Answer:
3.0 x10^-3 J
Explanation:
The potential energy of a spring is given by PE = (0.5)k*x^2
Where
K: Spring Constant = 60 N/m
x: displacement of the spring from its equilibrium position = 1cm = 0.01m
Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J
Answer:
Electric potential, E = 2100 volts
Explanation:
Given that,
Electric field, E = 3000 N/C
We need to find the electric potential at a point 0.7 m above the surface, d = 0.7 m
The electric potential is given by :


V = 2100 volts
So, the electric potential at a point 0.7 m above the surface is 2100 volts. Hence, this is the required solution.
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:
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Ohms Law: V = IR
V is the voltage in volts
I is the current in amps
R is the resistance in Ohms
Rearrange: R = V/I
R = (110)/(0.050)
R = 2200
There are 2200 Ohms of resistance in the circuit.