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3 years ago

Which is a characteristic of all waves?

Physics

2 answers:

juin [17]3 years ago

7 0

Answer:à

Explanation:waves carry energy in the direction in which they move

kramer3 years ago

4 0

Characteristics of all waves:

All waves carry energy.

All waves move through space.

All waves move through matter, from particle to particle.

All waves transfer energy parallel to the direction of wave mo tion.

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A 80 mW laser beam is polarized horizontally. It then passes through two polarizers. The axis of the first polarizer is oriented

inn [45]

The E(incident) for the first polarizer is E(incident)=-76.1930384332

The E(incident) for the second polarizer is E(incident)=12.340115991

Explanation:

Solving the problem:

Given

Transmitted data = 80 mW

First polarizer= 30 degree

Second polarizer= 60 degree

We have the formula,

E (transmitted) = E (incident)cos()

1. Through the first polarizer

E (transmitted) = E (incident)cos()

80=E(incident) cos(60 degree) (according to law subtract the given degree with 90 degree).

Then,

E(incident)=E(transmitted) cos()

E(incident) =80×cos 60 degree

E(incident)=-76.1930384332

The E(incident) for the first polarizer is E(incident)=-76.1930384332

2. Through the first polarizer

We have the formula,

E (transmitted) = E (incident)cos()

80=E(incident) cos(30 degree) (according to law subtract the given degree with 90 degree).

Then,

E(incident)=E(transmitted) cos()

E(incident) =80×cos 30 degree

E(incident) =12.340115991

The E(incident) for the second polarizer is E(incident)=12.340115991

4 0

3 years ago

You stand at the top of a deep well. To determine the depth, D, of the well you drop a rock from the top of the well and listen

Paladinen [302]

Answer:

(A)

$\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s \right )D+t_t^2v_s^2=0$

(B) D=54.71 m

Explanation:

Free Fall

When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula

$\displaystyle D=\frac{gt^2}{2}$

Solving for t

$\displaystyle t=\sqrt{\frac{2D}{g}}$

If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed

$\displaystyle t=\frac{D}{v_s}$

(A)

The total measured time is the sum of both times and it's given as $t_t=3.5\ seconds$

$\displaystyle t_t=\sqrt{\frac{2D}{g}}+\frac{D}{v_s}$

From this equation we'll manage to compute D

First, we isolate the square root

$\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}$

Let's square both sides

$\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}$

Multiplying by $v_s^2$

$\displaystyle \frac{2Dv_s^2}{g}=t_t^2v_s^2-2t_tDv_s+D^2$

Rearranging and factoring

$\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}$

Now, let's put in numbers:

$g=9.8\ m/s^2,\ v_s=345\ m/s,t_t=3.5\ sec$

$\displaystyle D^2-\left (\frac{2(345)^2}{9.8}+2(3.5)(345)\right )D+(12.25)345^2=0$

Computing all the coefficients:

$\displaystyle D^2-26,705.82D+1,458,056.25=0$

Solving for D, we have two possible solutions:

$D=54.71,\ D=26,651.11$

The second solution is called "extraneous", since it comes from squaring an equation, which can introduce non-valid (or external) solutions. It's impossible, given the conditions of the problem, that the well could be 26.5 km deep. So we'll keep the only solution as.

D=54.71 m

Let's prove our calculations by computing both times:

$\displaystyle t_1=\sqrt{\frac{2(54.71)}{9.8}}=3.34\ sec$