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kirill [66]
2 years ago
15

If you're driving towards the sun late in the afternoon, you can reduce the glare from the road by wearing sunglasses that permi

t only the passage of light that's
a. dispersed in a spherical plane.

b. polarized in a vertical plane.

c. dispersed in a vertical plane.

d. polarized in a horizontal plane.
Physics
2 answers:
Tcecarenko [31]2 years ago
8 0

I took the Penn Foster quiz and went through the study guide and the answer is A. POLARIZED IN A VERTICAL PLANE. Be sure to not just mark A its likely you dont have the same order of answers.

( it be great if you marked me brainliest;) )

zepelin [54]2 years ago
3 0
The correct option is B.
Sunglasses are designed to block polarized light. If you are driving toward the sun,  most of the sun's glare will be coming to you horizontally, thus, the best sunglasses to wear is the one, that will block the horizontal glare and allow only the vertical glare to pass through. 
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3 years ago
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Compute the dot product of the vectors u and v​, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and
OLga [1]

Answer:

\theta = 106.3 degree

Explanation:

As we know that

\vec w = -\hat i + 7\hat j

\vec v = 7\hat i - \hat j

also we know that

\vec v. \vec w = -14

it is given as

\vec v. \vec w = (-\hat i + 7\hat j).(7\hat i - \hat j)

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also we can find the magnitude of two vectors as

|v| = \sqrt{(-1)^2 + (7)^2}

|v| = \sqrt{50}

similarly we have

|w| = \sqrt{(7^2) + (-1)^2}

|w| = \sqrt{50}

now we know the formula of dot product as

\vec v. \vec w = |v||w| cos\theta

-14 = (\sqrt{50})^2cos\theta

\theta = cos^{-1}(\frac{-14}{50})

\theta = 106.3 degree

3 0
2 years ago
How much force is required to cause an object with a mass of 850 kg to accelerate at a rate of 2 meters per second squared (m/s^
arlik [135]

Answer:

How much force is required to cause an object with a mass of 850 kg to accelerate at a rate of 2 meters per second squared (m/s^2)?

Explanation:

<em>1700N </em>

<em> Mass multiplied by acceleration gives you the amount of force needed for it.</em>

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2 years ago
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