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3 years ago

Please answer ASAP!!

Engineering

2 answers:

Licemer1 [7]3 years ago

8 0

Answer:

Explanation:

that is what energy efficiency is

Dafna11 [192]3 years ago

4 0

The correct answer is a

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Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distanc

Andrei [34K]

Answer:

the answer is below

Explanation:

a) The conductivity of graphite (σ) is calculated using the formula:

$\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}$

where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012

Substituting:

$\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m$

b) f = 1 GHz = 10⁹ Hz.

$\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10} e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10} e=-1.5 \\\\z=\frac{-1.5}{log_{10} e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm$

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3 years ago

How to calculate maximum shear stress in the cross-section of a beam if the neutral axis is not on the thinnest part of the cros

likoan [24]

—By the maximum shear stress is then calculated by: where b = 2 (ro − ri) is the effective width of the cross section, Ic = π (ro4 − ri4) / 4 is the centroidal moment of inertia, and A = π (ro2 − ri2) is the area of the cross section.

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3 years ago

A rotator has a weight of 100lb with a centroidal radius of gyration of 9 in. Determine the moment of inertia about the center o

Ronch [10]

Answer:

8100 lbin²

Explanation:

Moment or inertia is expressed using the formula

I = mr²

M is the mass of the body

r is the radius of gyration

Given

W = 100lb

r = 9in

Required

Moment of inertia

I = Wr²

I = 100(9)²

I = 100×81

I = 8100lbin²

Hence the moment of inertia about the center of gravity for the rotator is

8100 lbin²

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3 years ago

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$