Answer:
you will pass but if your grade become lower and u get more absences you will fail
Answer:
electric field E = (1 /3 e₀) ρ r
Explanation:
For the application of the law of Gauss we must build a surface with a simple symmetry, in this case we build a spherical surface within the charged sphere and analyze the amount of charge by this surface.
The charge within our surface is
ρ = Q / V
Q ’= ρ V
'
The volume of the sphere is V = 4/3 π r³
Q ’= ρ 4/3 π r³
The symmetry of the sphere gives us which field is perpendicular to the surface, so the integral is reduced to the value of the electric field by the area
I E da = Q ’/ ε₀
E A = E 4 πi r² = Q ’/ ε₀
E = (1/4 π ε₀) Q ’/ r²
Now you relate the fraction of load Q ’with the total load, for this we use that the density is constant
R = Q ’/ V’ = Q / V
How you want the solution depending on the density (ρ) and the inner radius (r)
Q ’= R V’
Q ’= ρ 4/3 π r³
E = (1 /4π ε₀) (1 /r²) ρ 4/3 π r³
E = (1 /3 e₀) ρ r
Answer : 18.22 meters
Explanation:
1 yard. = 0.9144 meters
85.4 yards = 78.08976 meters
1 minute = 60 seconds
5 minutes = 300 seconds
Speed of Anisa = distance / time
Speed of Anisa = 78.08976 meters / 300 seconds
Speed of Anisa = 0.26029 meters / second.
Distance travelled in 70 seconds = speed * 70
Distance travelled in 70 seconds = 0.26029 * 70 = 18.22 meters
Answer:
The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
Explanation:
Given that,
Mass = 2.15 kg
Distance = 0.0895 m
Amplitude = 0.0235 m
We need to calculate the spring constant
Using newton's second law
Where, f = restoring force
Put the value into the formula
We need to calculate the kinetic energy of the mass
Using formula of kinetic energy
Here, 
Here, 
Put the value into the formula
Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
C. when the circuit is closed
