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kirill [66]
3 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]3 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]3 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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How do climate differences affect the movement at the Mariana Trench
vovangra [49]
It pushes the currents to opposite sides
8 0
3 years ago
The velocity of a body is given by the equation v= a + bx, where 'x' is displacement. The unit of b is .......
valina [46]

Answer:

s^ -1    ( or    1/sec)

Explanation:

Velocity is given in units of displacement / sec

like feet /sec   or   m/sec    

so b would have units of   s^-1

(or perhaps a more general term would be   time^-1)

8 0
1 year ago
A 47-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37 degrees above the
levacccp [35]

Tension in the rope due to applied force will be given as

F = 142 N

angle of applied force with horizontal is 37 degree

displacement along the floor = 6.1 m

so here we can use the formula of work done

W = F d cos\theta

now we can plug in all values above

W = 142 * 6.1 * cos37

W = 691.8 J

So here work done to pull is given by 691.8 J


8 0
3 years ago
A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre
Alisiya [41]

Answer:

0.546 \hat k

Explanation:

From the given information:

The force on a given current-carrying conductor is:

F = I ( \L  \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})

where the length usually in negative (x) direction can be computed as

\L ^ {\to }  = -x\hat i \\dL\limits ^ {\to }- dx\hat i

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})

F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)

F = I \int^3_1  - 9.0x^2 \ dx \hat k

F = I  (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k

F = I  (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ]  \hat k

where;

current I = 7.0 A

F = (7.0 \ A)  (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ]  \hat k

F = (7.0 \ A)  (9.0) \bigg [\dfrac{26}{3} \bigg ]  \hat k

F = 546 × 10⁻³ T/mT \hat k

F = 0.546 \hat k

4 0
3 years ago
In 1977 off the coast of Australia, the fastest speed by a vessel on the water
fenix001 [56]

Answer: 154.08 m/s

Explanation:

Average acceleration a_{ave} is the variation of velocity  \Delta V over a specified period of time  \Delta t:

a_{ave}=\frac{\Delta V}{\Delta t}}

Where:

a_{ave}=1.80 m/s^{2}

\Delta V=V_{f}-V_{o} being V_{o}=0 the initial velocity and V_{f} the final velocity

\Delta t=85.6 s

Then:

a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}

Since V_{o}=0:

a_{ave}=\frac{V_{f}}{\Delta t}}

Finding V_{f}:

V_{f}=a_{ave} \Delta t

V_{f}=(1.80 m/s^{2})(85.6 s)

Finally:

V_{f}=154.08 m/s

8 0
3 years ago
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