6.1 a medium pizza is 12 in across and costs $12. a large pizza is 16 in across and costs $18. which one is the better deal? justify your answer and state any assumptions.
6.2 a) the sun is roughly 100 times larger than the earth by diameter. about how many earths would you expect could “fit inside the sun”, using all of the available volume? b) actually, the ratio of linear scales is closer to 109 times larger for the sun. redo the prior calculation and quantify the percentage error associated with using the lower precision estimate. explain why it is not 9%.
6.3 the radii of mars and the earth are around 2,000 miles and 4,000 miles, respectively what do you expect for the ratio of their masses? look up the actual masses and explain any discrepancies.
6.4 the moon is about 1/4 the linear size of the earth (by radius). the gravitational acceleration on the moon is about 1/6 of that on the earth. what does this tell you about the moon’s composition relative to that of the earth?

Question

Accepted Answer

### 6.1
# Explanation:
## Step1: Calculate area of medium pizza
Assume pizza is circular. Area formula $A = \pi r^{2}$, diameter of medium pizza $d_{1}=12$ in, so radius $r_{1}=\frac{12}{2}=6$ in. Then $A_{1}=\pi	imes6^{2}=36\pi$ square - inches. Cost - per - area of medium pizza $C_{1}=\frac{12}{36\pi}=\frac{1}{3\pi}$ dollars per square - inch.
## Step2: Calculate area of large pizza
Diameter of large pizza $d_{2}=16$ in, so radius $r_{2}=\frac{16}{2}=8$ in. Then $A_{2}=\pi	imes8^{2}=64\pi$ square - inches. Cost - per - area of large pizza $C_{2}=\frac{18}{64\pi}=\frac{9}{32\pi}$ dollars per square - inch.
## Step3: Compare cost - per - area
Find a common denominator for $C_{1}$ and $C_{2}$. $C_{1}=\frac{1}{3\pi}=\frac{32}{96\pi}$ and $C_{2}=\frac{9}{32\pi}=\frac{27}{96\pi}$. Since $C_{2}<C_{1}$, the large pizza is a better deal.
# Answer:
The large pizza is the better deal. Assumptions: The pizza is circular and of uniform thickness.

### 6.2a
# Explanation:
## Step1: Recall volume formula for sphere
The volume of a sphere is $V=\frac{4}{3}\pi r^{3}$. If the diameter of the Sun is $D_{s}$ and of the Earth is $D_{e}$, and $D_{s} = 100D_{e}$, then $r_{s}=100r_{e}$ (where $r_{s}$ and $r_{e}$ are radii of the Sun and Earth respectively).
## Step2: Calculate ratio of volumes
$V_{s}=\frac{4}{3}\pi r_{s}^{3}$ and $V_{e}=\frac{4}{3}\pi r_{e}^{3}$. The ratio $\frac{V_{s}}{V_{e}}=\frac{\frac{4}{3}\pi r_{s}^{3}}{\frac{4}{3}\pi r_{e}^{3}}=(\frac{r_{s}}{r_{e}})^{3}$. Since $\frac{r_{s}}{r_{e}} = 100$, then $\frac{V_{s}}{V_{e}}=100^{3}=1000000$. So about 1 million Earths could fit inside the Sun.
# Answer:
About 1000000 Earths could fit inside the Sun.

### 6.2b
# Explanation:
## Step1: Calculate new volume ratio
If $\frac{r_{s}}{r_{e}} = 109$, then the actual volume ratio $\frac{V_{s}}{V_{e}}=(\frac{r_{s}}{r_{e}})^{3}=109^{3}=1295029$.
## Step2: Calculate percentage error
The estimated volume ratio was $100^{3}=1000000$. Error $E=\frac{	ext{Actual}-	ext{Estimated}}{	ext{Actual}}	imes100\%=\frac{1295029 - 1000000}{1295029}	imes100\%\approx22.77\%$. It is not 9% because the volume ratio is a cubic relationship. A 9% error in the linear scale (radius) gets magnified when cubed.
# Answer:
The percentage error is approximately $22.77\%$. The error is not 9% because volume is a cubic function of the radius.

### 6.3
# Explanation:
## Step1: Assume density is uniform
If we assume the planets are spheres and have uniform density $
ho$, mass $m=
ho V$ and $V = \frac{4}{3}\pi r^{3}$. Let $r_{M}$ be the radius of Mars and $r_{E}$ be the radius of Earth. Then $\frac{m_{M}}{m_{E}}=\frac{
ho V_{M}}{
ho V_{E}}=\frac{\frac{4}{3}\pi r_{M}^{3}}{\frac{4}{3}\pi r_{E}^{3}}=(\frac{r_{M}}{r_{E}})^{3}$. Given $r_{M} = 2000$ miles and $r_{E}=4000$ miles, $\frac{r_{M}}{r_{E}}=\frac{1}{2}$, so $\frac{m_{M}}{m_{E}}=(\frac{1}{2})^{3}=\frac{1}{8}$.
## Step2: Look up actual masses
The actual mass of Mars $m_{M}\approx6.42	imes10^{23}$ kg and the actual mass of Earth $m_{E}\approx5.97	imes10^{24}$ kg. The actual ratio $\frac{m_{M}}{m_{E}}\approx\frac{6.42	imes10^{23}}{5.97	imes10^{24}}\approx0.108$. The discrepancy is due to the fact that the assumption of uniform density is incorrect. Different planets have different compositions and densities.
# Answer:
Expected ratio: $\frac{1}{8}$. Actual ratio is approximately 0.108. The discrepancy is due to non - uniform densities of the planets.

### 6.4
# Explanation:
## Step1: Recall gravitational acceleration formula
The gravitational acceleration $g=\frac{GM}{r^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the celestial body, and $r$ is the radius. Let $r_{m}=\frac{1}{4}r_{e}$ (radius of Moon and Earth) and $g_{m}=\frac{1}{6}g_{e}$ (gravitational acceleration on Moon and Earth). From $g=\frac{GM}{r^{2}}$, we can express $M=\frac{gr^{2}}{G}$. Then $\frac{M_{m}}{M_{e}}=\frac{g_{m}r_{m}^{2}}{g_{e}r_{e}^{2}}$. Substituting $r_{m}=\frac{1}{4}r_{e}$ and $g_{m}=\frac{1}{6}g_{e}$, we get $\frac{M_{m}}{M_{e}}=\frac{\frac{1}{6}g_{e}(\frac{1}{4}r_{e})^{2}}{g_{e}r_{e}^{2}}=\frac{1}{6}	imes\frac{1}{16}=\frac{1}{96}$.
## Step2: Analyze composition
The lower gravitational acceleration on the Moon with a smaller linear size indicates that the average density of the Moon is lower than that of the Earth. This suggests that the Moon has a different composition, likely with less dense materials on average compared to the Earth.
# Answer:
The Moon likely has a lower average density than the Earth, indicating a different composition with less dense materials.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

6.1

Step1: Calculate area of medium pizza

Step2: Calculate area of large pizza

Step3: Compare cost - per - area

Step1: Recall volume formula for sphere

Step2: Calculate ratio of volumes

Step1: Calculate new volume ratio

Step2: Calculate percentage error

Answer:

6.2a