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SpyIntel [72]
3 years ago
5

Which types of electromagnetic waves have higher frequencies then the waves that make up ultraviolet light?

Physics
2 answers:
TiliK225 [7]3 years ago
5 0

Answer: Visible light makes up just a small part of the full electromagnetic spectrum. Electromagnetic waves with shorter wavelengths and higher frequencies include ultraviolet light, X-rays, and gamma rays.

vredina [299]3 years ago
5 0

Answer:

The answer is D

Explanation:

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The op amp in this circuit is ideal. R3 has a maximum value of 100 kΩ and σ is restricted to the range of 0.2 ≤ σ ≤ 1.0. a. Calc
Firlakuza [10]

I have attached the circuit image missing in the question.

Answer:

A) The range of vo is; -6.6V≤ vo ≤-1V

B) σ = 0.1861

Explanation:

A) First of all, Let VΔ be the voltage from the potentiometer contact to the ground.

Thus; [(0 - vg)/(2000)] +[(0 - vΔ)/(50,000)] = 0

So, [(- vg)/(2000)] +[(- vΔ)/(50,000)] = 0

Simplifying further; -25 vg - vΔ = 0

From the question, vg = 40mV = 0.04 V

So - 25(0.04) = vΔ

So: vΔ = - 1 V

Now, [vΔ/(σRΔ)] + [(vΔ - 0)/(50,000)] + [(vΔ - vo)/((1 - σ)RΔ))] = 0

So, multiplying each term by RΔ to get; [vΔ/(σ)] + [(vΔ x RΔ)/(50,000)] + [(vΔ - vo)/((1 - σ))] = 0

So RΔ = 100kΩ or 100,000Ω from the question.

So, substituting for RΔ, we get,

[vΔ/(σ)] + [2vΔ] + [(vΔ - vo)/((1 - σ))] = 0

Let's put the value of - 1 for vΔ as gotten before.

So, ( - 1/σ) - 2 + [(-1 - vo)/(1 - σ)] = 0

Now let's make vo the subject of the equation to get;

-1 - vo = (1 - σ)[2 + (1/σ)]

-1 - vo = 2 - 2σ + (1/σ) - 1

-vo = 1 + 2 - 2σ + (1/σ) - 1

-vo = 2 - 2σ + (1/σ)

vo = - 1 (2 - 2σ + (1/σ))

When σ = 0.2; vo = - 1(2 - 0.4 + 5) =

- 1 x 6.6 = - 6.6V

Also when σ = 1;

vo = - 1(2 - 2 + 1) = - 1V

Therefore, the range of vo is;

- 6.6V ≤ vo ≤ - 1V

B) it will saturate at vo = - 7V

So, from;

vo = - 1 (2 - 2σ + (1/σ))

-7 = - 1 (2 - 2σ + (1/σ))

Divide both sides by (-1)

7 = (2 - 2σ + (1/σ))

Now, subtract 2 from both sides to get; 5 = - 2σ + (1/σ)

Multiply each term by α to get;

5σ = - 2σ^(2) + 1

So 2σ^(2) + 5σ - 1 = 0

Solving simultaneously and picking the positive value , we get σ to be approximately 0.1861

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3 years ago
2Mg+2HCI TO 2MgCI+2H2
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Balanced is:

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The best and most correct answer among the choices provided by your question is the first choice or letter A.

<span>The minimum amount of energy that has to be added to start a reaction is the activation energy.</span>

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