Answer:
208 
Step-by-step explanation:
We can find the area of the net by adding up the area of each of the 6 rectangles that make up the net. Since two of each rectangle are the same, we only have to find the area of the 3 different sized rectangles and multiply each by 2.
Rectangle pairs are:
- Left rectangle and right rectangle
- Top rectangle and the rectangle above the bottom rectangle
- Bottom rectangle and the rectangle surrounded by all for sides
Now, let's solve the question.
Left rectangle:
6 x 4 = 24, rectangle has area of 24 squared cm
Top rectangle:
6 x 8 = 48, rectangle has area of 48 squared cm
Bottom rectangle:
4 x 8 = 32, rectangle has area of 32 squared cm
Add up the areas:
(24 x 2) + (48 x 2) + (32 x 2) = 208
The rectangle has a surface area of 208 squared cm
Answer:
A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:
A. closed at both ends
B. open at one end and closed at one end
C. open at both ends.
D. we cannot tell because we do not know the frequency of the sound.
The right choice is:
B. open at one end and closed at one end
.
Step-by-step explanation:
Given:
Length of the pipe, 
= 120 cm
Its wavelength 
= 480 cm
= 160 cm and 
= 96 cm
We have to find whether the pipe is open,closed or open-closed or none.
Note:
- The fundamental wavelength of a pipe which is open at both ends is 2L.
- The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.
So,
The fundamental wavelength:
⇒ 
It seems that the pipe is open at one end and closed at one end.
Now lets check with the subsequent wavelengths.
For one side open and one side closed pipe:
An odd-integer number of quarter wavelength have to fit into the tube of length L.
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
So the pipe is open at one end and closed at one end
.
