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3 years ago

What colors form the spectrum of visible light

Physics

1 answer:

Law Incorporation [45]3 years ago

3 0

Red, orange, yellow, green, blue, indigo, violet :)

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How many atoms are present in 3 molecules of chromium

iren2701 [21]

Answer:

1,8066 x 1024 atoms

Explanation:

1 mole of chromium contains 6,022 x 1023 atoms

so

3 moles of chromium contains 3 times as many atoms

3 x 6,022 x 1023 = 1,8 066 x 1024

5 0

3 years ago

What unit can be used to measure the mass of a grain of salt

Anna007 [38]

milligrams

Explanation:

The best unit to measure the mass of a grain of salt is the milligram.

This is about; a thousandth of a gram.

A grain of salt is a very small particle size that can still be visible with the eye.

It has a very low and small mass.

For substances like this, we use the milligram:

1000milligram = 1g

learn more:

Mass brainly.com/question/4736384

#learnwithBrainly

5 0

2 years ago

In the most common isotope of Hydrogen the nucleus is made out of a single proton. When this Hydrogen atom is neutral, a single

FinnZ [79.3K]

Answer:

The ratio of electric force to the gravitational force is $2.27\times 10^{39}$

Explanation:

It is given that,

Distance between electron and proton, $r=4.53\ A=4.53\times 10^{-10}\ m$

Electric force is given by :

$F_e=k\dfrac{q_1q_2}{r^2}$

Gravitational force is given by :

$F_g=G\dfrac{m_1m_2}{r^2}$

Where

$m_1$ is mass of electron, $m_1=9.1\times 10^{-31}\ kg$

$m_2$ is mass of proton, $m_2=1.67\times 10^{-27}\ kg$

$q_1$ is charge on electron, $q_1=-1.6\times 10^{-19}\ kg$

$q_2$ is charge on proton, $q_2=1.6\times 10^{-19}\ kg$

$\dfrac{F_e}{F_g}=\dfrac{kq_1q_2}{Gm_1m_2}$

$\dfrac{F_e}{F_g}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}$

$\dfrac{F_e}{F_g}=2.27\times 10^{39}$

So, the ratio of electric force to the gravitational force is $2.27\times 10^{39}$ . Hence, this is the required solution.

3 0

3 years ago

How can you keep sound out of a room

GREYUIT [131]

To keep<span> noise from entering your space, look for </span>sound<span> blockers</span>

8 0

3 years ago

Ch 31 HW Problem 31.63 10 of 15 Constants In an L-R-C series circuit, the source has a voltage amplitude of 116 V , R = 77.0 Ω ,

Degger [83]

Answer:

a. I = 0.76 A

b. Z = 150.74

c. RL₁ = 34.41 , RL₂ = 602.58

d. RL₂ = 602.58

Explanation:

V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V , Rc = 473 Ω

Using law of Ohm

V = I * R

I = Vc / Rc = 364 V / 473 Ω

I = 0.76 A

The impedance of the circuit in this case the resistance, capacitance and inductor

V = I * Z

Z = V / I

Z = 116 v / 0.76 A

Z = 150.74

The reactance of the inductor can be find using

Z² = R² + (RL² - Rc²)

Solve to RL'

RL = Rc (+ / -) √ ( Z² - R²)

RL = 473 (+ / -) √ 150.74² 77.0²

RL = 473 (+ / -) (129.58)

RL₁ = 34.41 , RL₂ = 602.58

The higher value have the less angular frequency

RL₂ = 602.58

ω = 1 / √L*C

ω = 1 / √ 602.58 * 473

f = 285.02 Hz

6 0

3 years ago