Answer:
the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.
Explanation:
We can answer this exercise using Gauss's law
Ф = ∫ e . dA = 
/ ε₀
field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field
From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.
Answer:
The weight of Earth's atmosphere exert is 
Explanation:
Given that,
Average pressure 
Radius of earth 
Pressure :
Pressure is equal to the force upon area.
We need to calculate the weight of earth's atmosphere
Using formula of pressure
Where, P = pressure
A = area
Put the value into the formula
Hence, The weight of Earth's atmosphere exert is
Say the initial point is (0,0)
The final point is
x = 200 + 135*cos(30) = 200 + 135*sqrt(3)/2 = 316.91 ft
y = 135*sin(30) = 135/2 = 67.5 ft
Resultant vector = (316.91, 67.5) - (0,0) = 316.91, 67.5) ft
Answer:
v= 2413.5 m/s
Explanation:
maximum change of speed of rocket
=(initial exhaust velocity)×ln [(initialmass/finalmass)]
let initial mass= m
final mass = m-m(4/5) = m/5
[since the 80% of mass which is fuel is exhausted]
V-0 = 1500 ln (1/0.2)
V= 1500×1.609 = 2413.5 m/s
therefore, its exhaust speed v= 2413.5 m/s
Answer:
Explanation:
First of all, we need to find the pressure exerted on the sphere, which is given by:
where
is the atmospheric pressure
is the water density
is the gravitational acceleration
is the depth
Substituting,
The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m
So the total area of the sphere is
And so, the inward force exerted on it is
