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tiny-mole [99]
2 years ago
8

2. You will need a magnifying glass and a small piece of scrap paper for this demonstration. a. Go outside on a sunny day and pu

t the piece of paper on a patch of asphalt, concrete, or a large flat rock. Make sure there is no other combustible material or debris in the vicinity. b. Hold the magnifying glass so it is close to the paper, with rays of sunlight shining through it. Move the magnifying glass side to side and close or away from the paper until the most concentrated ray of light is shining on the paper. This is the focal point for that magnifying glass. c. Measure the distance between the focal point and magnifying glass. This is the focal length for that magnifying glass. (Be careful; if you hold the magnifying glass in that place for long enough, the paper will catch fire. Do not ever do this on your skin or clothing, as it can cause serious burns.) d. Now take the magnifying glass inside and use it to look at the words in a book. What is the focal length now?
Physics
1 answer:
andrew-mc [135]2 years ago
4 0

The focal length of a magnifying glass is the distance between the focal point and optical centre of the magnifying glass.

<h3>Focal length</h3>

The focal length, f is the distance from a lens or mirror to the focal point, F.

This is the distance from a lens or mirror at which parallel light rays will meet for a converging lens or mirror or appear to diverge from for a diverging lens or mirror.

A magnifying glass is a converging lens which produces a enlarged, erect and virtual image when an object is placed between the focal point and optical centre.

A magnifying glass will bring to focus at a point sun rays which can cause the paper to catch fire if it is held in place for long.

This point at which the most concentrated ray of light is shining on the paper, is the focal point for that magnifying glass.

Therefore, the focal length of a magnifying glass is the distance between the focal point and optical centre of the magnifying glass.

Learn more about about focal length at: brainly.com/question/25779311

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A) 

     It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

S=S_{o}+v_{o}t+ \frac{at^2}{2} \\ 20= \frac{10t^2}{2} \\ t=2s
 
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B)
 
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v_{y}=v_{y_{o}}+gt \\ v_{y}=10\times2 \\ \boxed {v_{y}=20m/s}
  
     The angle of impact is given by:

cos(\theta) =\frac{v_{x}}{v_{y}}  \\ cos(\theta) = \frac{18}{20}  \\ cos(\theta) =0.9 \\ arccos(0.9)=\theta \\ \boxed {\theta \approx 25.84}


If you notice any mistake in my english, please let me know, because i am not native.

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2 years ago
A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre
pantera1 [17]

Answer:

Part a)

v = \sqrt{xg + \frac{MLg}{m}}

Part b)

t = 12 s

Explanation:

Part a)

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T = \frac{m}{L}xg + Mg

so the speed of the wave at that position is given as

v = \sqrt{\frac{T}{\mu}}

here we know that

\mu = \frac{m}{L}

now we have

v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}

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Part b)

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