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dalvyx [7]
3 years ago
6

The input to a harmonic generator is a periodic waveform v(t) = Ce^-alpha t, 0 < t < 1, T = 1 with alpha < 0. The outpu

t is to be derived through a filter which will pass only one frequency. If alpha can be adjusted, what value of alpha will give the maximum output through the filter for the second harmonic the third harmonic? the fifth harmonic?

Engineering
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer:

Please find attached

Explanation:

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a metal coin has certain properties that can be measured.which property of a coin is different on the moon that is on earth?
Sloan [31]

Answer:

Coins weigh less on the Moon.

Explanation:

Gravity is only 1/6th as strong on the Moon than it is on Earth. Where a nickle is about 5 grams on Earth, it is less than 1 gram on the Moon. Gravity is affected by the size of the planet or moon. The Moon is much less massive than the Earth.

8 0
3 years ago
The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centr
Sedaia [141]

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

(\sigma_ x)  = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}

Since d = 2r

Then:

(\sigma_ x)  = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}

(\sigma_ x)  = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}

(\sigma_ x)  = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}

(\sigma_ x)  = 11917000.38

(\sigma_ x)  = 11.917 \times 10^6 \ Pa

(\sigma_ x)  = 11.917 \ MPa

\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\  \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

\tau _{xy} =0

3 0
3 years ago
Does a general repair <br> manual have the answers to all repairs?
uranmaximum [27]

Explanation:

yes it has the answers to all repairs

4 0
3 years ago
The mean of hours that the average person watches television each day is 4.18 hours with a standard deviation of 1.19 hours. Fin
luda_lava [24]

Answer:

z = \frac{3-4.18}{1.19}=-0.992

z = \frac{5-4.18}{1.19}=0.689

And we can find this probability with this difference:

P(-0.992

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

Explanation:

For this case we can define the random variable X as "hours that a person watches television". For this case we don't have the distribution for X but we have the following parameters:

\mu = 4.18,\sigma =1.19

We can assume that the distribution for X is normal

X \sim N(\mu = 4.18 , \sigma =1.19)

And we want to find this probability:

P(3

And we can use the z score formula given by:

z=\frac[X- \mu}{\sigma}

And we can find the z score for each limit and we got:

z = \frac{3-4.18}{1.19}=-0.992

z = \frac{5-4.18}{1.19}=0.689

And we can find this probability with this difference:

P(-0.992

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

6 0
3 years ago
Application 1: Consider the following beam under combined loading of a distributed pressure and applied bending moment. Complete
Ivan

Answer:

Explanation:

kindly check he attached file below

7 0
4 years ago
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