A 62.3 N force pulls an object at an angle of 9 = 22.0° to its direction of horizontal motion. It moves a

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

3 years ago

Helppppppppppppppp asappppppppppppp

vichka [17]

The answer to the problem b.

8 0

3 years ago

What is the maximum number of electrons an oxygen atom can hold in its outer energy level?

gulaghasi [49]

The answer is 8.
Hope I helped.

6 0

3 years ago

A 78−kg skier is sliding down a ski slope at a constant velocity. The slope makes an angle of 21° above the horizontal direction

Feliz [49]

Answer:

274N 0.41

Explanation:

As he is sliding down in a constant speed then the force that accelerates him (weight) and the force that slows his down (friction) are equal.

then

<em>friction=mass x gravity x sin(21)</em>

Fr=78kg x 9.8m/s2 x sin(21)=274N

<em>friction= coefficient of kinetic friction x normal force of from the slope</em>

Fr= u x 78kg x 9.8m/s2 x cos(21)=274N

Fr= u x 78kg x 9.8m/s2 x cos(21)=274Nu=274/677=0.41

8 0

3 years ago

A particle with a charge of 5 × 10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a sta

creativ13 [48]

Answer:

r = 0.22m

Explanation:

To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.

Then, you have:

$F_c=F_e=ma_c$ (1)

m: mass of the particle = 20g = 20*10-3 kg

ac: centripetal acceleration = ?

q: charge of the particle = 5*10^-6C

Fe: electric force between the charges

The electric force is given by:

$F_e=k\frac{qq'}{r^2}$ (2)

r: radius of the orbit

q': charge of the particle at the center of the orbit = -5*10^-6C

Furthermore, the centripetal acceleration is:

$a_c=\frac{v^2}{r}$ (3)

v: speed of the particle = 7m/s

You replace the expressions (2) and (3) in the equation (1) and solve for r:

$k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}$

Finally, you replace the values of all parameters in the previous expression:

$r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m$

The radius of the circular trajectory is 0.22m

5 0

3 years ago