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Ray Of Light [21]
3 years ago
7

A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold 500 cm^3 of liquid. The mater

ial for the top and bottom costs 0.02 cent/cm^2, and the material for the sides costs 0.01 cent/cm^2.
a) Estimate the radius r and height h of the can that costs the least to manufacture. [suggestion: Express the cost C in terms of r.]
Physics
1 answer:
Rom4ik [11]3 years ago
4 0
The area of the top and bottom:
2πr²
Cost for top and bottom:
2πr²  x 0.02
= 0.04πr²

Area for side:
2πrh
Cost for side:
2πrh x 0.01
= 0.02πrh

Total cost:
C = 0.04πr² + 0.02πrh

We know that the volume of the can is:
V = πr²h
h = 500/πr²

Substituting this into the cost equation to get a cost function of radius:
C(r) = 0.04πr² + 0.02πr(500/πr²)
C(r) = 0.04πr² + 10/r

Now, we differentiate with respect to r and equate to 0 to obtain the minimum value:

0 = 0.08πr - 10/r²
10/r² = 0.08πr
r³ = 125/π

r = 3.41 cm
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3 years ago
According to the law of conservation of matter, what number must be the same on each side of a chemical equation?
olga2289 [7]

Answer:

4. The number of atoms of each element

Explanation:

<em><u>Why is 1 wrong?</u></em>

The number of substances (i.e. number of mols of substances) do not have to be the same. For instance, 2 mol of water reacts with 1 mol of oxygen to form 2 mol of water. Obviously the numer of substances is not conserved. Thus, it isn't a requirement.

<em><u>Why is 2 wrong?</u></em>

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<em><u>Why is 3 wrong?</u></em>

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<u><em>Why is 4 correct?</em></u>

The number of atoms are always conserved as atoms cannot be broken or created.

<em><u>Possible confusion...</u></em>

The number of atoms are always conserved but the number of molecules may not always be conserved.

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7 0
3 years ago
A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m
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Answer:

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Explanation:

Given that,

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Height = 1.8 m

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Using conservation of energy

K.E+P.E=P.E+K.E

\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2

\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}

Put the value into the formula

\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}

\dfrac{v_{1}^2}{2}=18.62

v_{1}=\sqrt{2\times18.62}

v_{1}=6.10\ m/s

Hence, The minimum speed when she leave the ground is 6.10 m/s.

6 0
3 years ago
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