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kifflom [539]
3 years ago
6

Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.

Physics
1 answer:
Shkiper50 [21]3 years ago
6 0

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

a)

We can solve this first part of the problem by applying the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

f=\frac{R}{2}

For this mirror, R = 20 cm, so its focal length is

f=\frac{20}{2}=+10 cm (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm

b)

We can solve this part by using the magnification equation:

M=-\frac{y'}{y}=\frac{q}{p}

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm

and the negative sign means that the image is inverted.

#LearnwithBrainly

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We see that the first displacement D_{1} = 30 km (blue arrow) is along the y-axis, but the second part of the ship's journey D_{2} = 20 km (red arrow) is at an angle with reference to y-axis.

So, we first find the components of the red arrow along X and Y.

Component of D_{2} along X-axis is given by  D_{2x}  = D_{2} Sin 30 = 10 km

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Net Displacement along X  D_{netX} = 10 km

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6 0
3 years ago
1. Is the sequence geometric? If so, identify the common ratio.
frozen [14]
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2)</span><span>Next Term (or nth term) = ar^n-1 
</span>
a = first term, i.e. 5 
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<span>n = .. </span>
<span>as you already have 1st , 2nd and 3rd terms</span>
<span>substituting now </span>
<span>T4= ar^n-1 </span>
<span>= 5*3^4-1 </span>
<span>= 5*3^3 </span>
<span>= 5*27 </span>
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<span>= 5*3^4 </span>
<span>= 5*81 </span>
<span>T5 = 405 </span>


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After the second meter,  96.6%  of what entered it emerges from it, and
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of the fiber.

==>  After 2 meters, the intensity has dwindled to  (0.966)² of its original level.
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==>  After  'x'  meters of fiber, the remaininglight intensity is (0.966) ^x-power
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If you shine 1,500 lumens into the front of the fiber, then after 'x' meters of
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lumens of light remaining.
 
=========================================

The genius engineers in the fiber design industry would not handle it this way.
When they look up the 'attenuation' of the cable in the fiber manufacturer's
catalog, it would say  "15dB per 100 meters".

What does that mean ?    Break it down:  15dB in 100 meters is <u>0.15dB per meter</u>.
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Look at this  ==>      10 log(0.966) =  <em><u>-0.15</u>  </em>  <==  loss per meter, in dB .

Armed with this information, the engineer ... calculating the loss in  'x'  meters of
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--  0.15 dB loss per meter

--  'x' meters of cable

--  0.15x dB of loss.

If  'x' happens to be, say,  72 meters, then the loss is  (72) (0.15) = 10.8 dB .

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Sorry. Didn't mean to ramble on. But I do stuff like this every day.
5 0
3 years ago
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Answer:

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D = 0.500 g/mL

V = ?

0.500 g/mL = \frac{125 g}{V}

V = 250 mL

3 0
3 years ago
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