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3 years ago

Which kinds of objects emit visible light in the electromagnetic spectrum?

Physics

2 answers:

SVETLANKA909090 [29]3 years ago

5 0

Answer: The correct answer is relatively hot objects.

Explanation:

Electromagnetic spectrum is a continuum of all the electromagnetic waves. They are arranged according to frequency and wavelength.

All objects emit wave according to their temperature.

The relatively hot objects emit the visible light or ultraviolet radiations having higher frequencies while the relatively cold objects emit wave of lower frequencies such as radio wave. The visible lights in the electromagnetic spectrum is seen by naked eyes.

The wavelengths of the visible light lies between 400 nm to 700 nm.

Therefore, the correct answer is relatively hot objects.

Triss [41]3 years ago

3 0

Answer : Relatively hot objects

Explanation : We know that, the temperature of the objects is inversely proportional to their wavelengths. The objects emitting radiation in the visible region have short wavelength and hence are relatively hotter.

We know the range of wavelength of the visible spectrum is from 400 nm to 780 nm.

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2 years ago

Initilal velocity of projectile at an angle of 22.5° is 10m/s. Then what is the magnitude of range of projectile?

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3 years ago

Determine

djyliett [7]

(i) The total capacitance for the circuit is 5 μF.

(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.

(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.

<h3>C1 and C2 are in series </h3>

$\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F$

<h3>C1 and C2 are parallel to C3</h3>

$C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F$

<h3>C(123) is series to C5 and C6</h3>

$\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F$

<h3>C7 and C8 are in series</h3>

$\frac{1}{C_{78}} = \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{78}} = \frac{2}{6} \\\\\frac{1}{C_{78}} =\frac{1}{3} \\\\C_{78} = 3 \ \mu F$

<h3>Total capaciatnce of the circuit</h3>

Ct + C(78) = 2 μF + 3 μF = 5 μF

<h3 /><h3>Total charge stored in the circuit</h3>

The total charge stored in the capacitor is calculated as follows;

Q = CV

Q = (5 x 10⁻⁶) x (20)

Q = 1 x 10⁻⁴ C

<h3>Charge stored in 3μF capacitor</h3>

Q = (3 x 10⁻⁶) x (20)

Q = 6 x 10⁻⁶ C

Learn more about capacitance of capacitor here: brainly.com/question/13578522

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