Answer:
The principle of momentum conservation states that if there no external force the total momentum of the system before and after the collision is conserved.
Since momentum is a vector, we should investigate the directions and magnitudes of initial and final momentum.
![\vec{P}_{initial} = \vec{P}_{final}\\\vec{P}_{initial} = m_1\vec{v}_1 + 0\\\vec{P}_{final} = m_1\vec{v}_1' + m_2 \vec{v}_2'](https://tex.z-dn.net/?f=%5Cvec%7BP%7D_%7Binitial%7D%20%3D%20%5Cvec%7BP%7D_%7Bfinal%7D%5C%5C%5Cvec%7BP%7D_%7Binitial%7D%20%3D%20m_1%5Cvec%7Bv%7D_1%20%2B%200%5C%5C%5Cvec%7BP%7D_%7Bfinal%7D%20%3D%20m_1%5Cvec%7Bv%7D_1%27%20%2B%20m_2%20%5Cvec%7Bv%7D_2%27)
If the first ball hits the second ball with an angle, we should separate the x- and y-components of the momentum (or velocity), and apply conservation of momentum separately on x- and y-directions.
As we know that as per Newton's II law we have
![F = \frac{dP}{dt}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7BdP%7D%7Bdt%7D)
here we will have
= change in momentum
= time interval in which momentum is changed
now in order to have least injury during jumping we need to have least force on the jumper
so in order to have least force we can say that the momentum must have to change in maximum time so that amount of force must be least
So we need to increase the time in which momentum of the system is changed
The layout of the stars in the sky is determined by the date, time of night, and your location (mainly latitude). So to pick the best star chart, you should go with the one that's closest to the present date and your location, then make allowance for what time it is. Everything in the sky moves about a degree every 4 minutes.
Answer:
Absolute pressure of the oil will be 102822.8 Pa
Explanation:
We have given height h = 31 cm = 0.31 m
Acceleration due to gravity ![g=9.8m/sec^2](https://tex.z-dn.net/?f=g%3D9.8m%2Fsec%5E2)
Specific gravity of oil = 0.600
So density of oil ![\rho =0.6\times 1000=600kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D0.6%5Ctimes%201000%3D600kg%2Fm%5E3)
We know that absolute pressure is given by
, here ![P_0=1.01\times 10^5Pa](https://tex.z-dn.net/?f=P_0%3D1.01%5Ctimes%2010%5E5Pa)
So absolute pressure will be equal to ![P=1.01\times 10^5+600\times 9.8\times 0.31=102822.8Pa](https://tex.z-dn.net/?f=P%3D1.01%5Ctimes%2010%5E5%2B600%5Ctimes%209.8%5Ctimes%200.31%3D102822.8Pa)
So absolute pressure of the oil will be 102822.8 Pa