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
Momentum = mass * velocity
I need parts A and B to explain the intuitions.
Answer:
49.85 V
Explanation:
u = 0, s = 5.62 cm, t = 1.15 x 10^-6 s
Let the electric field is E and voltage is V.
Use second equation of motion
s = ut + 1/2 a t^2
5.62 x 10^-2 = 0 + 0.5 a x (1.15 x 10^-6)^2
a = 8.5 x 10^10 m/s^2
m x a = q x E
E = m x a / q
E = (1.67 x 10^-27 x 8.5 x 10^10) / (1.6 x 10^-19)
E = 887.19 V/m
V = E x s
V = 887.19 x 5.62 x 10^-2 = 49.85 V
Answer:
<h3>The answer is option B</h3>
Explanation:
The wavelength of a wave can be found by using the formula
where
c is the speed of the wave
f is the frequency
From the question
c = 343 m/s
f = 466 Hz
We have
We have the final answer as
<h3>0.74 m</h3>
Hope this helps you
