Answer:
E = 1580594.95 N/C
Explanation:
To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:
(1)
dS: differential of the Gaussian surface
Qin: charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2
The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:
(2)
Qin is calculate by using the charge density:
(3)
Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.
The charge density is given by:
Next, you use the results of (3), (2) and (1):
Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:
hence, the electric field is 1580594.95 N/C
Part 1)
here we know that supply took 10 s to reach the ground
so here we will have
Part 2)
Here all the supply covered horizontal distance of 650 m in 10 s interval of time
so here we can say
200 joules of work energy are involved. That's all we need to know to answer the question. Once we know that 200 joules of work energy are involved, we don't care what was lifted, or how far, or how long it took, or how many people worked on it, or how much they were paid, or what was the distribution of their gender identities, or the ethnic diversity among the team. or what day each of them celebrates as their sabbath. Any other information besides the 200 joules is only there to distract us, and see whether we're paying attention.
Power = (work or energy) / (time to do the work or move the energy)
Power = (200 joules) / (5 seconds)
<em>Power = 40 watts</em>
Answer:
1,200 watts
Explanation:
1 watt = 1 Joule (J) of work / second
So, 3600 Joules of work / 3 seconds is:
3600 J / 3 seconds = 1,200 watts
