What is 6.42 written as a fraction in lowest terms?
1 answer:
You might be interested in
This question is Incomplete because it lacks the diagram showing the containers as well as the required options. Find attached to this answer the appropriate diagram.
Complete Question
Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufactures such dewars of different sizes. Shown here are two dewars, both of the same height but of different diameters. How much aluminium will be required to cover Dewar A compared to that required for Dewar B? (Note: The base and the lid are NOT made up of aluminium.)
A) 1/3 times
B) 3 times
C) 9 times
D) The same amount of aluminum will be required.
Answer:
B) 3 times
Step-by-step explanation:
From the question, we are told that we are to find the amount of aluminum to cover Dewar A and B , we are also told that the base and the lid are not covered with aluminum.
From this information, we can determine that we are to find the curved or Lateral surface area of the cylinder because the base and lid are not included
The curved surface area of a cylinder is calculated as 2πrh
We told that both cylinders have the same height. Hence,
For Dewar A
We have Diameter of 30 cm, radius = Diameter ÷ 2 = 30cm ÷ 2 = 15cm.
Height = h
The curved surface area = 2πrh
= 2 × π × 15 ×h
= 30πhcm²
For Dewar B
We have Diameter of 10 cm, radius = Diameter ÷ 2 = 10cm ÷ 2 = 5cm.
Height = h
The curved surface area = 2πrh
= 2 × π × 5 ×h
= 10πhcm²
When we compare the curved surface Dewar A to the curved surface area of Dewar B
Dewar A : Dewar B
30πhcm² : 10πhcm²
3(10πhcm²) : 10πhcm²
From the above comparison, we can see that 3 times more aluminum is required to cover Dewar A compared to Dewar B.
