Answer:
E = k Q₁ / r²
Explanation:
For this exercise that asks us for the electric field between the sphere and the spherical shell, we can use Gauss's law
Ф = ∫ E .dA = 
/ ε₀
where Ф the electric flow, qint is the charge inside the surface
To solve these problems we must create a Gaussian surface that takes advantage of the symmetry of the problem, in this almost our surface is a sphere of radius r, that this is the sphere of and the shell, bone
R <r <R_a
for this surface the electric field lines are radial and the radius of the sphere are also, therefore the two are parallel, which reduces the scalar product to the algebraic product.
E A = q_{int} /ε₀
The charge inside the surface is Q₁, since the other charge Q₂ is outside the Gaussian surface, therefore it does not contribute to the electric field
q_{int} = Q₁
The surface area is
A = 4π r²
we substitute
E 4π r² = Q₁ /ε₀
E = 1 / 4πε₀ Q₁ / r²
k = 1/4πε₀
E = k Q₁ / r²
Answer:
E) the flow of energy due to a temperature difference.
Explanation:
Heat can be described as the flow of energy due to a temperature difference.
Which is expressed mathematically as;
H = MCΔT
Where;
H is the quantity of heat in a body, measured in Joules
M is the mass of the body, measured in kg
C is the specific heat capacity of the body, J/kg.K
ΔT is change in temperature or temperature difference.
So, heat energy in any system flows from a hotter region to a colder region due to temperature difference.
E) the flow of energy due to a temperature difference.
Answer:
u = 13.67 m/s
Explanation:
given,
window height = 2 m
window is 7.5 m off the ground on its path up
total distance from the ground to pass the window = 2 + 7.5 = 9.5 m
time taken to go past the window = 1.30 s
using equation of motion
u = 13.67 m/s
hence, the initial velocity of the ball is equal to 13.67 m/s
The purpose of this lab is to determine whether the surface of an area would affect the coefficient of Friction. My classmates and I have learned a lot in this lab and that there could have been some errors in our lab because the strength of how a person pulls it might be a slight different than the normal force. I learned from this lab that the <span>surface area would have no effect on the coefficient of friction. </span>
