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Fiesta28 [93]
3 years ago
15

An object is 2.0 cm from a double convex lens with a focal length of 1.5 cm. Calculate the image distance

Physics
1 answer:
Tomtit [17]3 years ago
3 0

Answer:

0.857 cm

Explanation:

We are given that:

The focal length for a convex lens to be (f) = 1.5cm

The object distance (u) = - 2.0 cm

We are to determine the image distance (v) = ??? cm

By applying the lens formula:

\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}

By rearrangement and making (v) the subject of the above formula:

v = \dfrac{uf}{u-f}

replacing the given values:

v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}

v = \dfrac{-3.0}{(-3.5)}

v = 0.857 cm

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The heat coming from the sun warms the land more quickly than the sea. As a result of these, the air near the land warm up and rises and the cooler air from the sea moves in to replace the risen air. The correct answer is option A

There will be heat transfer from a region of higher temperature to the region of lower temperature. But in the case of land and sea breeze, the transfer of heat are the result of convectional current in nature. Because the land is a better absorber of heat and also has a lower specific heat capacity compare to sea, during the day, the heat coming from the sun warms the land more quickly than the sea. As a result of these, the air near the land warm up and rises.

The cooler air from the sea moves in to replace the risen air.

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Therefore, option A is correct

Learn more here : brainly.com/question/1114842

4 0
3 years ago
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Luis and Aisha conducted an experiment. They exerted different forces on four objects. Their results are shown in the table.
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Answer:

Object 3 has greatest acceleration.

Explanation:

Objects               Mass                                Force

1                            10 kg                               4 N              

2                           100 grams                       20 N

3                            10 grams                         4 N

4                             1 kg                                 20 N

Acceleration of object 1,

a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{4}{10}\\\\a_1=0.4\ m/s^2

Acceleration of object 2,

a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{20}{0.1}\\\\a_2=200\ m/s^2

Acceleration of object 3,

a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{4}{0.01}\\\\a_3=400\ m/s^2

Acceleration of object 4,

a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{20}{1}\\\\a_3=20\ m/s^2

It is clear that the acceleration of object 3 is 400\ m/s^2 and it is greatest of all. So, the correct option is (3).

4 0
3 years ago
Read 2 more answers
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