256 kPa because p-guage + p-absolute + p-atmospheric = 256
Answer:
C
Explanation:
The change in momentum of x has to be the opposite of the change in momentum of Y because the momentum is just transferred from one to another. But I'm still trying to figure it out how to calculate.
F - False.
The law of conservation of momentum states that the total momentum is conserved.
Answer:
h = 2.64 meters
Explanation:
It is given that,
Mass of one ball, ![m_1=3\ kg](https://tex.z-dn.net/?f=m_1%3D3%5C%20kg)
Speed of the first ball,
(upward)
Mass of the other ball, ![m_2=2\ kg](https://tex.z-dn.net/?f=m_2%3D2%5C%20kg)
Speed of the other ball,
(downward)
We know that in an inelastic collision, after the collision, both objects move with one common speed. Let it is given by V. Using the conservation of momentum to find it as :
![V=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7Bm_1v_1%2Bm_2v_2%7D%7Bm_1%2Bm_2%7D)
![V=\dfrac{3\times 20+2\times (-12)}{3+2}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B3%5Ctimes%2020%2B2%5Ctimes%20%28-12%29%7D%7B3%2B2%7D)
V = 7.2 m/s
Let h is the height reached by the combined balls of putty rise above the collision point. Using the conservation of energy as :
![mgh=\dfrac{1}{2}mV^2](https://tex.z-dn.net/?f=mgh%3D%5Cdfrac%7B1%7D%7B2%7DmV%5E2)
![h=\dfrac{V^2}{2g}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7BV%5E2%7D%7B2g%7D)
![h=\dfrac{7.2^2}{2\times 9.8}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B7.2%5E2%7D%7B2%5Ctimes%209.8%7D)
h = 2.64 meters
So, the height reached by the combined mass is 2.64 meters. Hence, this is the required solution.
Answer:
2.11eV
Explanation:
We know that speed of light is it's wavelength times frequency.
![\therefore f=v/\lambda\\=(3\times10^8m/s)/(589mm\times1m/1\times 10^9nm)\\=5.09\times10^1^4s^-1 \ or \ 5.09\times10^1^4Hz](https://tex.z-dn.net/?f=%5Ctherefore%20f%3Dv%2F%5Clambda%5C%5C%3D%283%5Ctimes10%5E8m%2Fs%29%2F%28589mm%5Ctimes1m%2F1%5Ctimes%2010%5E9nm%29%5C%5C%3D5.09%5Ctimes10%5E1%5E4s%5E-1%20%5C%20or%20%5C%205.09%5Ctimes10%5E1%5E4Hz)
Planck's constant is ![6.626\times 10^3^4Js](https://tex.z-dn.net/?f=6.626%5Ctimes%2010%5E3%5E4Js)
The energy gap is calculated by multyplying the light's frequency by planck's constant:
![E_c=5.09\times10^1^4s^-^1\times 6.626\times10^-^3^4Js\\\\=3.37\times 10^-^1^9J \ \ \ \ \ \ #1eV=1.06\times 10^-^1^9J\\\\=2.11eV](https://tex.z-dn.net/?f=E_c%3D5.09%5Ctimes10%5E1%5E4s%5E-%5E1%5Ctimes%206.626%5Ctimes10%5E-%5E3%5E4Js%5C%5C%5C%5C%3D3.37%5Ctimes%2010%5E-%5E1%5E9J%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%231eV%3D1.06%5Ctimes%2010%5E-%5E1%5E9J%5C%5C%5C%5C%3D2.11eV)
Hence, the energy gap is 2.11eV