Please please please please help soon!!

A certain kind of glass has an index of refraction of 1.640 for blue light of wavelength 440 nm and an index of 1.595 for red li

DIA [1.3K]

Answer:

0.52°

Explanation:

refractive index for blue light, nb = 1.640

Refractive index for red light, nr = 1.595

Angle of incidence, i = 30°

Let the angle of refraction for blue light is rb and the angle of refraction for red light is rR.

By use of Snell's law for blue light

$n_{b}=\frac{Sin i}{Sin r_{b}}$

$1.64=\frac{Sin 30}{Sin r_{b}}$

$1.64=\frac{0.5}{Sin r_{b}}$

$r_{b}=17.75^{\circ}$

By use of Snell's law for red light

$n_{R}=\frac{Sin i}{Sin r_{R}}$

$1.595=\frac{Sin 30}{Sin r_{R}$

$1.595=\frac{0.5}{Sin r_{R}}$

$r_{R}=18.27^{\circ}$

The angle between the two beams, $\theta =r_{R}-r_{b}$

θ = 18.27° - 17.75°

θ = 0.52°

8 0

3 years ago

A 60 g ball is dropped from rest from a height of 2.4 m. It bounces off the floor and rebounds to a maximum height of 1.9 m. If

kap26 [50]

Answer:

The force is 1.34 newtons and its direction is upward.

Explanation:

Choosing positive direction pointing towards the floor in this collision we're going to use the momentum-impulse theorem that states:

$J=\Delta p$ (1)

with $\Delta p=p_f-p_i$ the change in the momentum and J the impulse, with pi the initial momentum that is the momentum just before the collision and pf the final momentum th is the momentum just after the collision. The impulse J is also defined as:

$J=F_{avg}\Delta t$ (2)

with $F_{avg}$ the average force and $\Delta t$ the time the collision lasts

We can equate expressions (2) and (1):

$\Delta p=p_f-p_i=F_{avg}\Delta t$

Using the definition of linear momentum as mass (m) time velocity (v):

$mv_f-mv_i=F_{avg}\Delta t$

We can solve for Favg:

$F_{avg}=\frac{m(v_f-v_i)}{\Delta t}$ (3)

Now we should find the velocities vf and vi, we should do this using conservation of energy:

For the velocity the ball has just before reaches the floor:

$U_i=K_f$

With Ui the initial potential energy (there is not initial kinetic energy) and Kf the final kinetic energy (there is not final potential energy), then:

$mgh=\frac{mv_i^2}{2}$

solving for vi:

$v_i=\sqrt{2gh}=\sqrt{2*9.81*2.4}=6.86\frac{m}{s}$

For the velocity the ball has just after bounces the floor:

$K_i=U_f$

There is not initial potential energy because it's a floor level at this instant, and the there is not final kinetic energy because the ball has instantly zero velocity at its maximum height (hm), then:

$\frac{mf_i^2}{2}=mgh_m$