The charge of the copper nucleus is 29 times the charge of one proton:
the charge of the electron is
and their separation is
The magnitude of the electrostatic force between them is given by:
where
is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
Answer:
I = 4.28 [amp]
Explanation:
To solve this type of problems we must have knowledge of the law of ohm, which tells us that the voltage is equal to the product of resistance by current.
Initial data:
v = 1.5 [volt]
R = 0.35 [ohms]
v = I * R
therefore:
I = 1.5 / 0.35
I = 4.28 [amp]
Alright here the answer to number 2
Answer:
Option A
Lowering the amount of reactants
Explanation:
To reduce the rate of chemical reaction, one can reduce temperature or surface area. The addition of catalysts increases rate of reaction but decreasing the amount of reactants decreases rate of reaction. Therefore, from the choices provided, choice A is correct.
Answer:
Ф_cube /Ф_sphere = 3 /π
Explanation:
The electrical flow is
Ф = E A
where E is the electric field and A is the surface area
Let's shut down the electric field with Gauss's law
Фi = ∫ E .dA = 
/ ε₀
the Gaussian surface is a sphere so its area is
A = 4 π r²
the charge inside is
q_{int} = Q
we substitute
E 4π r² = Q /ε₀
E = 1 / 4πε₀ Q / r²
To calculate the flow on the two surfaces
* Sphere
Ф = E A
Ф = 1 / 4πε₀ Q / r² (4π r²)
Ф_sphere = Q /ε₀
* Cube
Let's find the side value of the cube inscribed inside the sphere.
In this case the radius of the sphere is half the diagonal of the cube
r = d / 2
We look for the diagonal with the Pythagorean theorem
d² = L² + L² = 2 L²
d = √2 L
we substitute
r = √2 / 2 L
r = L / √2
L = √2 r
now we can calculate the area of the cube that has 6 faces
A = 6 L²
A = 6 (√2 r)²
A = 12 r²
the flow is
Ф = E A
Ф = 1 / 4πε₀ Q/r² (12r²)
Ф_cubo = 3 /πε₀ Q
the relationship of these two flows is
Ф_cube /Ф_sphere = 3 /π
