A diver springs upward from a board that is 2.70 m above the water. At the instant she contacts the water her speed is 10.9 m/s
1 answer:
Answer:
vo=5.87m/s
Explanation:
Hello! In this problem we have a uniformly varied rectilinear movement.
Taking into account the data:
α =69.2
vf = 10m / s
h=2.7m
g=9.8m/s2
We know we want to know the speed on the y axis.
We calculate vfy
vfy = 10m / s * (sen69.2) = 9.35m / s
We can use the following equation.
We clear the vo (initial speed)
vo=5.87m/s
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Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:
Answer:
Explanation:
Since , we calculate the resistance rate by deriving this formula with respect to time:
Deriving what is left (remember that ):
So we have:
Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):