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3 years ago

Which phenomena support only the wave theory of light? Check all that apply.

Physics

2 answers:

eduard3 years ago

6 0

<h3>Answer;</h3>

Diffraction
interference

<h3>Explanation;</h3>

Light may have both wave or particle properties.

According to wave theory of light, light behaves like a wave. Light is an electromagnetic wave which means it does not require a material medium for transmission. Just like electromagnetic waves light possess both magnetic field and electric fields.

Light waves displays a transverse type of a wave in which it oscillates in a similar direction as that of the wave travel. Due to these characteristics of a wave light can undergo diffraction and also interference.

FrozenT [24]3 years ago

6 0

Answer:

Diffraction and interference

Explanation:

Light has a property that it exhibit both wave like and particle like behavior. It is an electromagnetic wave. It do not require any medium for its propagation.

Light shows diffraction and interference both. In interference of light wave, two waves combine to form an interference pattern. While in diffraction of light light bends through the obstacles.

So, the correct options are "diffraction and interference".

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How many mL of water would be displaced by 408 g of lead

Natasha2012 [34]

For this, we must know the density of lead. This is 11.36 grams per cubic centimeter. The mass present is 408 grams. Therefore, the volume occupied will be:
408 / 11.36 = 35.9 cubic centimeters. This is the volume of water that will be displaced.
Now, we know that one cubic centimeter is equivalent to one mL. Therefore,
35.9 ml of water will be displaced.

6 0

3 years ago

Suppose the electrons and protons in 1g of hydrogen could be separated and placed on the earth and the moon, respectively. Compa

MAXImum [283]

Answer:

The gravitational force is 3.509*10^17 times larger than the electrostatic force.

Explanation:

The Newton's law of universal gravitation and Coulombs law are:

$F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}$

Where:

G= 6.674×10^−11 N · (m/kg)2

k = 8.987×10^9 N·m2/C2

We can obtain the ratio of these forces dividing them:

$\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}$ --- (1)

The mass of the moon is 7.347 × 10^22 kilograms

The mass of the earth is 5.972 × 10^24 kg

And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C

Replacing these values in eq1:

$\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}$

Therefore

$\frac{F_{N}}{F_{C}}}}=3.509\times10^{17}$

This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field

7 0

3 years ago

Given the equation p2 = a3, what is the orbital period, in days, for the planet venus? (venus is located 0.72 au from the sun?)

melisa1 [442]

The correct answer is 223 days.

The relationship between the duration of revolution and the separation between the sun is shown by Kepler's third law. Using the notions of circular motion and the gravitational and centripetal forces, we may obtain this equation.

According to Kepler's third rule, the semi-major axis of an orbit is linked to the orbital period of a planet around the sun as follows:

p² = a³

where an is the semi-major axis/distance to the star and p is the orbital period in years.

It is said that a = 0.72 AU for Venus.

P= √(0.72 AU)^3 = 0.61 years.

365 days in a year = 222.9 ≈ 223 days.

To learn more about Kepler's third rule refer the link:
brainly.com/question/1608361

#SPJ4

5 0

1 year ago

A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. Wh

mixas84 [53]

Answer:

the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Explanation:

Given;

diameter of the ceiling fan, d = 90 cm = 0.9 m

angular speed of the fan, ω = 64 rpm

time taken for the fan to stop, t = 28 s

The distance traveled by the ceiling fan when it comes to a stop is calculated as;

$d = vt = \omega r\times t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m$

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

$v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s$

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

4 0

3 years ago

Given this measurement 2642 ft, how many significant figures are there?

Stolb23 [73]

4 sig figs. Just count

8 0

3 years ago