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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 certain cloud contains 220 water droplets per cubic centimeter. If 1cubic meter = 1,000,000cubic centimeters, how many drops a
lianna [129]
Wouldn't you just have to multiply 220 by 1,000,000? That would mean there are 220,000,000 water droplets in one cubic of the cloud.
3 0
3 years ago
A model train traveling at a constant speed around a circular track has a constant velocity
TiliK225 [7]
A model train traveling at a constant speed around a circular track has a constant velocity. FALSE.
Hope this helps you!

3 0
3 years ago
State galileo's law of inertia​
nekit [7.7K]

Answer:

Galileo's law of inertia states that; if no net force acts on an object, the object maintains its state of motion.

(The first law of motion is also known as Galileo's law of inertia)

8 0
3 years ago
A 2 kg toy cart and a 6 kg toy cart have a spring compressed between them. When the spring expands, it sends the 2 kg toy cart o
harkovskaia [24]

Answer:

The speed of second toy cart is 4 m/s.

(c) is correct option

Explanation:

Given that,

Mass of first toy cart = 2 kg

Mass of second toy cart = 6 kg

Speed of first toy cart = 12 m/s

We need to calculate the speed of second toy cart

Using formula of momentum

m_{1}v_{1}=m_{2}v_{2}

Where, m₁ = mass of first toy cart

m₂ = mass of second toy cart

v₁ = velocity of first toy cart

v₂ =  velocity of second toy cart

Put the value into th formula

2\times12=6\times v_{2}

v_{2}=\dfrac{2\times12}{6}

v_{2}=4\ m/s

Hence, The speed of second toy cart is 4 m/s.

(c) is correct option

4 0
3 years ago
A solid sphere of mass 2.50 kg and radius 0.120 m is at rest at the top of a ramp inclined 15.0°. It rolls to the bottom without
Lubov Fominskaja [6]

Answer:

KE= 30 J

Explanation:

Given that

m= 2.5 kg

r= 0.12 m

θ = 15°

h= 1.2 m

As we know that solid sphere rolls without slipping.It means that

v= ω r

v=Linear velocity

ω=Angular speed

r=radius

The total kinetic energy KE

KE=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

Moment of inertia of solid sphere I

I=\dfrac{2}{5}mr^2

Now from energy conservation

Energy at top = Energy at bottom

The potential energy at top = m g h

Potential energy at bottom = 0

Kinetic energy at top = 0

Kinetic energy at bottom = KE

m g h + 0 = 0 + KE

KE= 2.5 x 10 x 1.2                 ( take g =10 m/s²)

KE= 30 J

7 0
3 years ago
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