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Montano1993 [528]

4 years ago

3

Two nonlinear stages are cascaded. If the input/output characteristic of each stage is approximated by a third-order polynomial,

determine the P1dB of the cascade in terms of the P1dB of each stage.

1 answer:

Basile [38]4 years ago

3 0

Answer:

See explanations for step by step pricedure to answer.

Explanation:

Given that;

Two nonlinear stages are cascaded. If the input/output characteristic of each stage is approximated by a third-order polynomial, determine the P1dB of the cascade in terms of the P1dB of each stage.

There:

Pn = K * T * G * B * F

where:

Pn: output noise power (in W)

K: Boltzmann's costant

T: Absolute temperature(in K)

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3 years ago

A small submarine has a triangular stabilizing fin on its stern. The fin is 1 ft tall and 2 ft long. The water temperature where

Arturiano [62]

Answer:

$\mathbf{F_D \approx 1.071 \ lbf}$

Explanation:

Given that:

The height of a triangular stabilizing fin on its stern is 1 ft tall

and it length is 2 ft long.

Temperature = 60 °F

The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.

From these information given; we can have a diagrammatic representation describing how the triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.

The diagram can be found in the attached file below.

If we recall ,we know that;

Kinematic viscosity v = $1.2075 \times 10^{-5} \ ft^2/s$

the density of water ρ = 62.36 lb /ft³

$Re_{max} = \dfrac{Ux}{v}$

$Re_{max} = \dfrac{2.5 \ ft/s \times 2 \ ft }{1.2075 \times 10 ^{-5} \ ft^2/s}$

$Re_{max} = 414078.6749$

$Re_{max} = 4.14 \times 10^5$ which is less than < 5.0 × 10⁵

Now; For laminar flow; the drag on the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:

$dF_D = (\dfrac{0.664 \times \rho \times U^2 (2-x) dy}{\sqrt{Re_x}})^2$

where;

$(2-x) dy$ = strip area

$Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}$

Therefore;

$dF_D = (\dfrac{0.664 \times 62.36 \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})$

$dF_D = 1.136 \times(2-x)^{1/2} \ dy$

Let note that y = 0.5x from what we have in the diagram,

so , x = y/0.5

By applying the rule of integration on both sides, we have:

$\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy$

$\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-2y)^{1/2} \ dy$

Let U = (2-2y)

-2dy = du

dy = -du/2

$F_D = \int\limits^0_2 \ 1.136 \times(U)^{1/2} \ \dfrac{du}{-2}$

$F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du$

$F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 } }{\frac{1}{2}+1}]^0__2$

$F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} } ] ^0__2$

$F_D = -0.568 [0 - \dfrac{2}{3}(2)^{\frac{3}{2} } ]$

$F_D = -0.568 [- \dfrac{2}{3} (2.828427125)} ]$

$F_D = 1.071031071 \ lbf$

$\mathbf{F_D \approx 1.071 \ lbf}$

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3 years ago

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lara [203]

Answer:

Their air temperature will be within the same range of temperature.

Explanation:

Climatic condition varies with position on the globe. Climatic conditions is also affected by some natural occurrence like proximity to the ocean, topography of the region, elevation etc.

If both region are located on the same latitudinal distance from the equator, and both are close to the ocean, then there is a big chance of both regions having the same climatic condition and hence the same range of temperature.

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3 years ago

You do not need to remove the lead weights inside tires before recycling them.

almond37 [142]

There are NO “lead weights inside tires”!

Balancing of the entire wheel (rim plus tire ) is done by correctly fitting the tire to the rim to begin with (new tires have printed coloured marks identifying the heavy and light bits), and then by adding balancing weights to the metal rim, usually made of lead but not always.

If you’re recycling rims then yes, remove the weights and recycle those separately.

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3 years ago