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2 years ago

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"Suppose a horizontal laser beam is reflected off a plane mirror that is perfectly smooth and flat. At first, the mirror is angl

ed upward at 4 degrees, measured from the vertical. Then, the angle of the mirror is changed to 6 degrees upward. How would the angle formed by the incoming laser beam and the reflected laser beam change

1 answer:

Rashid [163]2 years ago

3 0

Answer:

The angle in between changed from 8 to 12 degrees

Explanation:

Solution:-

Angle between the incoming and reflected beam is always twice that of the inclination of the mirror.

Initially, Inclination angle θ = 4 degrees

Angle between incident and reflected beam is twice of the inclination of mirror. Hence,

Φ = 2*θ = 2*4 = 8 degrees.

Change, Inclination angle θ = 6 degrees

Angle between incident and reflected beam is twice of the inclination of mirror. Hence,

Φ = 2*θ = 2*6 = 12 degrees.

- The angle formed by the incoming laser beam and the reflected laser beam will change from 8 degrees to 12 degrees.

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4.A photon of green light strikes an unknown metal and an electron is emitted. The voltage is set to 2 volts. The electron canno

Anarel [89]

4) The correct answer is:
<span>B. An electron will be emitted in the second experiment, but it cannot be determined whether it will reach the second plate.

In fact, violet light has higher frequency than green light. This means that photons of violet light carry more energy than photons of green light (remember that the energy of a photon is proportional to its frequency: </span> $E=hf$ )<span>, so when they hit the surface of the metal, more energy is transferred to the electrons. The electron was already emitted with green light, so it must be emitted also with violet light, given the more energy transferred. The electron will also have more kinetic energy when hit by violet light, however, we cannot determine if it will reach the second plate, since we don't know how much energy has been used to extract the electron from the metal (in fact, we don't know the work function of the metal, i.e. the energy needed to extract the electron)

3) The correct answer is
</span><span>A. Violet light will cause electrons to be emitted at greater velocities than those removed by green light.

In fact, </span>violet light has higher frequency than green light. This means that photons of violet light carry more energy than photons of green light (remember that the energy of a photon is proportional to its frequency: $E=hf$ ), so when they hit the surface of the metal, more energy is transferred to the electrons. Therefore, the emitted electrons will have on average greater energy (and so, greater velocity) than those removed by green light.

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2 years ago

The wavelength of some red light is 700.5 nm. what is the frequency of this red light?

Alborosie

The frequency of the red light is 428 terahertz. To get the value of the red light's frequency, use the formula F = velocity/wavelength. The velocity of light is 3.00 x 10^8 m/s. For easier computation, convert 700.5 nanometers to meter. 1 nanometer is equal to 1 x 10^-9 meters. 700.5 nanometers is equal to 7.005 x 10^-7 meters. Divide the velocity 3.00 x 10^8m/s by wavelength 7.005 x 10^-7 meters. The result will be 4.28 x 10^14 Hertz or 428 terahertz.

4 0

2 years ago

A 14000N car traveling at 25m/s rounds a curve of radius 200m. Find the following: a. The centripetal acceleration of the car.

tamaranim1 [39]

Answer:

Explanation:

Given

Weight of car $W=14,000\ N$

mass of car $m=\frac{14,000}{9.8}=1428.57\ N$

velocity of car $v=25\ m/s$

radius $r=200\ m$

(a)Centripetal acceleration is given by

$a_c=\frac{v^2}{r}$

$a_c=\frac{25^2}{200}$

$a_c=3.125\ m$

(b)Force that provide centripetal acceleration

$F=F_c=\frac{mv^2}{r}$

$F=\frac{1428.57\times 25^2}{200}$

$F=4464.285\ N$

(c)Friction force between car and tires is given by

$=\mu N$

where $\mu$ =coefficient of static friction

N=normal reaction

Centripetal force will balance the friction force

$F_c=F_r$

$4464.285=\mu \times 1428.57\times 9.8$

$\mu =0.318$

6 0

2 years ago

Read 2 more answers

A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , $v_{i}=2 v_{e}$ . Here the radius is R, so find the escape velocity as follows,

$\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0$

$\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}$

$v_{e}^{2}=\frac{2 G M}{R}$

$v_{e}=\sqrt{\frac{2 G M}{R}}$

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as, $U_{i}=-\frac{G M m}{R}$

R tend to infinity when far away from the planet, so $v_{f}=0$

Then, kinetic energy at initial would be,

$k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}$

Similarly, kinetic energy at final would be,

$k_{f}=\frac{1}{2} m v_{f}^{2}$

Here, $v_{f}=\text { final velocity }$

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

$U_{i}+k_{i}=k_{f}+v_{f}$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}$

'm' and $\frac{1}{2}$ as common on both sides, so gets cancelled, we get as

$4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}$

We know, $v_{e}=\sqrt{\frac{2 G M}{R}}$ , it can be wriiten as $\left(v_{e}\right)^{2}=\frac{2 G M}{R}$ , we get

$4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

$v_{f}^{2}=3\left(v_{e}\right)^{2}$

Taking squares out, we get,

$v_{f}=\sqrt{3} v_{e}$

4 0

2 years ago

1. A 930-kg car traveling 56 km/h comes to a complete stop in 2.0 s. What is the

Juli2301 [7.4K]

The force exerted on the car during this stop is 6975N

<u>Explanation:</u>

Given-

Mass, m = 930kg

Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s

Time, t = 2s

Force, F = ?

F = m X a

F = m X s/t

F = 930 X 15/2

F = 6975N

Therefore, the force exerted on the car during this stop is 6975N

6 0

2 years ago