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3 years ago

Which part of the diagram indicates the amplitude of the wave?

Physics

2 answers:

vovikov84 [41]3 years ago

8 0

Answer : The rest position to the crest indicates the amplitude of the wave.

Explanation :

Amplitude of the wave : It is a wave that covers maximum amount of displacement on the medium from its rest position.

That means, the amplitude can be measured from rest position to the crest and also measured from rest position to trough.

Wavelength of the wave : It is the length of one complete wave cycle.

That means, the wavelength can be measured from crest to crest and trough to trough.

Hence, the rest position to the crest indicates the amplitude of the wave.

marysya [2.9K]3 years ago

7 0

Answer: The highest point upwards and lowest point downward

Explanation:

Amplitude is the point of maximum displacement.

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The correct answer is B.
In a velocity vs time graph, a line going up means an increase in velocity, or speed, which correlates to positive acceleration. Negative acceleration means slowing down.

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2 years ago

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A very large sheet of insulating material has had an excess of electrons placed on it to a surface charge density of –3.00nC/m2

lys-0071 [83]

Answer: sheet of charge

Explanation:

a )

Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.

b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge. The equi- potential surface is always perpendicular to electric field.

C ) Electric field which is almost uniform near the sheet of charge is equal t the following

E = σ / ε₀ where σ is charge density of surface and ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²

E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²

= .3389 x 10³

= 338.9 V / m

spacing between 1 V

= 1 / 338.9 m

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= 2.95 mm.

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2 years ago

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2 years ago

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] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I

Valentin [98]

Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

$\mathtt{=( 0.42\times 9000\times 1055.06) J}$

= 3988126.8 J

= 3.99 MJ

Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

i.e.

The mass of the coal that is burned per sec $=\dfrac{750}{3.99}$

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned $= \dfrac{1.1}{100} \times 187.97 \ lb/s$

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

= 1861 lb/hr

Hence, the net emissions rate of sulfur is 1861 lb/hr

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2 years ago