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3 years ago

Colors containing only one wavelength are called

Physics

2 answers:

Jobisdone [24]3 years ago

8 0

I think the answer you are looking for is "Pure Colors"

goblinko [34]3 years ago

6 0

The answer is: <span>Unsaturated </span>colors<span> such as pink, or purple variations such as magenta, are absent, for example, because they can be made </span>only<span> by a mix of multiple </span>wavelengths<span>. </span>Colors containing only one wavelength<span> are also </span>called<span> pure </span>colors<span> or spectral </span>colors<span>.</span>

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A student is preparing to bake a cake at home and places several ingredients on the countertop to use. First he preheats the ove

Evgen [1.6K]

Answer:

baking the cake batter

Explanation:

Baking the cake batter will indicate that chemical change has occurred here. What is a chemical change?

A chemical change is one in which a new kind of matter is formed.
It is usually accompanied by energy either evolution or absorption of energy in form of heat or light or both.
The process is irreversible.
When the batter bakes, a new substance different from the cake mix is obtainable.
We cannot get back the ingredient from this baked cake. It is impossible.
This is good indicator of chemical change.

7 0

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

8 0

2 years ago

State how a ray of light can travel from air into glass without bending?

Katyanochek1 [597]

Answer:

If you shine a beam of light (a bundle of parallel rays) through the air, it will travel in a straight line. Rays of light usually travel in straight lines until they hit something. If a ray of light hits the surface of a sheet of glass, some light will be reflected by the surface of the glass.

Explanation:

STREAM LALISA AND MONEY

6 0

3 years ago

A 2-kW electric heater takes 15 min to boil a quantity of water. If this is done once a day and power costs 10 cents per kWh, wh

frozen [14]

Answer:

Explanation:

Given data

power P= 2 kW

time t= 15 min to hours = 15/60= 1/4 h

cost of power consumption per kWh= 10 cent = $0.1

We are expected to compute the cost of operating the heater for 30 days

but let us computer the energy consumption for one day

Energy of heater for one day= 2* 1/4 = 0.5 kWh

the cost of operating the heater for 30 days= 0.5*0.1*30= $1.50

<u><em>Hence it will cost $1.50 for 30 days operation</em></u>

4 0

3 years ago

An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2

Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

$M=-\frac{q}{p}$

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

$M=-\frac{200 cm}{50 cm}=-4.0$

7 0

3 years ago