Answer:
In the air
Explanation:
There are three states of matter:
- Solids: in solids, the particles are tightly bond together by strong intermolecular forces, so they cannot move freely - they can only vibrate around their fixed position
- Liquids: in liquids, particles are more free to move, however there are still some intermolecular forces keeping them close to each other
- Gases: in gases, particles are completely free to move, as the intermolecular forces between them are negligible
For this reason, it is generally easier to compress/expand the volume of a gas with respect to the volume of a liquid.
In this problem, we are comparing water (which is a liquid) with air (which is a gas). From what we said above, this means that the change in volume is larger in the air rather than in the water.
Answer:
<em>The final speed of the second package is twice as much as the final speed of the first package.</em>
Explanation:
<u>Free Fall Motion</u>
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:
![v=gt](https://tex.z-dn.net/?f=v%3Dgt)
And the distance traveled downwards is:
![\displaystyle y=\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D%5Cfrac%7Bgt%5E2%7D%7B2%7D)
If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:
![\displaystyle t=\sqrt{\frac{2y}{g}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D%5Csqrt%7B%5Cfrac%7B2y%7D%7Bg%7D%7D)
Replacing into the first equation:
![\displaystyle v=g\sqrt{\frac{2y}{g}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%3Dg%5Csqrt%7B%5Cfrac%7B2y%7D%7Bg%7D%7D)
Rationalizing:
![\displaystyle v=\sqrt{2gy}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%3D%5Csqrt%7B2gy%7D)
Let's call v1 the final speed of the package dropped from a height H. Thus:
![\displaystyle v_1=\sqrt{2gH}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_1%3D%5Csqrt%7B2gH%7D)
Let v2 be the final speed of the package dropped from a height 4H. Thus:
![\displaystyle v_2=\sqrt{2g(4H)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%3D%5Csqrt%7B2g%284H%29%7D)
Taking out the square root of 4:
![\displaystyle v_2=2\sqrt{2gH}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%3D2%5Csqrt%7B2gH%7D)
Dividing v2/v1 we can compare the final speeds:
![\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%2Fv_1%3D%5Cfrac%7B2%5Csqrt%7B2gH%7D%7D%7B%5Csqrt%7B2gH%7D%7D)
Simplifying:
![\displaystyle v_2/v_1=2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%2Fv_1%3D2)
The final speed of the second package is twice as much as the final speed of the first package.
Answer:
b, a, c
Explanation:
The middle one has the shortest wavelength, then it's the top one and the last one has the longest wavelength.
Answer:
14.43° or 0.25184 rad
Explanation:
v = Speed of sound in air = 343 m/s
f = Frequency = 1240 Hz
d = Width in doorway = 1.11 m
Wavelength is given by
![\lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{343}{1240}\\\Rightarrow \lambda=0.2766\ m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bv%7D%7Bf%7D%5C%5C%5CRightarrow%20%5Clambda%3D%5Cfrac%7B343%7D%7B1240%7D%5C%5C%5CRightarrow%20%5Clambda%3D0.2766%5C%20m)
In the case of Fraunhofer diffraction we have the relation
![dsin\theta=\lambda\\\Rightarrow \theta=sin^{-1}\frac{\lambda}{d}\\\Rightarrow \theta=sin^{-1}\frac{0.2766}{1.11}\\\Rightarrow \theta=14.43^{\circ}\ or\ 0.25184\ rad](https://tex.z-dn.net/?f=dsin%5Ctheta%3D%5Clambda%5C%5C%5CRightarrow%20%5Ctheta%3Dsin%5E%7B-1%7D%5Cfrac%7B%5Clambda%7D%7Bd%7D%5C%5C%5CRightarrow%20%5Ctheta%3Dsin%5E%7B-1%7D%5Cfrac%7B0.2766%7D%7B1.11%7D%5C%5C%5CRightarrow%20%5Ctheta%3D14.43%5E%7B%5Ccirc%7D%5C%20or%5C%200.25184%5C%20rad)
The minimum angle relative to the center line perpendicular to the doorway will someone outside the room hear no sound is 14.43° or 0.25184 rad