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3 years ago

11. Technicians A and B are discussing

Engineering

1 answer:

algol [13]3 years ago

5 0

Answer:

C. Neither Technician A nor B

Explanation:

Just took the test

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Why water parameters of Buriganga river vary between wet and dry seasons? Explain.

Sedaia [141]

I honestly don’t know because I honestly don’t know what I mean

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3 years ago

A certain solar energy collector produces a maximum temperature of 100°C. The energy is used in a cyclic heat engine that operat

Gwar [14]

Answer:

$\eta _{max} = 0.2413 = 24.13%$

$\eta' _{max} = 0.5061 = 50.61%$

Given:

$T_{1max} = 100^{\circ} = 273 + 100 = 373 K$

operating temperature of heat engine, $T_{2} = 10^{\circ} = 273 + 10 = 283 K$

$T_{3max} = 300^{\circ} = 273 + 300 = 573 K$

Solution:

For a reversible cycle, maximum efficiency, $\eta _{max}$ is given by:

$\eta _{max} = 1 - \frac{T_{2}}{T_{1max}}$

$\eta _{max} = 1 - \frac{283}{373} = 0.24$

$\eta _{max} = 0.2413 = 24.13%$

Now, on re designing collector, maximum temperature, $T_{3max}$ changes to $300^{\circ}$ , so, the new maximum efficiency, $\eta' _{max}$ is given by:

$\eta' _{max} = 1 - \frac{T_{2}}{T_{3max}}$

$\eta _{max} = 1 - \frac{283}{573} = 0.5061$

$\eta _{max} = 0.5061 = 50.61%$

4 0

4 years ago

What does product integration refer to in an advanced manufacturing setting?

creativ13 [48]

Answer:

C!!

Explanation:

Combining all manufacturing processes to provide higher efficiency and fulfilling the requeriments.

6 0

4 years ago

Read 2 more answers

A particle travels along a straight line with a velocity v = (12 – 3t2) m/s. When t = 1 s, the particle is located 10 m to the l

arlik [135]

Answer:

The displacement from t = 0 to t = 10 s, is -880 m

Distance is 912 m

Explanation:

$v = (12 - 3t^2) m/s = ds/dt$ . . . . . . . . . . A

integrate above equation we get

$s = 12t - t^3 + C$

from information given in the question we have

t = 1 s, s = -10 m

so distance s will be

-10 = 12 - 1 + C,

C = -21

$s(t) = 12t - t^3 - 21$

we know that acceleration is given as

$a(t) = dv/dt = -6t$

[FROM EQUATION A]

Acceleration at t = 4 s, a(4) = -24 m/s^2

for the displacement from t = 0 to t = 10 s,

$s(10) - s(0) = (12*10 - 10^3 - 21) - (-21) = -880 m$

the distance the particle travels during this time period:

let v = 0,

$3t^2 = 12$

t = 2 s

Distance $= [s(2) - s(0)] + [s(2) - s(10)] = [1\times 2 - 2^3] + [(12\times 2 - 2^3) - (12\times 10 - 10^3)] = 912 m$

7 0

4 years ago

Nitrogen at an initial state of 300 K, 150 kPa and 0.2 m3 is compressed slowly in an isothermal process to a final pressure of 8

s344n2d4d5 [400]

Answer:

Work = - 4175.23 J

Heat Transfer = -4175.23 J

Explanation:

The work done in an isothermal process, is given by the following formula:

W = RT ln (P1/P2)

where,

W = Work done

R = Universal Gas Constant = 8.314 J/mol.k

T = Constant Temperature = 300 K

P1 = initial pressure = 150 KPa

P2 = final pressure = 800 KPa

using these values, we get:

W = (8.314 J/mol.K)(300 k) ln (150/800)

W = - 4175.23 J

Here, negative sign shows that work is done on the system.

In isothermal process, from 1st law of thermodynamics:

Heat Transfer = Q = W (Since change in internal energy is zero for isothermal processes)

W = - 4175.23 J

Here, negative sign shows that heat is transferred from the system to surrounding.

6 0

3 years ago

Read 2 more answers