The rear wheel on a clown’s bicycle has twice the radius of the front wheel. (a) When the bicycle is moving, is the linear speed
1 answer:
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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:
At 30 degrees, height = 14.39 m and distance = 49.83 m
At 45 degrees, height = 28.77 m and distance = 57.54 m
At 60 degrees, height = 43.16 m and distance = 49.83 m
Explanation:
Let ∝ be the angle with respect to the horizontal that the baseball is hit with.
The horizontal component of the velocity is vcos(∝) and the vertical component of the velocity is vsin(∝).
Ignore air resistant, only gravitational acceleration g = -9.81 m/s2 affect the ball vertically. We can use the following equation to calculate the time it takes to reach maximum height (at 0 speed)
So the vertical distance it travels within time t is
Similarly the horizontal distance it travels within time t is:
We can pre-calcualte the quantity
So
From here we can plug-in the angles values of 30, 45 and 60 degrees
At 30 degrees, height = 14.39 m and distance = 49.83 m
At 45 degrees, height = 28.77 m and distance = 57.54 m
At 60 degrees, height = 43.16 m and distance = 49.83 m