The test charge used to determine the electric field is very, very small so that its presence does not affect the electrical field because of the supply price. The electrical rate that produces the electric area is known as a source charge.
An electric-powered area is a physical area that surrounds electrically charged debris and exerts pressure on all differently charged particles inside the area, both attracting or repelling them. It also refers back to the physical field for a machine of charged particles.
Electric-powered discipline is described as the electrical force in keeping with unit fee. The path of the sector is taken to be the route of the force it might exert on a fine check price. the electric discipline is radially outward from a nice rate and radially in toward a bad point rate.
An electric discipline is a vicinity of space around an electrically charged particle or object in which an electric charge might sense force. An electric field is a vector amount and may be visualized as arrows going in the direction of or away from fees.
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kess dense material comes up. glycerin dencity= 37.8 / 30
syrup density = 80 / 60 so glycerin is less dense and comes up
Answer:
Explanation:
Given that,
Surface area A= 17m²
The speed at the top v" = 66m/s
Speed beneath is v' =40 m/s
The density of air p =1.29kg/m³
Weight of plane?
Assuming that,
the height difference between the top and bottom of the wind is negligible and we can ignore any change in gravitational potential energy of the fluid.
Using Bernoulli equation
P'+ ½pv'²+ pgh' = P'' + ½pv''² + pgh''
Where
P' is pressure at the bottom in N/m²
P" is pressure at the top in N/m²
v' is velocity at the bottom in m/s
v" is velocity at the top in m/s
Then, Bernoulli equation becomes
P'+ ½pv'² = P'' + ½pv''²
Rearranging
P' — P'' = ½pv"² —½pv'²
P'—P" = ½p ( v"² —v'²)
P'—P" = ½ × 1.29 × (66²-40²)
P'—P" = 1777.62 N/m²
Lift force can be found from
Pressure = force/Area
Force = ∆P ×A
Force = (P' —P")×A
Since we already have (P'—P")
Then, F=W = (P' —P")×A
W = 1777.62 × 17
W = 30,219.54 N
The weight of the plane is 30.22 KN
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 average induced emf in the loop is 0.20 V
Explanation:
Given:
Radius of loop 
m
Magnetic field 
T
Change in time 
sec
According to the faraday's law,
Induced emf is given by
Where 
magnetic flux
( here 
)
Where 
We neglect minus sign because it's shows lenz law
V
Therefore, the average induced emf in the loop is 0.20 V
