Answer:
It’s pretty self explanatory I hope this helps
Explanation:
You might see the word alloy described as a "mixture of metals", but that's a little bit misleading because some alloys contain only one metal and it's mixed in with other substances that are nonmetals (cast iron, for example, is an alloy made of just one metal, iron, mixed with one nonmetal, carbon). The best way to think of an alloy is as a material that's made up of at least two different chemical elements, one of which is a metal. The most important metallic component of an alloy (often representing 90 percent or more of the material) is called the main metal, the parent metal, or the base metal. The other components of an alloy (which are called alloying agents) can be either metals or nonmetals and they're present in much smaller quantities (sometimes less than 1 percent of the total). Although an alloy can sometimes be a compound (the elements it's made from are chemically bonded together), it's usually a solid solution (atoms of the elements are simply intermixed, like salt mixed with water).
Answer:
The second box
Explanation:
Work done is the force applied to move a body through a particular distance.
Work done = Force x distance
For the first box:
Weight = 12N
Distance = 1m
Work done = 12 x 1 = 12J
For the second box;
Weight = 1000N
Distance = 30cm = 0.3m
Work done = 1000 x 0.3 = 300J
The second box has a higher work done.
The answer is vaporization, then condensation.
Here is a set of flash card that can help with this subject.
Let's call
![f_n](https://tex.z-dn.net/?f=f_n)
the frequency of the nth-harmonic and
![f_{n+1}](https://tex.z-dn.net/?f=f_%7Bn%2B1%7D)
the frequency of the (n+1)th harmonic, wish is the next harmonic.
Since the frequency of the nth-harmonic is n times the fundamental frequency f1:
![f_n = n f_1](https://tex.z-dn.net/?f=f_n%20%3D%20n%20f_1)
then the difference between two successive harmonics is equal to the fundamental frequency of the tube:
![f_{n+1}-f_n = (n+1)f_1 - nf_1 = nf_1 + f_1 - nf _1 = f_1](https://tex.z-dn.net/?f=f_%7Bn%2B1%7D-f_n%20%3D%20%28n%2B1%29f_1%20-%20nf_1%20%3D%20nf_1%20%2B%20f_1%20-%20nf%20_1%20%3D%20f_1)
so, by using 350 Hz and 280 Hz as successive harmonics, we find the fundamental frequency of the tube:
![f_1 = 350 Hz - 280 Hz = 70 Hz](https://tex.z-dn.net/?f=f_1%20%3D%20350%20Hz%20-%20280%20Hz%20%3D%2070%20Hz)
The wavelength of the first harmonic is twice the length of the tube:
![\lambda = 2 L=2 \cdot 1.70 m=3.40 m](https://tex.z-dn.net/?f=%5Clambda%20%3D%202%20L%3D2%20%5Ccdot%201.70%20m%3D3.40%20m)
And since we know both frequency and wavelength, we can find the speed of the wave in the tube, which is the speed of sound in the gas in the tube: