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ExtremeBDS [4]
1 year ago
14

Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufa

ctures 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.)

Mathematics
1 answer:
Nitella [24]1 year ago
6 0

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.

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The area of the triangle shown is 40.0 cm2.
kompoz [17]

Answer:

The height of the triangle is 6.4cm

Step-by-step explanation:

- If the triangle has a side that measures 12.5cm, it means that two of its sides are equal in length.

Which is equivalent to the description of an isosceles triangle.

To find the height, use the equation of the area.

Area = (Base * height) / 2

- We know the value of the area and its shorter side that could be the base.

40cm = (12.5 cm * h) / 2

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80cm / 12.5cm = h

h = 6.4cm

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If candies are sold at 3 pcs for php.2.00.how many candies can melai get if she has php20.00?​
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1 year ago
Solve the following multiplication and division problems. a. 8 T. 1,398 lb. 14 oz. × 6 b. 349 lb. 6 oz. ÷ 130 c. 6 T. 294 lb. ÷
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Answer:  a) 278382 oz

b) 43 oz

c) 64.1 oz

Step-by-step explanation:

a) 8 T. 1,398 lb. 14 oz. × 6

First we need to change it in 'oz'.

As we know that

1\ ton=32000\ oz\\\\1\ lb=16\ oz

so, it becomes,

8\times 32000+1398\times 16+14\ oz\\\\=278382\ oz

8 T. 1,398 lb. 14 oz. × 6 becomes

278382\times 6\\\\=1670292\ oz

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It becomes,

349\times 16+6\ oz\\\\=5590\ oz\\\\\text{ at last it becomes}\\\\=\frac{5590}{130}\\\\=43\ oz

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First it becomes,

6\times 32000+294\times 16\ oz\\\\=196704\ oz

At last it becomes,

\frac{196704}{3071}\\\\=64.1\ oz

Hence, a) 278382 oz

b) 43 oz

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