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3 years ago

A student measures the speed of yellow light in water to be 2.00x10^8

Physics

1 answer:

max2010maxim [7]3 years ago

7 0

NOTE: The given question is incomplete.

<u>The complete question is given below.</u>

A student measures the speed of yellow light in water to be 2.00 x 10⁸ m/s. Calculate the speed of light in air.

Solution:

Speed of yellow light in water (v) = 2.00 x 10⁸ m/s

Refractive Index of water with respect to air (μ) = 4/3

Refractive Index = Speed of yellow light in air / Speed of yellow light in water

Or, The speed of yellow light in air = Refractive Index × Speed of yellow light in water

or, = (4/3) × 2.00 x 10⁸ m/s

or, = 2.67 × 10⁸ m/s ≈ 3.0 × 10⁸ m/s

Hence, the required speed of yellow light in the air will be 3.0 × 10⁸ m/s.

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Photoelectric effect:

Lynna [10]

Answer:

A. K = 0.546 eV

B. cooper and iron will not emit electrons

Explanation:

A. This is a problem about photoelectric effect. Then you have the following equation:

$K=h\nu-\Phi=h\frac{c}{\lambda} -\Phi$ (1)

K: kinetic energy of the ejected electron

Ф: Work function of the metal = 2.48eV

h: Planck constant = 4.136*10^{-15} eV.s

λ: wavelength of light = 410nm - 750nm

c: speed of light = 3*10^8 m/s

As you can see in the equation (1), higher the wavelength, lower the kinetic energy. Then, the maximum kinetic energy is obtained with the lower wavelength (410nm). Thus, you replace the values of all variables :

$K=(4.136*10^{-15}eV)\frac{3*10^8m/s}{410*10^{-9}m}-2.48eV\\\\K=0.546eV$

B. First you calculate the energy of the photon with wavelengths of 410nm and 750nm

$E_1=(4.136*10^{-15}eV)\frac{3*10^{8}m/s}{410*10^{-9}m}=3.02eV\\\\E_2=(4.13610^{-15}eV)\frac{3*10^{8}m/s}{750*10^{-9}m}=1.6544eV$

You compare the energies E1 and E2 with the work functions of the metals and you can conclude:

sodium = 2.3eV < E1

cesium = 2.1 eV < E1

cooper = 4.7eV > E1 (this metal will not emit electrons)

iron = 4.5eV > E1 (this metal will not emit electrons)

5 0

3 years ago

What force is required to push a block (mass m) up an inclined plane that makes an angle of θ with the horizon at a constant vel

Verizon [17]

Answer:

b. mg ( μ · cos θ + sin θ)

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

We have the following parallel forces acting on the block (in parallel direction to the direction of movement):

F = applied force.

Fr = friction force.

wx = parallel component of the weight.

According to Newton´s second law:

∑F = m · a

Where "m" is the mass of the block and "a" its acceleration.

Then:

F - Fr - wx = m · a

Since the block is to be pushed at a constant velocity, the acceleration is zero. Then:

F - Fr - wx = 0

F = Fr + wx

The applied force has to be equal to the friction force plus the parallel component of the weight to push the block at constant velocity.

The friction force is calculated as follows:

Fr = μ · N

Where N is the normal force and μ is the coefficient of friction.

Notice that the normal force is of the same magnitude as the perpendicular component of the weight, wy.

Let´s apply Newton´s second law in the perpendicular direction to show this:

∑F = m · a

N - wy = m · a

The acceleration of the block in the perpendicular direction is zero. Then:

N - wy = 0

N = wy

And wy can be obtained by trigonometry (see figure):

wy = W · cos θ

N = wy = mg · cos θ

The parallel component of the weight is calculated using trigonometry (see figure):

wx = W · sin θ

wx = mg · sin θ

Then the applied force will be:

F = Fr + wx

F = μ · N + mg · sin θ (N = wy = mg · cos θ)

F = μ · mg · cos θ + mg · sin θ

F = mg ( μ · cos θ + sin θ)

The correct answer is the b.

8 0

3 years ago

What kind of motion is the motion of the “flying bully”? How much and what is

lesantik [10]

<span>The flying bully is a move used in the Superhero Movie "Hancock", it is not a real motion in our universe. However, the direction would be towards the target object and the acceleration would be maximal.</span>

7 0

3 years ago

a ball of mass 0.80 kg moving at a speed of 2.5 m/s along a straight line collided with a mass 2.5 kg which was initially statio

Likurg_2 [28]

The speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

The given parameters;

mass of the ball, m₁ = 0.8 kg
speed of the ball, u₁ = 2.5 m/s
mass of the object at rest, m₂ = 2.5 kg
final velocity of the object at rest, v₂ = 1 m/s

Let the final velocity of the 0.8 kg ball immediately after collision = v₁

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.8 x 2.5) + (2.5 x 0) = (0.8)v₁ + 2.5(1)

2 = 2.5 + (0.8)v₁

-0.5 = (0.8)v₁

$v_1 = \frac{-0.5}{0.8} \\\\v_1 = -0.625 \ m/s$

Thus, the speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

Learn more here: brainly.com/question/7694106

5 0

2 years ago

WILL MARK BRAINLIEST PLEASE HELP AND SHOW ALL WORK

jok3333 [9.3K]

Answer:

Same reading.

Explanation:

Assume that after the string breaks the ball falls through the liquid with constant speed. If the mass of the bucket and the liquid is 1.20 kg, and the mass of the ball is 0.150 kg,

A.) Before the string break, the total weight = weight of the can + weight of the water.

According to Archimedes' Principle which state that: “A body immersed in a liquid loses weight by an amount equal to the weight of the liquid displaced.” Archimedes principle also states that: “When a body is immersed in a liquid, an upward thrust, equal to the weight of the liquid displaced, acts on it

B.) After the string break.

The scale will have the same reading as before the string break.

6 0

3 years ago