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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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An example of an aerobic activity would be A, Cross-country skiing
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Complete the sentence below using one of the following
astra-53 [7]

Answer:

equilibrium

Explanation:

acceleration will remain constant

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What type of motion is shown with this graph? (5 points)
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It should be Constant speed. The line goes straight & doesn’t change within the graph.
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Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th
elena-14-01-66 [18.8K]

The peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

<h3>Relationship between electric and magnetic field</h3>

The relationship between electric and magnetic field at a given peak electric field is given as;

c = (E₀) / (B₀)

where;

  • c is speed of light
  • E₀ is the peak electric field
  • B₀ is the peak magnetic field

B₀ = E₀ / c

B₀ = (2.9) / (3 x 10⁹)

B₀ = 9.67 x 10⁻¹⁰ T

Thus, the peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

Learn more about peak magnetic field here: brainly.com/question/24487261

8 0
2 years ago
You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 77.2 kg hop on board for a ride
lions [1.4K]

To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.

By Hook's law we know that force is defined as,

F= kx

Where,

k = spring constant

x = Displacement change

PART A) For the case of the spring constant we can use the above equation and clear k so that

k= \frac{F}{x}

k = \frac{mg}{x}

k= \frac{77.2*9.8}{0.0637}

k = 11876.92N/m

Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m

PART B) In the case of speed we can obtain it through the period, which is given by

T = \frac{2\pi}{\omega}

Re-arrange to find \omega,

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{2.14}

\omega = 2.93rad/s

Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to

\omega^2 = \frac{k}{m}

m = \frac{k}{\omega^2}

m = \frac{ 11876.92}{2.93}

m = 4093.55Kg

Therefore the mass of the trailer is 4093.55Kg

PART C) The frequency by definition is inversely to the period therefore

f = \frac{1}{T}

f = \frac{1}{2.14}

f = 0.4672 Hz

Therefore the frequency of the oscillation is 0.4672 Hz

PART D) The time it takes to make the route 10 times would be 10 times the period, that is

t_T = 10*T

t_T = 10 *2.14s

t_T = 21.4s

Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s

5 0
4 years ago
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