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Sauron [17]
3 years ago
7

Which of the following experiments or phenomena provided evidence for the particle nature of light?

Physics
2 answers:
Marysya12 [62]3 years ago
8 0
<h3><u>Answer;</u></h3>

Expulsion of electrons with varying frequencies of light observed in the photoelectron effect.

<h3><u>Explanation</u>;</h3>
  • <em><u>The photoelectric effect supports a particle theory of light in that it behaves like an elastic collision between two particles, the photon of light and the electron of the metal.</u></em>
  • Albert Einstein observed the photoelectric effect in which ultraviolet light forces a surface to release electrons when the light hits. He explained the reaction by defining light as a stream of photons, or energy packets.
Dima020 [189]3 years ago
6 0

Answer:

Expulsion of electrons with varying frequencies of light observed in the photoelectron effect.

Explanation:

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Structures on a bird feather act like a diffraction grating having 7000 7000 lines per centimeter. What is the angle of the firs
Anestetic [448]

Answer:

θ = 28.9°

Explanation:

We are given;

Wavelength; λ = 602nm = 602 x 10^(-9) m

Lines per centimetre = 7000 /cm = 700000 /m

Thus, the distance between 2 adjacent lines is;

d = 1/700000 = 1.43 x 10^(-6) m

The angle at which diffracted light is formed is given by the formula

mλ = d sinθ

Where;

m is the mth order of the diffraction

λ is the wavelength of the incident light

d is the distance separating the centres of 2 adjacent slits

θ is the angle at which diffraction occurs

From the question, m is 1 because it says first order.

Thus, plugging in the relevant values into mλ = d sinθ, we have;

1 x 602 x 10^(-9) = 1.43 x 10^(-6) sinθ

sinθ = 602 x 10^(-9)/(1.43 x 10^(-6))

sinθ = 0.42098

θ = sin^(-1) 0.42098

θ = 28.9°

8 0
3 years ago
A person on a cruise ship is doing laps on the promenade deck. On one portion of the track the person is moving north with a spe
Aleksandr-060686 [28]

T<u>he direction of motion</u> of the person relative to the water is <u>16.7° north of east.</u>

Why?

We can solve the problem by applying the Pitagorean Theorem, where the first speed (to the north) and the second speed (to the east) corresponds to two legs of the right triangle formed with them. (north and east directions are perpendicular each other)

We can calculate the angle that give the direction using the following formula:

Tan(\alpha)=\frac{NorthSpeed}{EastSpeed}\\\\Tan(\alpha)^{-1}=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}\\\\\alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}

Now, substituting the given information we have:

alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}

\alpha =Tan(\frac{3.6\frac{m}{s} }{12\frac{m}{s} })^{-1}\\\\\alpha =Tan(0.3)^{-1}=16.69\°(North-East)=16.7\°(North-East)

Hence, we have that <u>the direction of motion</u> of the person relative to the water is 16.7° north of east.

Have a nice day!

7 0
2 years ago
A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

H_{max}= \dfrac{v^2 sin^2(\theta)}{g}

now,

H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}

H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}

now, equating both the equations

\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}

\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}

   v₂² = 31061.79

   v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0
3 years ago
Which property of light remains unchanged when it enters a different medium?
Amiraneli [1.4K]
The FREQUENCY of light remains unchanged once it leaves the source.
3 0
3 years ago
Read 2 more answers
A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th
gregori [183]

Answer: The center of gravity is 1.1338 m away from the left side of the barbell

Explanation:

Length of the barbell = 1.90 m

The distance center of gravity from left = x

Mass on the left side = 25 kg

The distance center of gravity from right = 1.90 - x

Mass on the right side = 37 kg

At the balance point: m_1x_1=m_2x_2

25 kg\times x=37 kg\times (1.90-x)

x=1.1338 m

The center of gravity is 1.1338 m away from the left side of the barbell

7 0
3 years ago
Read 2 more answers
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