At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
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
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 visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
