3. If the distance of the screen is moved from 100. cm to 200. cm the area of light would in-crease from 150. cm^2 to

A. 37.5 cm2

B. 75 cm2

C. 300 cm2

D. 600 cm24.

Light is projected onto a screen 75.0 cm from a light source. The light sensor records a reading of 4,436 lux at this distance. If the screen is moved from 75.0 cm to 150. cm, the light sensor reading would change to

A. 17,700 lux

B. 8,870 lux

C. 2,218 lux

D. 1,110 lux

5. If you use a screen in the last problem and tilt it the light reading would go down, this is be-cause.

A. In science when we tilt things they always get lower

B. As we tilt the screen the area of light increases and spreads the light over a lager area.

C. As we tilt the serene the area of light decreases and makes the light more concentrated.

3) C

4 D

5) C

Explanation:

3) given that

Initial distance of the screen = 100cm

Initial area = 150 cm^2

Final distance = 200 cm

The intensity of light is inversely proportional to the square of the distance. That is

Intensity of light I = 1/d2

And also I = P/A

1/d^2 = P/A

P = A/d^2

P1 = P2

150/100 = A/200

1.5 = A/200

A = 1.5 × 200

A = 300 cm^2

4.) Light is projected onto a screen 75.0 cm from a light source. The light intensity = 4436 lux

If the screen is moved from 75.0 cm to 150. cm, the light sensor reading will be

Using inverse square law

I = 1/d^2

I×d^2 = constant. Therefore,

4436 × 75^2 = I × 150^2

I = 24952500/22500

I = 1109 lux

5.) We can express the relationship between luminosity, brightness, and distance with a simple formula.

As we tilt the serene the area of light decreases and makes the light more concentrated.

search it up!

explanation:

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.