Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Answer:
The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm
Explanation:
The given parameters of the basketball are;
The diameter of the basketball = 23 cm (9 inches)
The distance of the light bulb from the closest side of the basketball = 3 m
The distance from the ball to the wall = 4 m
The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m
The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)
Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115
The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m
Therefore;
((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115
∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115
((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm
The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm
The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
