Answer:
the awnser is momentum i did this before
Explanation:
At certain altitude, the temperature of air decrease, The air becomes saturated and water vapour molecules starts condensing.
As the altitude of air increase, the atmospheric pressure decrease due to which the temperature of the air decrease. The water molecules in the atmosphere start condensing, which saturate the air (that is air can no hold water molecules), due to which the water vapour molecules starts condensing and falls on the earth in the form of rain.
The correct answer is amplify each other
If they move at the same speed with the same frequency in the the same direction, then they will amplify each other due to constructive interference.
Answer:
Acceleration, ![a=0.105\ m/s^2](https://tex.z-dn.net/?f=a%3D0.105%5C%20m%2Fs%5E2)
Explanation:
It is given that,
Mass of Gary, m = 80.5 kg
Mass of his bicycle, m' = 13.7 kg
Net unbalanced force acting on him, F = 9.86 N
Let a is the value of acceleration. Both cycle and Gary will move with same acceleration. Using the second law of motion to find it as :
![F=ma](https://tex.z-dn.net/?f=F%3Dma)
![F=(m+m')a](https://tex.z-dn.net/?f=F%3D%28m%2Bm%27%29a)
![a=\dfrac{F}{m+m'}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7BF%7D%7Bm%2Bm%27%7D)
![a=\dfrac{9.86}{80.5+13.7}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B9.86%7D%7B80.5%2B13.7%7D)
![a=0.10467\ m/s^2](https://tex.z-dn.net/?f=a%3D0.10467%5C%20m%2Fs%5E2)
or
![a=0.105\ m/s^2](https://tex.z-dn.net/?f=a%3D0.105%5C%20m%2Fs%5E2)
So, the value of acceleration of the Gary and bicycle is
. Hence, this is the required solution.
Answer:
The sphere's volume charge density is 2.58 μC/m³.
Explanation:
Given that,
Radius of sphere R= 8.40 cm
Electric field ![E= 2.04\times10^{3}\ N/C](https://tex.z-dn.net/?f=E%3D%202.04%5Ctimes10%5E%7B3%7D%5C%20N%2FC)
Distance r= 16.8 cm
We need to calculate the sphere's volume charge density
Using Gauss's law
![\int{\vec{E}\cdot\vec{da}}=\dfrac{Q_{enc}}{\epsilon_{0}}](https://tex.z-dn.net/?f=%5Cint%7B%5Cvec%7BE%7D%5Ccdot%5Cvec%7Bda%7D%7D%3D%5Cdfrac%7BQ_%7Benc%7D%7D%7B%5Cepsilon_%7B0%7D%7D)
![E\times 4\pi r^2=\dfrac{1}{\epsilon_{0}}\times\dfrac{4}{3}\piR^3\rho](https://tex.z-dn.net/?f=E%5Ctimes%204%5Cpi%20r%5E2%3D%5Cdfrac%7B1%7D%7B%5Cepsilon_%7B0%7D%7D%5Ctimes%5Cdfrac%7B4%7D%7B3%7D%5CpiR%5E3%5Crho)
![\rho=\dfrac{3\times E\times\epsilon_{0}r^2}{R^3}](https://tex.z-dn.net/?f=%5Crho%3D%5Cdfrac%7B3%5Ctimes%20E%5Ctimes%5Cepsilon_%7B0%7Dr%5E2%7D%7BR%5E3%7D)
Put the value into the formula
![\rho=\dfrac{3\times2.04\times10^{3}\times8.85\times10^{-12}\times(16.8\times10^{-2})^2}{(8.40\times10^{-2})^3}](https://tex.z-dn.net/?f=%5Crho%3D%5Cdfrac%7B3%5Ctimes2.04%5Ctimes10%5E%7B3%7D%5Ctimes8.85%5Ctimes10%5E%7B-12%7D%5Ctimes%2816.8%5Ctimes10%5E%7B-2%7D%29%5E2%7D%7B%288.40%5Ctimes10%5E%7B-2%7D%29%5E3%7D)
![\rho=2.58\times10^{-6}\ C/m^3](https://tex.z-dn.net/?f=%5Crho%3D2.58%5Ctimes10%5E%7B-6%7D%5C%20C%2Fm%5E3)
![\rho=2.58\ \mu C/m^3](https://tex.z-dn.net/?f=%5Crho%3D2.58%5C%20%5Cmu%20C%2Fm%5E3)
Hence, The sphere's volume charge density is 2.58 μC/m³.