Covalent bonds form when two or more atoms share electrons
To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as
The potential energy, ![U = 13.0 J](https://tex.z-dn.net/?f=U%20%3D%2013.0%20J)
The volume, ![V = 6.00 mm^3 = 6.00*10^{-9}m^3](https://tex.z-dn.net/?f=V%20%3D%206.00%20mm%5E3%20%3D%206.00%2A10%5E%7B-9%7Dm%5E3)
The potential energy per unit volume is defined as the energy density.
![u = \frac{U}{V}](https://tex.z-dn.net/?f=u%20%3D%20%5Cfrac%7BU%7D%7BV%7D)
![u= \frac{(13.0 J)}{(6.00*10^{-9} m^3)}](https://tex.z-dn.net/?f=u%3D%20%5Cfrac%7B%2813.0%20J%29%7D%7B%286.00%2A10%5E%7B-9%7D%20m%5E3%29%7D)
![u= 2.167109 J/m^3](https://tex.z-dn.net/?f=u%3D%202.167109%20J%2Fm%5E3)
The energy density related with electric field is given by
![u = \frac{1}{2} \epsilon_0 E^2](https://tex.z-dn.net/?f=u%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cepsilon_0%20E%5E2)
Here, the permitivity of the free space is
![\epsilon_0 = 8.85*10^-{12} C^2/N \cdot m^2](https://tex.z-dn.net/?f=%5Cepsilon_0%20%3D%208.85%2A10%5E-%7B12%7D%20C%5E2%2FN%20%5Ccdot%20m%5E2)
Therefore, rerranging to find the electric field strength we have,
![E = \sqrt{\frac{2u}{\epsilon_0}}](https://tex.z-dn.net/?f=E%20%3D%20%5Csqrt%7B%5Cfrac%7B2u%7D%7B%5Cepsilon_0%7D%7D)
![E = \sqrt{\frac{2(2.167109)}{8.85*10^{-12}}}](https://tex.z-dn.net/?f=E%20%3D%20%5Csqrt%7B%5Cfrac%7B2%282.167109%29%7D%7B8.85%2A10%5E%7B-12%7D%7D%7D)
![E = 2.211010 V/m](https://tex.z-dn.net/?f=E%20%3D%202.211010%20V%2Fm)
Therefore the electric field is 2.21V/m
Therefore, if the block moves from its position of maximum spring stretch to maximum spring compression in 0.25 s, the time required for a full cycle is twice as much; T = 0.5 s.
Answer:
Relation between initial speed of bullet and height h is given as
![v = \frac{m + M}{m}\sqrt{2gh}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bm%20%2B%20M%7D%7Bm%7D%5Csqrt%7B2gh%7D)
Explanation:
As we know that system of block and bullet swings up to height h after collision
So we have
![(m + M)gh = \frac{1}{2}(m + M)v_1^2](https://tex.z-dn.net/?f=%28m%20%2B%20M%29gh%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28m%20%2B%20M%29v_1%5E2)
so we have
![v_1 = \sqrt{2gh}](https://tex.z-dn.net/?f=v_1%20%3D%20%5Csqrt%7B2gh%7D)
so speed of the block + bullet just after the impact is given by above equation
Now we also know that there is no force on the system of bullet + block in the direction of motion
So we can use momentum conservation
![mv = (m + M)v_1](https://tex.z-dn.net/?f=mv%20%3D%20%28m%20%2B%20M%29v_1)
now we have
![v = \frac{m + M}{m}\sqrt{2gh}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bm%20%2B%20M%7D%7Bm%7D%5Csqrt%7B2gh%7D)