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3 years ago

Which phenomena support only the wave theory of light? Check all that apply.

Physics

2 answers:

eduard3 years ago

6 0

<h3>Answer;</h3>

Diffraction
interference

<h3>Explanation;</h3>

Light may have both wave or particle properties.

According to wave theory of light, light behaves like a wave. Light is an electromagnetic wave which means it does not require a material medium for transmission. Just like electromagnetic waves light possess both magnetic field and electric fields.

Light waves displays a transverse type of a wave in which it oscillates in a similar direction as that of the wave travel. Due to these characteristics of a wave light can undergo diffraction and also interference.

FrozenT [24]3 years ago

6 0

Answer:

Diffraction and interference

Explanation:

Light has a property that it exhibit both wave like and particle like behavior. It is an electromagnetic wave. It do not require any medium for its propagation.

Light shows diffraction and interference both. In interference of light wave, two waves combine to form an interference pattern. While in diffraction of light light bends through the obstacles.

So, the correct options are "diffraction and interference".

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3 years ago

What are the difficulties with the capture hypothesis of the Moon’s origin?

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Answer:

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3 years ago

A boat is traveling 60m/s north and hits a current that is 14m/a west what is the boats direction

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2 years ago

A car move with uniform acceleration along a straight line pqr .Its speed at p and r are 5m/s and 25m/s respectively if pq:qr is

topjm [15]

Answer:

Answer: Option d.

Explanation:

Accelerated Motion

When an object changes its spped in the same amounts in the same times, the acceleration is constant and its value is

$\displaystyle a=\frac{v_f-v_o}{t}$

Where $v_f, v_o, t$ are the final speed, initial speed, and time taken to change them, respectively

From the above equation we can know

$v_f=v_o+at$

The distance traveled is computed as

$\displaystyle x=v_ot+\frac{at^2}{2}$

The question talks about a car moving in a straight line with constant acceleration. It goes through the points p,q,r such as

$v_p=5\ m/s, v_r=25\ m/s$

The ratio of the distances traveled in each segment is

$\displaystyle \frac{x_1}{x_2}=\frac{1}{2}$

being $x_1$ the distance from p to q and $x_2$ the distance from q to r

It means that

$x_2=2x_1$

From the equation for speed

$v_q=v_p+at_1\ \ \ ..........[1]$

$v_r=v_q+at_2\ \ \ ..........[2]$

Replacing [1] into [2]

$v_r=v_p+at_1+at_2$

$v_r=v_p+a(t_1+t_2)$

Solving for a

$\displaystyle a=\frac{v_r-v_p}{t_1+t_2}\ \ \ .........[3]$

We now write the equation for both distances .

$\displaystyle x_1=v_pt_1+\frac{at_1^2}{2}$

$\displaystyle x_2=v_qt_2+\frac{at_2^2}{2}$

Using [1] again

$\displaystyle x_2=(v_p+at_1)t_2+\frac{at_2^2}{2}$

Since

$x_2=2x_1$

We have

$\displaystyle (v_p+at_1)t_2+\frac{at_2^2}{2}=2\left (v_pt_1+\frac{at_1^2}{2}\right )$

Operating

$\displaystyle v_pt_2+at_1t_2+\frac{at_2^2}{2}=2v_pt_1+at_1^2$

Rearranging

$\displaystyle v_pt_2-2v_pt_1=at_1^2-at_1t_2-\frac{at_2^2}{2}$

Factoring both sides

$\displaystyle v_p(t_2-2t_1)=\frac{a}{2}\left (2t_1^2-2t_1t_2-t_2^2 \right )$

Replacing the equation [3] for a :

$\displaystyle 2v_p(t_2-2t_1)=\frac{v_r-v_p}{t_1+t_2}\left (2t_1^2-2t_1t_2-t_2^2 \right )$

Replacing $v_p=5,\ v_r=25,\ v_r-v_p=20$ , and operating the denominator

$\displaystyle 10(t_2-2t_1)\left (t_1+t_2 \right )=20\left (2t_1^2-2t_1t_2-t_2^2 \right )$

Operating and simplifying, we get a second-degree equation

$\displaystyle t_2^2+t_1t_2-2t_1^2=0$

Factoring

$(t_2-t_1)(t_2+2t_1)=0$

The only positive and valid answer is

$t_2=t_1$

Or equivalently

$\displaystyle \frac{t_1}{t_2}=1$

The option d. is correct

6 0

3 years ago