The water glasses are filled to the same height and have the same radius.

Question

Fill in the blanks:

(4 points)
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
a) It would takeResponse area times filling up glass B and pouring into glass A to fill up glass A.

b) If the height of the glasses doubled, then the volumes would increase timesResponse area. (Hint: Use your formulas to substitute V=13πr2h
and V=πr2h
)

c) If the radius of the glasses doubled, then the volumes would increase timesResponse area. (Hint: Use your formulas to substitute V=13πr2h
and V=πr2h
)

d) If the radius and the height both doubled, then the volumes would increase timesResponse area. (Hint: Use your formulas to substitute V=13πr2h
and V=πr2h
)

Answer 1

To fill in the blanks, we need to understand how volume changes as we modify the height and radius of the glasses.

1. Volume formula: The volume \( V \) of a cylinder is given by \( V = \pi r^2 h \).

a) To fill up glass A with glass B, since both have the same height and radius, we can just pour the contents directly. The response should be based on their same volume, hence:

Response: 1 time

b) If the height of the glasses doubled (i.e., \( h \) becomes \( 2h \)), the volume would increase by a factor of:

Response: 2 times

(Original volume of glass = \( \pi r^2 h \); New volume = \( \pi r^2 (2h) = 2\pi r^2 h \))

c) If the radius of the glasses doubled (i.e., \( r \) becomes \( 2r \)), the volume would increase by a factor of:

Response: 4 times

(Original volume of glass = \( \pi r^2 h \); New volume = \( \pi (2r)^2 h = 4\pi r^2 h \))

d) If both the radius and height are doubled (i.e., \( r \) becomes \( 2r \) and \( h \) becomes \( 2h \)), the volume would increase by a factor of:

Response: 8 times

(Original volume of glass = \( \pi r^2 h \); New volume = \( \pi (2r)^2 (2h) = 8\pi r^2 h \))

So, filling in the blanks based on the analysis above:

a) It would take 1 times filling up glass B and pouring into glass A to fill up glass A.

b) If the height of the glasses doubled, then the volumes would increase 2 times.

c) If the radius of the glasses doubled, then the volumes would increase 4 times.

d) If the radius and the height both doubled, then the volumes would increase 8 times.