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3 years ago

What is the difference between transverse and longitudinal waves?

2 answers:

nataly862011 [7]3 years ago

8 0

<span> Displacement of the medium perpendicular to the direction of propagation of the wave. that would be your answer</span>

Deffense [45]3 years ago

3 0

<span>For transverse waves the displacement of the medium is perpendicular to the direction of propagation of the wave. A ripple on a pond and a wave on a string are easily visualized transverse waves. </span>

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A parallel beam of light in air makes an angle of 43.5 ∘ with the surface of a glass plate having a refractive index of 1.68. Yo

aniked [119]

Answer:

a) 46.5º b) 64.4º

Explanation:

To solve this problem we will use the laws of geometric optics

a) For this part we will use the law of reflection that states that the reflected and incident angle are equal

θ = 43.5º

This angle measured from the surface is

θ_r = 90 -43.5

θ_s = 46.5º

b) In this part the law of refraction must be used

n₁ sin θ₁ = n₂. Sin θ₂

sin θ₂ = n₁ / n₂ sin θ₁

The index of air refraction is n₁ = 1

The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface

θ_s = 90 - θ

θ_s = 90- 43.5

θ_s = 46.5º

sin θ₂ = 1 / 1.68 sin 46.5

sin θ₂ = 0.4318

θ₂ = 25.6º

The angle with respect to the surface is

θ₂_s = 90 - 25.6

θ₂_s = 64.4º

measured in the fourth quadrant

3 0

2 years ago

Find the voltage change when: a. An electric field does 12 J of work on a 0.0001-C charge. b. The same electric field does 24 J

kondaur [170]

Explanation:

Given that,

(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :

$V=\dfrac{W}{q}$

$V=\dfrac{12}{0.0001}$

$V=12\times 10^4\ Volt$

(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :

$V=\dfrac{W}{q}$

$V=\dfrac{24}{0.0002}$

$V=12\times 10^4\ Volt$

Therefore, this is the required solution.

7 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

2 years ago

What material parameters determine resistivity?

sleet_krkn [62]

Answer:

Resistivity $\rho =\frac{RA}{l}$

It depends upon cross sectional area and length of material

Explanation:

The resistance of any material is given by $R=\frac{\rho l}{A}$ , here $\rho$ is the resistivity of material , l is length of material and A is cross sectional area

So resistivity $\rho =\frac{RA}{l}$

So resistuivity of any material depends upon area of cross section and length of material

If cross sectional area will be more then resistivity will be more. And is length of the material will be more then resistivity will be less

3 0

2 years ago

Please help and check all that apply and I will mark brainliest if it’s correct

Yuri [45]

A syncline is visable

3 0

3 years ago