A 3 N force pushes on a object for 20 meters. Find the work done

A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of

andre [41]

Answer:

The angle of incidence is greater than the angle of refraction

Explanation:

Refraction occurs when a light wave passes through the boundary between two mediums.

When a ray of light is refracted, it changes speed and direction, according to Snell's Law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where :

$n_1$ is the index of refraction of the 1st medium

$n_2$ is the index of refraction of the 2nd medium

$\theta_1$ is the angle of incidence (the angle between the incident ray and the normal to the boundary)

$\theta_2$ is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

In this problem, we have a ray of light passing from air into clear plastic. We have:

$n_1=1.00$ (index of refraction of air)

$n_2=1.50$ approx. (index of refraction in clear plastic)

Snell's Law can be rewritten as

$sin \theta_2 =\frac{n_1}{n_2}sin \theta_1$

And since $n_2>n_1$ , we have

$\frac{n_1}{n_2}$

And so

$\theta_2$

Which means that

The angle of incidence is greater than the angle of refraction

6 0

2 years ago

A hydrocarbon compound z decolourises bromine liquid

Leno4ka [110]

Answer:

...

Explanation:

................................................

8 0

2 years ago

An elevator has the mass of 3 tons of power needed to raise the elevator 50m in 15 s? PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE

olasank [31]

Answer:

Power=?,time=15,mass=3,height=50,g=10.

Power=mgh/t......3×10×50=1500/15=100Watts

7 0

2 years ago

Read 2 more answers

If a pendulum has a period of 4 seconds at the north pole with gravity force 9.83 ms^-2 and a 3.97m pendulum length, how could t

sladkih [1.3K]

Answer:

The same pendulum could be adjusted to have the same period, in the equator must have a length of 3.949m.

Explanation:

Tnp= 4 sec

gnp= 9.83 m/sec²

Lnp= 3.97m

Tequ= 4 sec

gequ= 9.78 m/sec²

Lequ=?

Lequ= (Lnp* gequ) / gnp

Lequ= 3.949 m

5 0

2 years ago

Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1

Nonamiya [84]

Answer:

$a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

Magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$

Explanation:

$t_1=2\ \text{s}$

$v_x=1\ \text{m/s}$

$v_y=3\ \text{m/s}$

$t_2=2.5\ \text{s}$

$v_x=4\ \text{m/s}$

$v_y=3\ \text{m/s}$

Average acceleration in the different axes

$a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2$

$a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2$

The components of the acceleration is $a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

The magnitude of acceleration

$a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2$

Direction

$\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}$

The magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$ .

7 0

2 years ago