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
C. 2-
electrons have a negative charge, thus adding two electrons to an atom will give it a charge of 2 times -1 or -2
For rectilinear motions, derived formulas all based on Newton's laws of motion are formulated. The equation for acceleration is
a = (v2-v1)/t, where v2 and v1 is the final and initial velocity of the rocket. We know that at the end of 1.41 s, the rocket comes to a stop. So, v2=0. Then, we can determine v1.
-52.7 = (0-v1)/1.41
v1 = 74.31 m/s
We can use v1 for the formula of the maximum height attained by an object thrown upwards:
Hmax = v1^2/2g = (74.31^2)/(2*9.81) = 281.42 m
The maximum height attained by the model rocket is 281.42 m.
For the amount of time for the whole flight of the model rocket, there are 3 sections to this: time at constant acceleration, time when it lost fuel and reached its maximum height and the time for the free fall.
Time at constant acceleration is given to be 1.41 s. Time when it lost fuel covers the difference of the maximum height and the distance travelled at constant acceleration.
2ax=v2^2-v1^2
2(-52.7)(x) = 0^2-74.31^2
x =52.4 m (distance it covered at constant acceleration)
Then. when it travels upwards only by a force of gravity,
d = v1(t) + 1/2*a*t^2
281.42-52.386 = (0)^2+1/2*(9.81)(t^2)
t = 6.83 s (time when it lost fuel and reached its maximum height)
Lastly, for free falling objects, the equation is
t = √2y/g = √2(281.42)/9.81 = 7.57 s
Therefore, the total time= 1.41+6.83+7.57 = 15.81 s
