Answer:
The correct answer is B
Explanation:
Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity
Φ
= ∫ E. dA =
/ ε₀
For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product
∫ E dA =
/ ε₀
The area of a sphere is
A = 4π r²
E 4π r² =
/ ε₀
E = (1 /4πε₀
) q / r²
Having the solution of the problem let's analyze the points:
A ) r = 3R / 4 = 0.75 R.
In this case there is no charge inside the Gaussian surface therefore the electric field is zero
E = 0
B) r = 5R / 4 = 1.25R
In this case the entire charge is inside the Gaussian surface, the field is
E = (1 /4πε₀
) Q / (1.25R)²
E = (1 /4πε₀
) Q / R2 1 / 1.56²
E₀ = (1 /4π ε₀
) Q / R²
= Eo /1.56
²
= 0.41 Eo
C) r = 2R
All charge inside is inside the Gaussian surface
=(1 /4π ε₀
) Q 1/(2R)²
= (1 /4π ε₀
) q/R² 1/4
= Eo 1/4
= 0.25 Eo
D) False the field changes with distance
The correct answer is B
Answer:
2.2 s
Explanation:
Hi!
Let's consider the origin of the coordinate system at the ground, and consider that the clam starts with zero velocity, the equation of motion of the clam is given by

We are looking for a time t for which x(t) = 0

Solving for t:

Rounding at the first decimal:
t = 2.2 s
Answer:
B. A repulsive force of 8.0*10^3 N.
Explanation:
As we know by Coulomb's law that the electrostatic force between two charges is given as

here we know that


r = 3.0 m
now we have


since both charges are similar charges so they will repel each other by the force we calculated above so correct answer will be
B. A repulsive force of 8.0*10^3 N.
Frost will disturb the smooth flow of air over the wing, unpleasantly
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may avoid the airplane from becoming flying at normal departure speed.