Answer:
Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:
lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:
To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):
=![\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7B4%5Cpi%20e_%7B0%7D%20%7D%20%5B%5Cfrac%7B1%7D%7BR%7D%20-%5Cfrac%7Br%5E%7B2%7D-R%5E%7B2%7D%20%20%7D%7B2R%5E%7B3%7D%20%7D%20%5D)
∴NOTE: Graph is attached
Answer:
Yes, it is reasonable to neglect it.
Explanation:
Hello,
In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):
After that, we compute the potential energy 1.00 m above the reference point:
Then, we compute the average kinetic energy at the specified temperature:
Whereas 
stands for the Avogadro's number for which we have:
In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.
Regards.
Answer:
8 KJ/ s
Explanation:
Heat pumps Transfer thermal energy through absorbing of heat that comes from cold region and then release to warmer area by utilizing external power.
The coefficient of performance known as COP provide the ratio of both heating and cooling that are supplied to required work.
✓QH=The rate at which heat is produced = ?
✓COP= Coefficient of performance of a residential heat pump = 1.6
✓ W(in)= power consumption= 5KW
QH=The rate at which heat is produced=[Coefficient of performance of a residential heat pump] × [power consumption]
= 1.6 × 5KW
=8 KJ/ s
Generally, rings form from moons, asteroids, or comets that have disintegrated due to a collision or because they got too close to their planet (Roche Limit). ... Most of the debris, however, will not have enough energy to overcome the planet's gravity and will remain in orbit around the planet.
