Answer:
0.009 N, repulsive
Explanation:
The electrostatic force between two electric charges is given by:
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
In this problem, we have
are the two charges
r = 4.5 m is their separation
Substituting into the equation, we find
Moreover, the force is repulsive. In fact, the following rules apply:
- When two charges have same sign, they repel each other
- When two charges have opposite signs, they attract each other
Answer:
what is the hunting age? i've never heard of it
As we know that electrostatic force between two charges is given as
here we know that electrostatic repulsion force is balanced by the gravitational force between them
so here force of attraction due to gravitation is given as
here we can assume that both will have equal charge of magnitude "q"
now we have
now we have
Answer:
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
Explanation:
For this exercise let's use the electric field expression
E = k q / r²
where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee
let's calculate the field for each charge
Q = 24 pC = 24 10⁻¹² C
E₁ = 9 10⁹ 24 10⁻¹² / 0.20²
E₁ = 5.4 N / C
Q = 32 pC = 32 10⁻¹² C
E₂ = 9 10⁹ 32 10⁻¹² / 0.2²
E₂ = 7.2 N / C
let's find the difference between these two fields
ΔE = E₂ -E₁
ΔE = 7.2 - 5.4
ΔE = 1.8 N / C
the minimum detection field is
E_minimum = 0.77 N / C
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
