Question

Zia is building a plastic model rocket that has the combined shape of a cone and a cylinder as shown. additionally, the cylinder has a hemisphere hollowed out of its bottom. the plastic for the cone weighs 1.4 grams per cubic centimeter and the plastic for the cylinder weights only 0.8 grams per cubic centimeter.
(a) the volume of plastic that remains in the cylinder after it has been hollowed out to the nearest cubic centimeter.
(b) what has a greater total weight, the plastic that makes up the cone or the plastic that makes up the cylinder after it has been hollowed out?

Answer 1

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:

                                  $V_L =\pi \frac{d^2}{4} * L$

- Determine the volume of complete cylindrical body as follows:

   

                                 $V_L = \pi \frac{(6)^2}{4} * 10\\\\V_L = 90\pi cm^3$

- Where the volume of hemisphere with diameter ( d = 6 cm ) is given by:

                                 $V_h = \frac{\pi }{12}*d^3$

- Determine the volume of hemisphere gouged out as follows:

                                 $V_h = \frac{\pi }{12}*6^3\\\\V_h = 18\pi cm^3$

- Apply the principle of super-position and subtract the volume of hemisphere from the cylinder as follows to the nearest ( cm^3 ):

                               $V = V_L - V_h\\\\V = 90\pi - 18\pi \\\\V = 226 cm^3$

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:

   

                               $V_c = \frac{\pi }{12} *d^2 * h$

- Determine the volume of cone:

                              $V_c = \frac{\pi }{12} *6^2 * 5\\\\V_c = 15\pi cm^3$

- The mass of plastic for the cylinder and the cone can be evaluated using their respective densities and volumes as follows:

                             $m_i = p_i * V_i$

- The mass of plastic used to make the cylinder ( after removing hemispherical amount ) is:

                           $m_L = p_L * V\\\\m_L = 0.8 * 226\\\\m_L = 180.8 g$

- Similarly the mass of plastic used to make the cone would be:

                           $m_c = p_c * V_c\\\\m_c = 1.4 * 15\pi \\\\m_c = 65.973 g$

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 ).