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fredd [130]
3 years ago
11

When people do "the wave" at a sports stadium, just like the particles of a real wave, each person _____.

Physics
1 answer:
mel-nik [20]3 years ago
7 0
Sounds like C. A wave happens bellow the water and simply is energy pushing its way to something
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What is the net force on this object?
damaskus [11]

Upward and downward forces cancel out. Net force is 8 newtons to the right

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A soap box derby car at the top of a ramp has more _________ energy and at the finish line at the bottom of the ramp it has more
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4 0
3 years ago
At what angle should the axes of two polaroids be placed so as to reduce the intensity of the incident unpolarized light to 13.
prohojiy [21]

Answer:

θ = 66.90°

Explanation:

we know that

I= \frac{I_0}{2}cos^2\theta

I= intensity of polarized light =1

I_o= intensity of unpolarized light = 13

putting vales we get

1= \frac{13}{2}cos^2\theta

⇒\theta=cos^{-1} \sqrt{\frac{1}{6.5} }

therefore θ = 66.90°

5 0
3 years ago
The eye is actually a multiple-lens system, but we can approximate it with a single-lens system for most of our purposes. When t
hram777 [196]

Explanation:

Given that,

The optical power of the equivalent single lens is 45.4 diopters.

(a) The relationship between the focal length and the focal length is given by:

f=\dfrac{1}{P}

f=\dfrac{1}{45.4}

f = 0.022 m

or

f = 2.2 cm

(b) We need to find how far in front of the retina is this "equivalent lens" located. It is given by using lens formula as :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Here, u = infinity

\dfrac{1}{v}=\dfrac{1}{2.2}

v = 2.2 cm

So, at 2.2 cm in front of the retina is this "equivalent lens" located.

Hence, this is the required solution.

5 0
3 years ago
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