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kirill [66]
3 years ago
10

a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde

d. find the radius and height of the cone with maximum volume that can be formed in this way.
Physics
1 answer:
Svet_ta [14]3 years ago
3 0

The expression for the radius and height of the cone can be obtained from

the property of a function at the maximum point.

  • The \ radius, \ of \ the \ base \ of \ the \ cone \ is \ \sqrt{ \dfrac{3}{4}} \times radius \ of \ circular \ sheet \ metal
  • The height of the cone is half the length of the radius of the circular sheet metal.

Reasons:

The part used to form the cone = A sector of a circle

The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.

Volume \ of \ a \ cone = \dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot h

  • Volume \ of \ a \ cone, \, V = \dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}

θ/360·2·π·s = 2·π·r

Where;

s = The radius of he circular sheet metal

h = s² - r²

  • \dfrac{dV}{dr} = \dfrac{d}{dr}  \left(\dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}\right) = \dfrac{\pi \cdot (3 \cdot r^2 \cdot  s^2 - 4 \cdot r^4)}{\sqrt{(s^2- r^2)}} = 0

3·r²·s² - 4·r⁴ = 0

3·r²·s² = 4·r⁴

3·s² = 4·r²

\underline{\left  \right. The \ radius, \, r =\sqrt{ \dfrac{3}{4}} \cdot s}

\underline{The \ height, \, h =\sqrt{s^2 - \dfrac{3}{4}\cdot s^2} = \dfrac{s}{2}}}

Learn more here:

brainly.com/question/14466080

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A. How many calories are needed to raise the temperature of 1 gram of water by 1 °C?
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A) 1 cal

B) 80 cal

C) 540 cal

Explanation:

A)

The amount of heat energy needed to raise the temperature of a certain mass of a substance is given by

Q=mC\Delta T

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C is the specific heat capacity

\Delta T is the change in temperature

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m = 1 g is the mass of water

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\Delta T=1^{\circ}C is the change in temperature

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B)

For a solid substance at its melting point, the amount of heat needed to melt completely the substance is given by

Q=m\lambda_f

where

m is the mass of the substance

\lambda_f is the specific latent heat of fusion of the substance

In this problem:

- The ice is already at melting point, 0 °C

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So, the heat needed is

Q=(1)(80)=80 cal

C)

For a liquid substance at its boiling point, the amount of heat needed to boil completely the substance is given by

Q=m\lambda_v

where

m is the mass of the substance

\lambda_v is the specific latent heat of vaporization of the substance

In this problem:

- The water is already at boiling point, 100 °C

- Mass of the water: m=1g

- Specific latent heat of vaporization of water: \lambda_v=540 cal/g

So, the heat needed is

Q=(1)(540)=540 cal

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