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
Answer:
D: The side of Magnet A that's attracted to Magnet B's south pole must be Magnet A's north pole
Explanation:
D: The side of Magnet A that's attracted to Magnet B's south pole must be Magnet A's north pole because
1) opposite poles attract each other
2) similar poles repel each other
3)magnetic lines of force start at the north pole and end at the south pole
Answer:
constant at the speed of light
Answer:
The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
Explanation:
We are given that
Angular acceleration, 
Diameter of the wheel, d=21 cm
Radius of wheel,
cm
Radius of wheel, 
1m=100 cm
Magnitude of total linear acceleration, a=
We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.
Tangential acceleration,


Radial acceleration,
We know that

Using the formula

Squaring on both sides
we get






Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s