A=deltav/t
Vf=20m/s
Vi=0m/s
(20-0)/30
=0.667 m/s^2
Answer:
the maximum current is 500 A
Explanation:
Given the data in the question;
the B field magnitude on the surface of the wire is;
B = μ₀i / 2πr
we are to determine the maximum current so we rearrange to find i
B2πr = μ₀i
i = B2πr / μ₀
given that;
diameter d = 2 mm = 0.002 m
radius = 0.002 / 2 = 0.001 m
B = 0.100 T
we know that permeability; μ₀ = 4π × 10⁻⁷ Tm/A
so we substitute
i = (0.100)(2π×0.001 ) / 4π × 10⁻⁷
i = 500 A
Therefore, the maximum current is 500 A
Answer:
476.387 Hz
714.583 Hz
Explanation:
L = Length of tube
v = Speed of sound in air = 343 m/s
Frequency for a closed tube is given by

The frequency is 476.387 Hz
If it was one third full 

The frequency is 714.583 Hz
Answer:
Option B seems to be the appropriate choice,
Explanation:
The given question is incomplete. Please find the attachment of the complete question.
- From either the principle or law of the Coulombs, there would have been either pressure (force) through attachment or repulsion amongst charges.
- It is evident from the analysis presented that perhaps the bee becomes positively (+) charged and would have been drawn either by negatively (-) charged roots of the plants or stem.
As a consequence, the vegetation is electric current polarized because as activated bee pursued.
Answer:
A fumbled ball may be recovered and advanced by either team
hope it helps