Answer:
Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all - option A
Explanation:
Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all; the reason being that: no magnetic field is being produced by a charge that is static. Only a moving charge can produce a magnetic effect. And the magnet can not have any torque due to its own magnetic lines of force.
Answer:
6840 N
Explanation:
The force acting on the car can be found by using Newton's second law:
F = ma
where
F is the net force on the car
m is the mass of the car
a is its acceleration
For the car in this problem,
m = 1800 kg
Substituting,
Explanation:
Fluid gauge pressure is:
P = ρgh
where ρ is the fluid density and h is the depth of the fluid.
P = (1000 kg/m³) (9.8 m/s²) (1642 m)
P = 16,091,600 Pa
Rounded to four significant figures, the gauge pressure is 16.09 MPa.
Answer:
The new period of rotation using the new spring would be less than the period of rotation using the original spring
Explanation:
Generally the period of rotation of the mass is mathematically represented as
Here I is the moment of inertia of the mass about the rotation axis and k is the spring constant
Now looking at the equation we can tell that T is inversely proportional to the square root of the spring constant which means that for a larger spring constant the time period would be lesser
