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3 years ago

What is the sig fig of 200

1 answer:

8 0

Assuming you don't have a decimal after the zeros, there is only one sig fig. Zeros never count unless they're between other numbers or hold precision like 2.0, because it shows how accurate the measurement was

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As the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor:___________.

sergij07 [2.7K]

Answer:

B) approaches infinity

Explanation:

The capacitive reactance of an AC capacitor is given by;

$X_C = \frac{1}{\omega C } = \frac{1}{2\pi f C}$

Where;

C is the capacitance

f is the frequency of the ac voltage

$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi (0) C} \\\\X_C = \frac{1}{0} \\\\X_C = \ infinity$

Therefore, as the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor approaches infinity.

The correct option is (B) approaches infinity

7 0

3 years ago

tekilochka [14]

The answer is 25.37 hope this helps

5 0

3 years ago

The object has a redshift of 7.6 and the JWST observes the object at a wavelength of 2 microme-

olasank [31]

b. The wavelength of light emitted by the object is 233 nm

c. The type of radiation originally emitted by the object is ultraviolet radiation.

To find the wavelength, we need to know what redshift is.

<h3>What is redshift?</h3>

Redshift is the increase in wavelength and the corresponding decrease of frequency and photon energy of electromagnetic radiation.

Redshift is given by z = λ'/λ - 1 where

λ' = observed wavelength and
λ = emitted wavelength.

Making λ subject of the formula, we have

λ = λ'/(1 + z)

Given that has a redshift of 7.6 and the JWST observes the object at a wavelength of 2 micrometres (mid-infrared light).

So,

z = 7.6 and
λ' = 2μm

<h3>(b) What is the wavelength of the light emitted by the object?</h3>

Substituting the values of the variables into the equation, we have

λ = λ'/(1 + z)

λ = 2μm/(1 + 7.6)

λ = 2μm/8.6

λ = 0.233 μm

λ = 233 nm

So, the wavelength of light emitted by the object is 233 nm

<h3>c. What type of radiation was originally emitted by the object?</h3>

Since the wavelength is 233 nm and the wavelength is in the range of ultraviolet radiation 200 nm - 315 nm.

So, the type of radiation originally emitted by the object is ultraviolet radiation.

Learn more about redshift here:

brainly.com/question/27915180

#SPJ1

8 0

3 years ago

Explain why the "X" chromosome expresses more traits than the "y" chromosome.

wel

Explanation:

The X chromosome (females have 2 of them while males have 1) is five times larger than the Y chromosome and has 10 times the number of genes.

5 0

3 years ago

In an open system, such as a campfire, matter can?

-Dominant- [34]

I am positive with my and answer and i think yes

7 0

3 years ago