Find the surface area of the composite solid. Leave your answer in terms of pi

Mathematics

1 answer:

aev [14]3 years ago

3 0

We can separate this solid into two pieces: a cylinder and a cone. I will be finding the surface area's separately, then adding them together in the end.

Cylinder: The formula for the surface area of a cylinder is 2πrh+2πr². HOWEVER, since the cone and cylinder are connected, they don't count as part of the surface area, we don't need the 2 in 2πr². The radius, or r, is 5 cm. The height, or h, is 12 cm. Then, we use those numbers and plug it in the formula.

2πrh+πr²

2π(5)(12)+π(5)²

120π+25π

145π cm²

Cone: The surface area of the cone without its base is πrL where r is the radius (5) and L is the lateral side (13)
π·5·13=65π

Total: We add the total SA of the cylinder and cone together to find the total: 145π+65π = 210π cm²

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The average density of the material in intergalactic space is approximately 2.5 × 10–27 kg/m3. what is the volume of a lead samp

Lesechka [4]

Answer:

e. $1.8\times 10^{-6}m^3$

Step-by-step explanation:

It is given that,

The density of intergalactic space material is $2.5 \times 10^{-27}$ kg per cubic meter.

And the volume of intergalactic space material is $8.0\times 10^{24} m^3$

So the mass of that much intergalactic space material is,

$m= \rho\times v$ , where 'm' is the mass, and 'v' is the volume.

Putting the values we get,

$m=2.5 \times 10^{-27}\times 8.0 \times 10^{24}$

$m=20\times 10^{(-27+24)}=20\times 10^{-3} kg$

It is also given that the mass of lead is the same as the mass of the intergalactic space material. Therefore, the mass of lead is $20 \times 10^{-3}=2 \times 10^{-2}kg$