Answer: Yes both gases would have the same entropy.
Explanation:
The formula for the change in the entropy is as follows,
![\Delta S = \frac{Q}{T}](https://tex.z-dn.net/?f=%5CDelta%20S%20%3D%20%5Cfrac%7BQ%7D%7BT%7D)
Here, \Delta S is the change in the entropy, Q is the heat transfer and T is the temperature.
If the temperature of the system increases then there will be increase in the entropy as the randomness of the system increases.
In the given problem, if the both gases were initially at the same absolute temperature. Then there will be same entropy change in both gases.
Therefore, yes both gases would have the same entropy.
Answer:
1.After 11 hours, the cruise ship will be _233.75__nautical miles from the lighthouse.
2.At the start of the journey, the cruise ship was _10.5__
nautical miles from the lighthouse.
3.The cruise ship is traveling at a speed of_21.25_ nautical miles per hour.
Explanation:
Distance covered in 2 hours= 95.5-53NM
=42.5NM
therefore,
![Speed= Distance/time................(1)](https://tex.z-dn.net/?f=Speed%3D%20Distance%2Ftime................%281%29)
put value of distance and time in equation (1)
therefore,
![speed= 42.5/2hour](https://tex.z-dn.net/?f=speed%3D%2042.5%2F2hour)
speed=21.25NM/hour
Therefore distance covered in 11 hours= speed* time
![Distance=21.25*11](https://tex.z-dn.net/?f=Distance%3D21.25%2A11)
=233.75NM
2. Since distance covered in 2 hours is 42.5
therefore, at starting of the journey the cruise distance from light house is
= 53NM-42.5NM
=10.5NM
3. Distance covered in 2 hours= 95.5-53NM
=42.5NM
therefore,
![Speed= Distance/time................(1)](https://tex.z-dn.net/?f=Speed%3D%20Distance%2Ftime................%281%29)
put value of distance and time in equation (1)
therefore,
![speed= 42.5/2hour](https://tex.z-dn.net/?f=speed%3D%2042.5%2F2hour)
speed=21.25NM/hour
Therefore Cruise ship is traveling at speed of 21.25NM/hour
Answer and Explanation:
FFD is the distance between the film on which the image is obtained and the center of the anode tube. The magnification and resolution of the image depends on the FFd By varying the FFD we can change the magnification and resolution of the image. The standard FFD is about 100 centimeters.
New studies have found that by changing the FFD to 130 cm the radiation dosage reduces while the image quality remains practically the same.
Answer:
When a baseball is pitched, hit and flies in the air, one or more of the physical principles formulated over 300 years ago by Sir Isaac Newton act on it. Folklore tells how the mathematician and physicist first realized the law of gravity while observing a falling apple. Had Newton watched a baseball game instead, he might have formulated all three laws of motion by the seventh-inning stretch.
I think it is right.
A) ![x=\pm \frac{A}{2\sqrt{2}}](https://tex.z-dn.net/?f=x%3D%5Cpm%20%5Cfrac%7BA%7D%7B2%5Csqrt%7B2%7D%7D)
The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):
(1)
where k is the spring constant.
The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:
(2)
where x is the displacement, m the mass, and v the speed.
We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:
![U=\frac{1}{3}K](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B3%7DK)
Using (2) we can rewrite this as
![U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B3%7D%28E-U%29%3D%5Cfrac%7B1%7D%7B3%7DE-%5Cfrac%7B1%7D%7B3%7DU%5C%5CU%3D%5Cfrac%7BE%7D%7B4%7D)
And using (1), we find
![U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2](https://tex.z-dn.net/?f=U%3D%5Cfrac%7BE%7D%7B4%7D%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7DkA%5E2%7D%7B4%7D%3D%5Cfrac%7B1%7D%7B8%7DkA%5E2)
Substituting
into the last equation, we find the value of x:
![\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dkx%5E2%3D%5Cfrac%7B1%7D%7B8%7DkA%5E2%5C%5Cx%3D%5Cpm%20%5Cfrac%7BA%7D%7B2%5Csqrt%7B2%7D%7D)
B) ![x=\pm \frac{3}{\sqrt{10}}A](https://tex.z-dn.net/?f=x%3D%5Cpm%20%5Cfrac%7B3%7D%7B%5Csqrt%7B10%7D%7DA)
In this case, the kinetic energy is 1/10 of the total energy:
![K=\frac{1}{10}E](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B10%7DE)
Since we have
![K=E-U](https://tex.z-dn.net/?f=K%3DE-U)
we can write
![E-U=\frac{1}{10}E\\U=\frac{9}{10}E](https://tex.z-dn.net/?f=E-U%3D%5Cfrac%7B1%7D%7B10%7DE%5C%5CU%3D%5Cfrac%7B9%7D%7B10%7DE)
And so we find:
![\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dkx%5E2%20%3D%20%5Cfrac%7B9%7D%7B10%7D%28%5Cfrac%7B1%7D%7B2%7DkA%5E2%29%3D%5Cfrac%7B9%7D%7B20%7DkA%5E2%5C%5Cx%5E2%20%3D%20%5Cfrac%7B9%7D%7B10%7DA%5E2%5C%5Cx%3D%5Cpm%20%5Cfrac%7B3%7D%7B%5Csqrt%7B10%7D%7DA)