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pishuonlain [190]
3 years ago
8

Lidia plans an experimental investigation to see how the thickness of a lens affects the point where a beam of light is focused.

She plans to shine light through a lens and measure the distance to a screen that is placed where the beam focuses.
What are the controls in Lidia’s investigation?

the material and thickness of the lens
the type of light and the material of the lens
the distance to the screen and the type of light
the thickness of the lens and the distance to the screen
Physics
1 answer:
Nostrana [21]3 years ago
4 0

Answer:

Option B

the type of light and the material of the lens

Explanation:

Since the experiment focuses on investigating how thickness of a lens affects the point where a beam of light, it means thickness of lens here are indepedent variable since Lydia has control over the thickness and can alter it. However, the point where beam of light focuses on is depedent variable since Lydia will just observe the point but can't change it. To ensure that other external factors don't affect the results, Lydia has to ensure that the type of material irrespective of thickness remain constant and also the source of light must be kept constant. Therefore, the controls will be the type of light and the material of the lens

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CHECK COMPLETE QUESTION

The cube of wood having an edge dimension of 19.7cm and a density of 647kg/m3 floats on water

(a) What is the distance from the horizontal top surface of the cube to the water level answer in cm

(b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface answer in kg

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b)2.78 kg

Explanation:

Given:

Let us calculate the volume first, we were given dimension as 19.7cm=0.197m

Volume is (0.197 meters)³ = 0.00764m³

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weight = 4.947 x 9.8 N/kg = 48.44N

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which is equal to the weight of volume of the displaced water. Or the mass, the calculation is the same.

We know that density of fresh water at 20ºC is 998 kg/m³

Then we can calculate the volume of displaced water as

4.947 kg / 998 kg/m³ = 0.00496 m³

We know that the displaced water has a shape of a rectangular solid with 0.197 meters on the two horizontal dimensions, and h as the height to the surface then

V = 0.197²h = 0.00496

0.00496= 0.197²h

h = 0.1278 meters or 12.78 cm

Then the the distance exposed, would be 19.7–12.78 = 6.29 cm

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mass of displaced water is 0.00774m³ x 998 kg/m³ = 7.724 kg

Added mass is the mass of the displaced water – mass of block

= 7.724–4.947 = 2.78 kg

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