What flows in electricity? A. Electrons B. Protons C. Neutrons

The index of refraction for red light in water is 1.331 and that for blue light is 1.340. If a ray of white light enters the wat

kobusy [5.1K]

Answer:

Angle of refraction for red light is $43.01^{\circ}$

Angle of refraction for blue light is $42.68^{\circ}$

Explanation:

It is given refractive index for red light is $\mu _{red}=1.331$

Refractive index of blue light $\mu _{blue}=1.340$

Angle of incidence $i=65.30^{\circ}$

According to law of refraction $\mu =\frac{sini}{sinr}$

For red light $1.331 =\frac{sin65.30^{\circ}}{sinr}$

$1.331 =\frac{0.908}{sinr}$

$sinr=0.682$

$r=43.01^{\circ}$

Therefore angle of refraction for red light is $43.01^{\circ}$

Similarly for blue light $1.340 =\frac{sin65.30^{\circ}}{sinr}$

$1.340 =\frac{0.908}{sinr}$

$sinr=0.677$

r = $42.68^{\circ}$

Therefore angle of refraction for blue light is $42.68^{\circ}$

6 0

3 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago

18. A simple series circuit has a current of 2 A

QveST [7]

I think it is;
Current = vtotal/requ
2A = vtotal/0.5
2*0.5= vtotal
1=vtotal

5 0

2 years ago

Blank is overlapping food chains to better show the way energy moves through an ecosystem

Komok [63]

A Food web is multiple overlapping food chains.

8 0

3 years ago

In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 6.7 rev/

Alexxandr [17]

Answer: 7.095 rev

Explanation: since the motion of the washer is of a constant angular acceleration, then Newton's laws of motion applies.

Recall that

ω = ω0 + αt

Where α = angular acceleration = ?

ω = final angular velocity = 6.7rev/s

ω0 = initial angular velocity = 0 rev/s

t = time taken =8.8s

Since the motion of the dryer starts from rest, ω0 = 0, we have that

ω = αt

6.7 = α(8.8)

α = 6.7/8.8

α = 0.76 rev/s²

This value α = 0.76 rev/s² is a constant value and it is the same all through the motion.

When the lid of the washer was lifted, the motion was of a constant deceleration.

Recall that

θ = ω0t + αt²/2

Where θ = angular displacement

Since the washer is now decelerating, the final velocity (ω= 6.7 rev/s) of the accelerating motion now because the initial velocity of the decelerating motion.

Hence ω0 = 6.7 rev/s, α = - 0.76 rev/s² ( negative sign to denote deceleration) and t = 16.5s

By slotting in the parameters into the formulae, we have that

θ = 6.7 (16.5) + (-0.76)(16.5)²/2

θ = 110.55 - 103.455

θ = 7.095 rev

3 0

4 years ago