A will be the fastest and c the slowest because of the dip it has a is a straight line fastest way to get from a to b is a straight line b is the second fastest and d is last
Answer: 10.58 C has flowed during the lightning bolt
Explanation:
Given that;
Time of flow t = 1.2 × 10⁻³
perpendicular distance r = 21 m
Magnetic field B = 8.4 x 10⁻⁵ T
Now lets consider the expression for magnetic field;
B = u₀I / 2πr
the current flow is;
I = ( B × 2πr ) / u₀
so we substitute
I = ( (8.4 x 10⁻⁵) × 2 × 3.14 × 21 ) / 4π ×10⁻⁷
= 0.01107792 / 0.000001256
= 8820 A
Hence the charge flows during lightning bolt will be;
q = It
so we substitute
q = 8820 × 1.2 × 10⁻³
q = 10.58 C
therefore 10.58 C has flowed during the lightning bolt
Answer:
B) electrons transferred from sphere to rod.
(2) 1.248 x 10¹¹ electrons were transferred
Explanation:
Given;
initial charge on the plastic rod, q₁ = 15nC
final charge on the plastic rod, q₂ = - 5nC
let the charge acquired by the plastic rod = q
q + 15nC = -5nC
q = -5nC - 15nC
q = -20 nC
Thus, the plastic rod acquired excess negative charge from the metal sphere.
Hence, electrons transferred from sphere to rod
B) electrons transferred from sphere to rod.
2) How many charged particles were transferred?
1.602 x 10⁻¹⁹ C = 1 electron
20 x 10⁻⁹ C = ?
= 1.248 x 10¹¹ electrons
Thus,1.248 x 10¹¹ electrons were transferred
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
