The unmagnetized pieces of iron would be randomly pointing to directions, this is true because although influenced with the magnetic domain, the direction of the unmagnetized iron field of attraction is not uniform or does not have preferred direction.
<h2>The option ( c ) is correct </h2>
Explanation:
As the frequency of oscillation of any oscillator is doubled
The velocity of sound v = νλ
here ν is the frequency and λ is the wavelength
Now if ν becomes double , the wavelength λ becomes one half . The velocity of sound remains the same in the same medium .
Thus option ( c ) is correct
Answer:
Approximately
, assuming that the gravitational field strength is
.
Explanation:
Let
denote the required angular velocity of this Ferris wheel. Let
denote the mass of a particular passenger on this Ferris wheel.
At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:
- Weight of the passenger (downwards),
, and possibly - Normal force
that the Ferris wheel exerts on this passenger (upwards.)
This passenger would feel "weightless" if the normal force on them is
- that is,
.
The net force on this passenger is
. Hence, when
, the net force on this passenger would be equal to
.
Passengers on this Ferris wheel are in a centripetal motion of angular velocity
around a circle of radius
. Thus, the centripetal acceleration of these passengers would be
. The net force on a passenger of mass
would be
.
Notice that
. Solve this equation for
, the angular speed of this Ferris wheel. Since
and
:
.
.
The question is asking for the angular velocity of this Ferris wheel in the unit
, where
. Apply unit conversion:
.
Answer:
c. 40200 J
Explanation:
Assume gravitational constant g = 9.8m/s2. The weight of the 2000kg vehicle is

In addition to the friction averaging at 500N, the total force is
F = 20000 + 500 = 20100 N
The work required to generate this force over a distance of 2m would be
F*s = 20500 * 2 = 40200 J
So c.40200 J is the correct answer