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3 years ago

What happens to a light ray passing through the center of a lens?

Physics

2 answers:

lara31 [8.8K]3 years ago

7 0

What happens when a light ray passes through the focal point, and then reflects off of a convex mirror?

The reflected ray will run parallel to the center of curvature.

The reflected ray will run parallel to the principal axis.

The reflected ray will pass through the center of curvature.

The reflected ray will pass through a point between the center of curvature and the focal point.

The reflected ray will pass through a point between the focal point and the vertex.

Is it out of these choices?

Readme [11.4K]3 years ago

6 0

Theoretically, the light ray will not change direction. The center of a lens is called the optical center. If the ray of light hits the lens even just a little bit off center, the light ray is refracted. Refraction happens when the incident and exit angles of the ray are different. The optical center is guaranteed to not have any difference in angle, which allows the light ray to pass through directly without changing direction.

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Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i

leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are $\dfrac{\lambda}{2}$

Thus; for both speakers; the wavelength of the sound is:

$\dfrac{\lambda}{2} = (10+30) cm$

$\dfrac{\lambda}{2} = (40) cm$

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{8}+ \Delta \phi_o$

$\pi \ rad = \dfrac{ \pi}{4}+ \Delta \phi_o$

$\Delta \phi_o = \pi -\dfrac{ \pi}{4}$

$\Delta \phi_o = \dfrac{ 4\pi - \pi}{4}$

$\Delta \phi_o = \dfrac{ 3\pi}{4} \ rad$

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad$

$\Delta \phi = \dfrac{3 \pi}{4} \ rad$

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

$A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)$

$A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)$

$A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)$

A = 0.765a

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2 years ago

a force of 50 newtons pulls a rope attached to a 150 newton sled across a horizontal surface at a constant velocity of 5 meters

lakkis [162]

Answer:

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3 years ago

How many states of matter do you think are in the photo? Describe at least two.

laila [671]

Answer:

Three, the liquid in the man's cup, the stove is solid, and the air around them is a gas

Explanation:

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3 years ago

Question 3 of 15

omeli [17]

Answer:

B) Degrees

Explanation:

The directions of the vectors are often defined in terms of due East, due North, due West and due South. A direction exactly in between of North and East can be described as Northeast, similarly we can describe directions in terms of Northwest, Southeast and South west.

From these, the direction of a vector can be easily expressed in degrees, which is measured counter clockwise about its tail from due East. Considering that we can say that East is at 0° , North is at 90° , West is at 180 and South is at 270° counter clockwise rotation from due East.

So, we know that the direction of a vector lying somewhere between due East i.e 0° and due North i.e 90°, will be measured in degrees, which will have a value between 0°-90°

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3 years ago

in a softball game, a batter hits the ball at the velocity of 27m/s and angle of 40 shown below. What is the maximum range of th

Nata [24]

Answer:

R = 73.25 m

Explanation:

We have,

Initial speed of the ball is 27 m/s

It is projected at an angle of 40 degrees

The maximum range of the ball is given by :

$R=\dfrac{u^2\sin2\theta}{g}$

Plugging all the values we get :

$R=\dfrac{(27)^2\sin2(40)}{9.8}\\R=73.25\ m$

So, the maximum range of the ball is 73.25 m

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3 years ago