Answer:
2.64 x 10⁻⁶T
Explanation:
The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;
B = (μ₀ I) / (2π r) ----------------(i)
B is magnetic field
I is current through the wire
r is the distance from the wire
μ₀ is the magnetic constant = 4π x 10⁻⁷Hm⁻¹
From the question;
I = 0.7A
r = 0.053m
Substitute these values into equation (i) as follows;
B = (4π x 10⁻⁷ x 0.7) / (2π x 0.053)
B = 2.64 x 10⁻⁶T
Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T
The mechanical advantage of a machine is the ratio of the force produced by the machine to the force applied to it. Therefore, we may calculate the applied force using:
Mechanical advantage = force by machine / force applied
6 = 2 / force applied
Force applied = 1/3
Thus, the distance that the effort must move will be 1/3 inch
We have no idea. There are probably many things involved ... the planet's mass,
the availability of small bodies in the neighborhood that can be captured, etc.
Using the pressure law (P1 x V1)/ T1 = (P2 x V2)/ T2 where P1= the initial pressure V1= initial volume T1= initial temperature and P2= the final pressure V2= the final volume T2 = the final temperature and temperature is always in kelvin
The second object, the one that had twice the force applied to it, would move twice as far, I believe.
