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dem82 [27]
2 years ago
12

What is the mass of a cylinder of lead with a radius of 2 centimeters and a height of 6 centimeters, given that the density of l

ead is 11.4g/cm^3?
Mathematics
1 answer:
ycow [4]2 years ago
7 0

The mass of lead cylinder is: 859.10 grams

Step-by-step explanation:

In order to find the mass of a cylinder we have to find the volume first

Given

Radius = r = 2 cm

Height = h = 6 cm

The volume of cylinder is given by:

V = \pi r^2h

Putting the values

V = 3.14 * (2)^2 * 6\\= 3.14*4*6\\=75.36\ cm^3

Now,

Density of lead per cm^3 = 11.4g

Mass\ of\ cylinder = Volume*Density\\= 75.36*11.4\\=859.10\ grams

The mass of lead cylinder is: 859.10 grams

Keywords: Volume, Density

Learn more about volume at:

  • brainly.com/question/2115122
  • brainly.com/question/2116906

#LearnwithBrainly

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Given: △ABC, AB=5sqrt2 <br> m∠A=45°, m∠C=30°<br> Find: BC and AC
Marysya12 [62]

BC is 10 units and AC is 5+5\sqrt{3} units

Step-by-step explanation:

Let us revise the sine rule

In ΔABC:

  • \frac{AB}{sin(C)}=\frac{BC}{sin(A)}=\frac{AC}{sin(B)}
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Let us use this rule to solve the problem

In ΔABC:

∵ m∠A = 45°

∵ m∠C = 30°

- The sum of measures of the interior angles of a triangle is 180°

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∴ 45 + m∠B + 30 = 180

- Add the like terms

∴ m∠B + 75 = 180

- Subtract 75 from both sides

∴ m∠B = 105°

∵ \frac{AB}{sin(C)}=\frac{BC}{sin(A)}

∵ AB = 5\sqrt{2}

- Substitute AB and the 3 angles in the rule above

∴ \frac{5\sqrt{2}}{sin(30)}=\frac{BC}{sin(45)}

- By using cross multiplication

∴ (BC) × sin(30) = 5\sqrt{2} × sin(45)

∵ sin(30) = 0.5 and sin(45) = \frac{1}{\sqrt{2}}

∴ 0.5 (BC) = 5

- Divide both sides by 0.5

∴ BC = 10 units

∵ \frac{AB}{sin(C)}=\frac{AC}{sin(B)}

- Substitute AB and the 3 angles in the rule above

∴ \frac{5\sqrt{2}}{sin(30)}=\frac{AC}{sin(105)}

- By using cross multiplication

∴ (AC) × sin(30) = 5\sqrt{2} × sin(105)

∵ sin(105) = \frac{\sqrt{6}+\sqrt{2}}{4}

∴ 0.5 (AC) = \frac{5+5\sqrt{3}}{2}

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Learn more:

You can learn more about the sine rule in brainly.com/question/12985572

#LearnwithBrainly

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