(d) Suppose you use a spring to launch a payload horizontally from the asteroid so that the payload ends up far from the asteroi
1 answer:
Answer:
ks= 133.2 N/m
Explanation:
- Assuming that we can neglect the gravitational potential energy of the mass, and that no other forces acting on the payload, total mechanical energy must be conserved.
- This energy, at any time, is part elastic potential energy (stored in the spring) and part kinetic energy.
- When the spring is initially compressed, the payload is at rest, so all energy is elastic potential.
- Once the spring has returned to its natural state, all this elastic potential energy must have been turned into kinetic energy.
- If the payload is launched horizontally, and no gravity is present,this means that its final speed will be horizontal only also, according to Newton's First Law.
- So, we can write the following equation:
- where ΔU = -1/2*k*(Δx)² (2)
- and ΔK = 1/2*m*v² (3)
- Replacing in (2) and (3) by the givens, and simplifying, we can find the stiffness ks as follows:
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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: