8. In soft magnetic materials such as iron, what happens when an external magnetic field is removed?
a. The domain alignment persists.
b. The orientation of domains fluctuates.
c. The material becomes a hard magnetic material.
d. The orientation of domains changes, and the material returns to an unmagnetized state.
9. According to Lenz’s law, if the applied magnetic field changes,
a. the induced field attempts to keep the total field strength constant.
b. the induced field attempts to increase the total field strength.
c. the induced field attempts to decrease the total field strength.
d. the induced field attempts to oscillate about an equilibrium value.
10. The direction of the force on a current-carrying wire in an external magnetic field is
a. perpendicular to the current only.
b. perpendicular to the magnetic field only.
c. perpendicular to the current and to the magnetic field.
d. parallel to the current and to the magnetic field
Answer: East of North
Explanation:
We have the following data:
Speed of the wind from East to West:
Speed of the bee relative to the air:
If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).
Now, we need to find the direction the bee should fly directly to the flower (due North):
Clearing :
4. The Coyote has an initial position vector of .
4a. The Coyote has an initial velocity vector of . His position at time is given by the vector
where is the Coyote's acceleration vector at time . He experiences acceleration only in the downward direction because of gravity, and in particular where . Splitting up the position vector into components, we have with
The Coyote hits the ground when :
4b. Here we evaluate at the time found in (4a).
5. The shell has initial position vector , and we're told that after some time the bullet (now separated from the shell) has a position of .
5a. The vertical component of the shell's position vector is
We find the shell hits the ground at
5b. The horizontal component of the bullet's position vector is
where is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for :