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3 years ago

Please help me bro come on

Engineering

2 answers:

melomori [17]3 years ago

6 0

Answer:

12

Explanation:

5 • 3 = 15
3 • 3 = 9
4 • 3 = 12

ankoles [38]3 years ago

5 0

The answer is 12! hope you got it right

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An amplifier with 40 dB of small-signal, open-circuit voltage gain, an input resistance of 1 MO, and an output resistance of 100

Vanyuwa [196]

convert 40db to standard gain

AL=10^40/20=100

calculate total voltage gain

=AL×RL/RL+Ri

=83.33

38.41 DB

calculate power

Pi=Vi^2/Ri Po=Vo^2/RL

power gain= Po/Pi

=13.90×10^6

3 0

3 years ago

Using the Rayleigh criterion, calculate the minimum feature size that can be resolved in a system with a 0.18 NA lens when g-lin

Vladimir79 [104]

Answer:

# for a g line, R = 1.211 μm

# for an I-line, R = 1.013 μm

# for a g line, R = 0.726 μm

# for an I-line, R = 0.243 μm

# for a g line, R = 0.605 μm

# for an I-line, R = 0.608 μm

Explanation:

We know that;

Rayleigh Resolution R = 0.5 × λ/NA

for a g line, λ = 436 nm

for an I-line λ = 365 nm

Now when NA = 0.18

# for a g line, λ = 436 nm

R = 0.5 × 436/0.18 = 1.211 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.18 = 1.013 μm

when NA = 0.30

# for a g line, λ = 436 nm

R = 0.5 × 436/0.30 = 0.726 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 = 0.243 μm

when NA = 0.36

# for a g line, λ = 436 nm

R = 0.5 × 436/0.36 = 0.605 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 = 0.608 μm

6 0

3 years ago

A house is losing heat at a rate of 1700 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. Exp

Furkat [3]

Answer:

1700kJ/h.K

944.4kJ/h.R

944.4kJ/h.°F

Explanation:

Conversions for different temperature units are below:

1K = 1°C + 273K

1R = T(K) * 1.8

= (1°C + 273) * 1.8

1°F = (1°C * 1.8) + 32

Q/delta T = 1700kJ/h.°C

T (K) = 1700kJ/h.°C

= 1700kJ/K

T (R) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8R

= 944.4kJ/h.R

T (°F) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8°F

= 944.4kJ/h.°F

Note that arithmetic operations like subtraction and addition of values do not change or affect the value of a change in temperature (delta T) hence, the arithmetic operations are not reflected in the conversion. Illustration: 5°C - 3°C

= 2°C

(273+5) - (273+3)

= 2 K

5 0

3 years ago

**Please Help. ASAP**

natima [27]

Answer:

The answer is below

Explanation:

$\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at$

$\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)$

$x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}$

$x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}$

$x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}$

$E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\ \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}$

$ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\$

4 0

3 years ago

At steady state, air at 200 kPa, 325 K, and mass flow rate

Vera_Pavlovna [14]

Letra A

A letra

A.
Thank

3 0

3 years ago