Answer:
(A) - V=11.8098 cm^3, The actual volume of a bouncy ball is less than the estimation.
(B) -[tex]m=\rho V \Rightarrow w=mg[/tex]
(C) - Yes, the [tex]A_{paper}\geq SA_{box}[/tex]
Step-by-step explanation:
Given the three part question.
(A) - Given that 1000 bouncy balls fill a box (dimensions: 27cm by 27cm by 16.2cm) in its entirety. Estimate the volume of one bouncy ball.
(B) - Describe a method in which you could find the weight of one bouncy ball
(C) - Given a square piece of wrapping paper (24 by 24 in), is this enough to cover the entire box?
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Part (A):
(1) - Compute the volume of the box
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Volume of a Box:}}\\\\V=l\times w\times h\end{array}\right}[/tex]
[tex]V_{tot.}=l\times w\times h\\\\\Longrightarrow V_{tot.}=27\times 27\times 16.2\\\\\therefore \boxed{V_{tot.}=11809.8 \ cm^3}[/tex]
(2) - Use the volume we just found to estimate the volume of one bouncy ball
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Volume of One Bouncy Ball:}}\\\\V_{1 \ ball}=\frac{V_{tot.}}{1000} \end{array}\right}[/tex]
[tex]V_{1 \ ball}=\frac{V_{tot.}}{1000}\\\\\Longrightarrow V_{1 \ ball}=\frac{11809.8}{1000} \\\\\therefore \boxed{V_{1 \ ball} \approx 11.8098 \ cm^3}[/tex]
Thus, the estimated volume of one ball is found. Now, is this the actual volume of one ball? No, the actual volume of 1 ball would be less than the value we just computed. This is because there is space left between each ball. I have attached an image for you to visualize.
Part (B):
One way you could find the weight of one balls is by find the ball's mass using the following formula. You would have to look up the density of the bouncy ball's material.
[tex]mass=density\times volume \rightarrow \boxed{m=\rho V}[/tex]
Then using the value you find for mass you could then use the following formula to determine 1 ball's weight.
[tex]weight=mass \times gravity \rightarrow \boxed{w=mg}[/tex]
Part (C):
(1) - Compute the surface area of the box
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Surface Area of a Rectangular Box:}}\\\\SA=2(wl+hl+hw)\end{array}\right}[/tex]
[tex]SA_{box}=2(wl+hl+hw)\\\\\Longrightarrow SA_{box}=2((16.2)(27)+(27)(27)+(27)(16.2))\\\\\Longrightarrow SA_{box}=2(437.4+729+437.4)\\\\\Longrightarrow SA_{box}=2(1603.8)\\\\\therefore \boxed{SA_{box}=3207.6 \ cm^2}[/tex]
(2) - Compute the surface area of the square piece of wrapping paper
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{ Area of a Square:}}\\\\A=s^2\end{array}\right}[/tex]
[tex]A_{paper}=s^2\\\\\Longrightarrow A_{paper}=(24)^2\\\\\therefore \boxed{A_{paper}=576 \ in^2}[/tex]
(3) - The surface area of the box and wrapping paper are in different units. So, we must do a unit conversion
[tex]\boxed{1 \ cm^2 =0.155 \ in^2}[/tex]
[tex]\frac{3207.6 \ cm^2}{1} \times \frac{0.155 \ in^2}{1 \ cm^2}=\boxed{497.178 \ in^2} =SA_{box}[/tex]
(4) - The wrapping paper will completely cover the box if [tex]A_{paper}\geq SA_{box}[/tex]
[tex]576 \ in^2\geq 497.178 \ in^2 \therefore \text{The wrapping paper will cover the box}[/tex]
Thus, all parts have been answered.
