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2 years ago

Solve the inequality below.Use the drop-down menus to describe the solution and its graph.

Engineering

1 answer:

Sloan [31]2 years ago

3 0

Answer:

make paragraph and solution

Explanation:

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To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at TH and Tc, would you incr

alukav5142 [94]

$\ T_{c}$ has greater effect.

Explanation:

$\eta_{\max }=1-\frac{T_{c}}{T_{A}}$

$T_{c}\\$ = Temperature of cold reservoir

$T_{H}$ = Temperature of hot reservoir

when $T_{c}$ is decreased by 't',

$\eta_{\text {incre }}$ = $1-\frac{\left(\tau_{c}-t\right)}{T_{H}}$

$=n \ + \frac{t}{T_{n}}$ $-(i)$

when ${T_{H}}$ is increased by 'T'

$\eta_{i n c}=\frac{n+\frac{t}{T_{H}}}{\left(1+\frac{k}{T_{H}}\right)}-(ii)$

$\eta_{\text {incre }} \ T_{c}>\eta_{\text {incre }} T_{\text {H }}$

7 0

2 years ago

An oscilloscope display grid or scale is called?

zaharov [31]

Answer:

An oscilloscope display grid or scale is called a graticule.

Explanation:

5 0

2 years ago

Read 2 more answers

An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at

polet [3.4K]

Answer:

q = 1.73 W

Explanation:

given data

small end = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

thermal conductivity of asbestos = 0.173 W/mK

solution

first we will get here side of cross section that is express as

$S = S1 + \frac{S2-S1}{L} x$ ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + $\frac{10-5}{15} x$

S = $\frac{0.15 + x}{3}$ m

and

now we get here Area of section at distance x is

area A = S² ...............2

area A = $(\frac{0.15 + x}{3})^2$ m²

and

now we take here small length dx and temperature difference is dt

so as per fourier law

heat conduction is express as

heat conduction q = $\frac{-k\times A\ dt}{dx}$ ...............3

put here value and we get

heat conduction q = $-k\times (\frac{0.15 + x}{3})^2 \ \frac{dt}{dx}$

it will be express as

$q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)$

now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K

$q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}$

solve it and we get

q (30) = (0.173) × (600 - 300)

q = 1.73 W

5 0

2 years ago

Print the two strings, firstString and secondString, in alphabetical order. Assume the strings are lowercase. End with newline.

CaHeK987 [17]

Hai!

Please name what kind of Script your Using then I would love to help.

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5 0

2 years ago

Describe How water is tested for color in platinum cobalt method please answer

Vikentia [17]

Answer:

The color of water is measured by comparing the water to platinum cobalt color standards representing APHA Standard Color Units. Tests 0, 20, 50, 80, 110, 140, 170, and 200 APHA color units

Explanation:

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2 years ago