Answer:48 V
Explanation:
Given
Three charged particle with charge
Electric Potential is given by
Distance of 
from 
similarly 
Potential at is
![V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]](https://tex.z-dn.net/?f=V_%7Bnet%7D%3Dk%5B%5Cfrac%7Bq_1%7D%7Bd_1%7D%2B%5Cfrac%7Bq_2%7D%7Bd_2%7D%2B%5Cfrac%7Bq_3%7D%7Bd_3%7D%5D)
![V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}](https://tex.z-dn.net/?f=V_%7Bnet%7D%3D9%5Ctimes%2010%5E9%5B%5Cfrac%7B50%7D%7B10%7D-%5Cfrac%7B80%7D%7B12%7D%2B%5Cfrac%7B70%7D%7B10%7D%5D%5Ctimes%2010%5E%7B-9%7D)
No, one more group must get the same result as the other group. This will make it more reliable and reproducible.
Answer:
The correct option is A = 1960 N/m²
Explanation:
Given that,
Mass m= 20,000kg
Area A = 100m²
Pressure different between top and bottom
Assume the plane has reached a cruising altitude and is not changing elevation. Then sum the forces in the vertical direction is given as
∑Fy = Wp + FL = 0
where
Wp = is the weight of the plane, and
FL is the lift pushing up on the plane.
Let solve for FL since the mass of the plane is given:
Wp + FL = 0
FL = -Wp
FL = -mg
FL = -20,000× -9.81
FL = 196,200N
FL should be positive since it is opposing the weight of the plane.
Let Equate FL to the pressure differential multiplied by the area of the wings:
FL = (Pb −Pt)⋅A
where Pb and Pt are the static pressures on bottom and top of the wings, respectively
FL = ∆P • A
∆P = FL/A
∆P = 196,200 / 100
∆P = 1962 N/m²
∆P ≈ 1960 N/m²
The pressure difference between the top and bottom surface of each wing when the airplane is in flight at a constant altitude is approximately 1960 N/m². Option A is correct
The only way of telling about dark energy is our observation of how the universe has been expanding. It basically works the opposite as gravity, pushing things away from it. Thus, the closest answer would be D. The shape of galaxies in cluster galaxies.
