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
.
Answer:
Step-by-step explanation:
Use Pythagoras’ theorem
square of hypotenuse = sum of squares of two other sides
let the missing side be d
30^2 = 20^2 + d^2
600 = 400 + d^2
d^2 = 200
d = square root of 200
d = 14.142 cm
1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3
. Find the
dimensions of the box that requires the least amount of cardboard.
Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize
A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make
the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute
this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing
something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax
or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t
hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From
these, we obtain x
2y = 8 = xy2
. This forces x = y = 2, which forces z = 1. Calculating second
derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum
for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices
of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small
so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither
closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage
something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each
of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus,
moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
Answer:
$406,300
Step-by-step explanation:
$2,700 + $3,600 = $6,300
$400,000 + $6,300 = $406,300
It should be 92 sorry if its wrong
