Respuesta :
Answer:
226 cm^3
The mass of plastic used to make cylinder is greater
Step-by-step explanation:
Given:-
- The density of cone material, ρc = 1.4 g / cm^3
- The density of cylinder material, ρl = 0.8 g / cm^3
Solution:-
- To determine the volume of plastic that remains in the cylinder after gouging out a hemispherical amount of material.
- We will first consider a solid cylinder with length ( L = 10 cm ) and diameter ( d = 6 cm ). The volume of a cylinder is expressed as follows:
[tex]V_L =\pi \frac{d^2}{4} * L[/tex]
- Determine the volume of complete cylindrical body as follows:
[tex]V_L = \pi \frac{(6)^2}{4} * 10\\\\V_L = 90\pi cm^3\\[/tex]
- Where the volume of hemisphere with diameter ( d = 6 cm ) is given by:
[tex]V_h = \frac{\pi }{12}*d^3[/tex]
- Determine the volume of hemisphere gouged out as follows:
[tex]V_h = \frac{\pi }{12}*6^3\\\\V_h = 18\pi cm^3[/tex]
- Apply the principle of super-position and subtract the volume of hemisphere from the cylinder as follows to the nearest ( cm^3 ):
[tex]V = V_L - V_h\\\\V = 90\pi - 18\pi \\\\V = 226 cm^3[/tex]
Answer: The amount of volume that remains in the cylinder is 226 cm^3
- The volume of cone with base diameter ( d = 6 cm ) and height ( h = 5 cm ) is expressed as follows:
[tex]V_c = \frac{\pi }{12} *d^2 * h[/tex]
- Determine the volume of cone:
[tex]V_c = \frac{\pi }{12} *6^2 * 5\\\\V_c = 15\pi cm^3[/tex]
- The mass of plastic for the cylinder and the cone can be evaluated using their respective densities and volumes as follows:
[tex]m_i = p_i * V_i[/tex]
- The mass of plastic used to make the cylinder ( after removing hemispherical amount ) is:
[tex]m_L = p_L * V\\\\m_L = 0.8 * 226\\\\m_L = 180.8 g[/tex]
- Similarly the mass of plastic used to make the cone would be:
[tex]m_c = p_c * V_c\\\\m_c = 1.4 * 15\pi \\\\m_c = 65.973 g[/tex]
Answer: The total weight of the cylinder ( m_l = 180.8 g ) is greater than the total weight of the cone ( m_c = 66 g ).
The volume of the remaining plastic in the cylinder is large, which
makes the weight much larger than the weight of the cone.
Responses:
- (a) Volume of the remaining plastic in the cylinder is 226 cm³
- (b) The weight of the cylinder is greater than the weight of the cone.
How can the weight and volume be evaluated?
Density of the plastic for the cone = 1.4 g/cm³
Density of the plastic used for the cylinder = 0.8 g/cm³
From a similar question, we have;
Height of the cylinder = 10 cm
Diameter of the cylinder = 6 cm
Height of the cone = 5 cm
(a) Radius of the cylinder, r = 6 cm ÷ 2 = 3 cm
Volume of a cylinder = π·r²·h
Volume of a hemisphere = [tex]\mathbf{\frac{2}{3}}[/tex] × π× r³
Volume of the cylinder after it has been hollowed out, V, is therefore;
- [tex]V = \mathbf{\pi \times r^2 \times h - \frac{2}{3} \times \pi \times r^3}[/tex]
Which gives;
[tex]V = \pi \times 3^2 \times 10 - \frac{2}{3} \times \pi \times 3^3 \approx \mathbf{ 226}[/tex]
- Volume of the cylinder after it has been hollowed out, V ≈ 226 cm³
(b) Volume of the cone = [tex]\mathbf{\frac{1}{3}}[/tex] × π × 3² × 5 ≈ 47.1
Mass of the cone = 47.1 cm³ × 1.4 g/cm³ ≈ 66 g
Mass of the hollowed cylinder ≈ 226 cm³ × 0.8 g/cm³ = 180.8 g
- The mass and therefore, the weight of the plastic that makes up the hollowed cylinder is greater than the weight of the plastic that makes up the cone.
Learn more about volume and density of solids here:
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