Answer:
N= 3
Explanation:
For this exercise we must use Faraday's law
E = - dФ / dt
Ф = B . A = B Acos θ
tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives
E = - A cos θ (B - B₀) / t
The angle enters the magnetic field and the normal to the area is zero
cos 0 = 1
A = π r²
In the length of the wire there are N turns each with a length L₀ = 2π r
L = N (2π r)
r = L / 2π N
we substitute
A = L² / (4π N²)
The magnetic field produced by a solenoid is
B = μ₀ N/L I
for which
B₀ = μ₀ N/L I
The final field is zero, because the current is zero
B = 0
We substitute
E = - (L² / 4π N²) (0 - μ₀ N/L I) / t
E = μ₀ L I / (4π N t)
N = μ₀ L I / (4π t E)
The electromotive force is E = 0.80 mV = 0.8 10⁻³ V
let's calculate
N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]
N = 320 10⁻⁷ / 9.6 10⁻⁶
N = 33.3 10⁻¹
N= 3
Answer:
A_resulting = 0.2 m
Explanation:
Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.
With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is
A_res = 2A
A_resultant = 2 .01
A_resulting = 0.2 m
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.
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To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.
The volume of a tank is given by
Where
d = Diameter
h = Height
Considering that there are two stages, let's define the initial and final volume as,
We know as well by definition that
Then we have for the statement that
Replacing the previous data
Solving to get h,
Therefore the change is
Therefore te change in the height of the water in the tank is 0.37mm
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.
At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .
At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .
At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .
