Refer to the diagram shown below.
Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.
The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m
The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²
The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s
Answer: 1.375 s
Part (a): Magnetic dipole moment
Magnetic dipole moment = IA, I = Current, A = Area of the loop
Then,
Magnetic dipole moment = 2.6*π*0.15^2 = 0.184 Am^2
Part (b): Torque acting on the loop
T = IAB SinФ, where B = Magnetic field, Ф = Angle
Then,
T = Magnetic dipole moment*B*SinФ = 0.184*12*Sin 41 = 1.447 Nm
Think of it like this, gravity has to pull harder on the heavier object to make them fall at the same rate , but doesn't have to pull as hard for the lighter object , thus is why sometimes heavier objects fall faster then lighter ones
With acceleration

and initial velocity

the velocity at time <em>t</em> (b) is given by




We can get the position at time <em>t</em> (a) by integrating the velocity:

The particle starts at the origin, so
.



Get the coordinates at <em>t</em> = 8.00 s by evaluating
at this time:


so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).
Get the speed at <em>t</em> = 8.00 s by evaluating
at the same time:


This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:
