will be stop.Because temperature of the both body will become at the same and this stage is called thermal equilibrium.

So an office worker claim is correct.

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A belt drive was designed to transmit the power of P=7.5 kW with the velocity v=10m/s. The tensile load of the tight side is twi

Leviafan [203]

Answer:

F₁ = 1500 N

F₂ = 750 N

$F_{e}$ = 500 N

Explanation:

Given :

Power transmission, P = 7.5 kW

= 7.5 x 1000 W

= 7500 W

Belt velocity, V = 10 m/s

F₁ = 2 F₂

Now we know from power transmission equation

P = ( F₁ - F₂ ) x V

7500 = ( F₁ - F₂ ) x 10

750 = F₁ - F₂

750 = 2 F₂ - F₂ ( ∵F₁ = 2 F₂ )

∴F₂ = 750 N

Now F₁ = 2 F₂

F₁ = 2 x F₂

F₁ = 2 x 750

F₁ = 1500 N , this is the maximum force.

Therefore we know,

$F_{max}$ = 3 x $F_{e}$

where $F_{e}$ is centrifugal force

$F_{e}$ = $F_{max}$ / 3

= 1500 / 3

= 500 N

8 0

3 years ago

Gray cast iron, with an ultimate tensile strength of 31 ksi and an ultimate compressive strength of 109 ksi, has the following s

suter [353]

Using an appropriate failure theory, find the factor of safety in each case. State the name of the theory that you are using the theory is max stress theory.

<h3>Wat is the max stress theory?</h3>

The most shear strain concept states that the failure or yielding of a ductile fabric will arise whilst the most shear strain of the fabric equals or exceeds the shear strain fee at yield factor withinside the uniaxial tensile test.”

Stress states at various critical locations are f= 2.662.

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1 year ago

Tensile Strength (MPa) Number-Average Molecular Weight (g/mol)

IceJOKER [234]

Answer:

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

Explanation:

The question can be well structured in a table format as illustrated below:

Tensile Strength (MPa) Number- Average Molecular Weight (g/mol)

82 12,700

156 28,500

The tensile strength and number-average molecular weight for two polyethylene materials given above.

Estimate the number-average molecular weight that is required to give a tensile strength required above. Using the data given find TS (infinity) in MPa.

<u>SOLUTION:</u>

We know that :

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

where;

$T_S$ = Tensile Strength

$T_{S \infty}$ = Tensile Strength (Infinity)

$M_n$ = Number- Average Molecular Weight (g/mol)

SO;

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

From equation (1) ; collecting the like terms; we have :

$T_{S \infty} =82+ \dfrac{A}{12700}$

From equation (2) ; we have:

$T_{S \infty} =156+ \dfrac{A}{28500}$

So; $T_{S \infty} = T_{S \infty}$

Then;

$T_{S \infty} =82+ \dfrac{A}{12700} =156+ \dfrac{A}{28500}$

Solving by L.C.M

$\dfrac{82(12700) + A}{12700} =\dfrac{156(28500) + A}{28500}$

$\dfrac{1041400 + A}{12700} =\dfrac{4446000 + A}{28500}$

By cross multiplying ; we have:

$({4446000 + A})* {12700} ={28500} *({1041400 + A})$

$(5.64642*10^{10} + 12700A) =(2.96799*10^{10}+ 28500A)$

Collecting like terms ; we have

$(5.64642*10^{10} - 2.96799*10^{10} ) =( 28500A- 12700A)$

$2.67843*10^{10} = 15800 \ A$

Dividing both sides by 15800:

$\dfrac{ 2.67843*10^{10} }{15800} =\dfrac{15800 \ A}{15800}$

A = 1695208.861

From equation (1);

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

Replacing A = 1695208.861 in the above equation; we have:

$82= T_{S \infty} - \dfrac{1695208.861}{12700}$

$T_{S \infty}= 82 + \dfrac{1695208.861}{12700}$

$T_{S \infty}= \dfrac{82(12700) +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{1041400 +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{2736608.861 }{12700}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

From equation(2);

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

Replacing A = 1695208.861 in the above equation; we have:

$156= T_{S \infty} - \dfrac{1695208.861}{28500}$

$T_{S \infty}= 156 + \dfrac{1695208.861}{28500}$

$T_{S \infty}= \dfrac{156(28500) +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{4446000 +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{6141208.861}{28500}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

We are to also estimate the number- average molecular weight that is required to give a tensile strength required above.

If the Tensile Strength (MPa) is 82 MPa

Definitely the average molecular weight will be = 12,700 g/mol

If the Tensile Strength (MPa) is 156 MPa

Definitely the average molecular weight will be = 28,500 g/mol

But;

Let us assume that the Tensile Strength (MPa) = 181 MPa for example.

Using the same formula:

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

Then:

$181 = 215.481- \dfrac{1695208.861 }{M_n}$

Collecting like terms ; we have:

$\dfrac{1695208.861 }{M_n} = 215.481- 181$

$\dfrac{1695208.861 }{M_n} =34.481$

1695208.861= 34.481 $M_n$

Dividing both sides by 34.481; we have:

$M_n$ = $\dfrac{1695208.861}{34.481}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

5 0

3 years ago

What is the primary function of NCEES?

joja [24]

National Council of Examiners for engineering and surveying a nonprofit organization

7 0

3 years ago

Consider a continuous-time LTI system whose frequency response is I x sin(4w) H(jw) = -x h(t)e-jwtdt = w If the input to this sy

Helga [31]

Answer:

I believe this is your question.

See the hand worked solution attached.

8 0

3 years ago