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4 years ago

Light incident on a surface at an angle of 45° undergoes diffused reflection. At what angle will it reflect?

2 answers:

8 0

<span>The correct answer is option D. i.e. at any angle between -90 and 90 degrees. In the diffused reflection ,the surface is rough and the direction of light in not the same as the incident light. It can be reflected at any angle. </span>

Korvikt [17]4 years ago

3 0

Answer:

D. at any angle between -90° and 90°

Explanation:

First of all, we need to understand the difference between specular reflection and diffuse reflection:

- Specular reflection occurs when the surface hit by the light is perfectly smooth. In this case, the law of reflection states that the angle at which the light is reflected is equal to the angle of incidence of the light (measured relative to the normal to the surface)

- Diffuse reflection occurs when the surface hit by the light is rough. In this case, the law of reflection is still valid for every ray of light hitting the surface; however, due to the imperfections of the surface, each ray of light hits the surface at a different angle of incidence. This means that the angle of reflection for each ray will be different, so light is reflected in any direction between -90 degrees and 90 degrees, which corresponds to the minimum and the maximum angle of reflection measured relative the normal to the surface.

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A ball is released from the top of a hill. How fast is the ball going when it reaches the base of the hill? Approximate g as 10

irina [24]

First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.

5 0

3 years ago

Read 2 more answers

The cooking time for a roast scales like the 2/3rds power of the mass. Based on scaling laws, how much longer does a 20-lb roast

fgiga [73]

Answer:

2.726472 s more or 1.5874 times more time is taken than 10-lb roast.

Explanation:

Given:

- The cooking time t is related the mass of food m by:

t = m^(2/3)

- Mass of roast 1 m_1 = 20 lb

- Mass of roast 2 m_2 = 10 lb

Find:

how much longer does a 20-lb roast take than a 10-lb roast?

Solution:

- Compute the times for individual roasts using the given relation:

t_1 = (20)^(2/3) = 7.36806 s

t_2 = (10)^(2/3) = 4.641588 s

- Now take a ration of t_1 to t_2, to see how many times more time is taken by massive roast:

t_1 / t_2 = (20 / 10)^(2/3)

- Compute: t_1 / t_2 = 2^(2/3) = 1.5874 s

- Hence, a 20-lb roast takes 1.5874 times more seconds than 10- lb roast.

t_2 - t_1 = 2.726472 s more

3 0

3 years ago

What is the momentum of the wooden ball after the Collision

Rzqust [24]

Look it up on google copy and paste

7 0

3 years ago

The frequency of a wavelength ?

serious [3.7K]

Answer:

$f=0.1$

Explanation:

Use the equation:

$f=\frac{1}{T}$ (where T is the period)

$f=\frac{1}{9}$ (9 comes from the period of a full wavelength)

$f=0.1$

6 0

2 years ago

A hockey puck sliding along frictionless ice with speed v to the right collides with a horizontal spring and compresses it by 1.

patriot [66]

Answer:

The spring's maximum compression will be 2.0 cm

Explanation:

There are two energies in this problem, kinetic energy $K= \frac{mv^{2}}{2}$ and elastic potential energy $U= \frac{kx^{2}}{2}$ (with m the mass, v the velocity, x the compression and k the spring constant. ) so the total mechanical energy at every moment is the sum of the two energies:

$E=K+U$

Here we have a situation where the total mechanical energy of the system is conserved because there are no dissipative forces (there's no friction), so:

$E_{i}=E_{f}$

$K_{i}+U_{i}=K_{f}+U_{f}$

Note that at the initial moment where the hockey puck has not compressed the spring all the energy of the system is kinetic energy, but for a momentary stop all the energy of the system is potential elastic energy, so we have:

$K_{i} = U_{f}$

$\frac{mv^{2}}{2}=\frac{kx^{2}}{2}$ (1)

Due conservation of energy the equality (1) has to be maintained, so if we let k and m constant x has to increase the same as v to maintain the equality. Therefore, if we increase velocity to 2v we have to increase compression to 2x to conserve the equality. This is 2(1.0) = 2.0 cm

7 0

3 years ago