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tester [92]
3 years ago
15

Light striking a metal surface causes electrons to be emitted from the metal via the photoelectric effect.In a particular experi

ment to study the photoelectric effect, the frequency of the incident light and the temperature of the metal are held constant. Assuming that the light incident on the metal surface causes electrons to be ejected from the metal, what happens if the intensity of the incident light is increased?Check all that apply.A. The work function of the metal decreases.B. The number of electrons emitted from the metal per second increases.C. The maximum speed of the emitted electrons increases.D. The stopping potential increases.
Physics
1 answer:
ANEK [815]3 years ago
5 0

The number of electrons emitted from the metal per second increases if the intensity of the incident light is increased.

Answer: Option B

<u>Explanation:</u>

As a result of photoelectric effect, electrons are emitted by the light incident on a metal surface. The emitted electrons count and its kinetic energy can measure as the function of light intensity and frequency. Like physicists, at the 20th century beginning, it should be expected that the light wave's energy (its intensity) will be transformed into the kinetic energy of emitted electrons.

In addition, the electrons count emitting from metal must vary with light wave frequency. This frequency relationship was expected because the electric field oscillates due to the light wave and the metal electrons react to different frequencies. In other words, the number of electrons emitted was expected to be frequency dependent and their kinetic energy should be dependent on the intensity (constant wavelength) of light.

Thus, the maximum in kinetic energy of electrons emitted increases with increase in light's frequency and is experimentally independent of light intensity. So, the number of emitted electrons is proportionate to the intensity of the incident light.

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False as oxygen is the second most abundant and nitrogen is the most abundant at 78%.
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A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic
defon

Answer:

0.52 Nm

Explanation:

A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T

Angle between the plane of loop and magnetic field = 30 Degree

Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree

θ = 60°

Torque = N i A B Sinθ

Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60

Torque = 0.52 Nm

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3 years ago
Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo
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Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567  x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

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2 years ago
How are solar energy, wind energy and hydroelectric energy related?
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A racing car travels on a circular track of radius 158 m, moving with a constant linear speed of 19.1 m/s. Find its angular spee
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Answer:

\omega=0.12\frac{rad}{s}

Explanation:

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\omega=\frac{2\pi}{T}(1)

Here T is the period, that is, the time taken to complete onee revolution:

T=\frac{2\pi r}{v}(2)

Replacing (2) in (1):

\omega=\frac{2\pi}{\frac{2\pi r}{v}}=\frac{v}{r}\\\omega=\frac{19.1\frac{m}{s}}{158m}\\\omega=0.12\frac{rad}{s}

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