Answer:
4 smaller disks
Explanation:
We are given;
Mass of smaller and larger disks = M
Radius of smaller disk = R
Radius of larger disk = 4R
Formula for moment of inertia about cylinder axis is:
I = ½MR²
Thus;
For small disk, I_small = ½MR²
For large disk, I_large = ½M(2R)² = 2MR²
We are told that moment of inertia of System A consists of two of the larger disks. Thus;
I_A = 2 × I_large = 2 × 2MR²
I_A = 4MR²
We are also told that System B consists of one of the larger disks and a number of the smaller disks. Thus;
I_B = I_large + n(I_small)
Where n is the number of smaller disks.
I_B = 2MR² + n(½MR²)
I_B = MR²(2 + n/2)
We are told that the moment of inertia for system A equals the moment of inertia for system B. Thus;
I_A = I_B
So;
4MR² = MR²(2 + n/2)
MR² will cancel out to give;
4 = 2 + n/2
Multiply through by 2 to give;
8 = 4 + n
n = 8 - 4
n = 4
Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force. We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.
If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as 588 newtons or as
132.3 pounds. That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.
If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is
y(t) = y₀ + M sin(2π t/15) .
The vertical speed of the deck is y'(t) = M (2π/15) cos(2π t/15)
and its vertical acceleration is y''(t) = - (2πM/15) (2π/15) sin(2π t/15)
= - (4 π² M / 15²) sin(2π t/15)
= - 0.1755 M sin(2π t/15) .
There's the important number ... the 0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.
The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of 0.1755 x amplitude).
At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of 65kg, when in reality it's only 60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.
Now I'm going to wave my hands in the air a bit:
Apparent weight = (apparent mass) x (real acceleration of gravity)
(Apparent mass) = (65/60) = 1.08333 x real mass.
Apparent 'gravity' = 1.08333 x real acceleration of gravity.
The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.
0.08333 G = 0.1755 M
The 'M' is what we need to find.
Divide each side by 0.1755 : M = (0.08333 / 0.1755) G
'G' = 9.0 m/s²
M = (0.08333 / 0.1755) (9.8) = 4.65 meters .
That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .
Answer:
The decibel of the remaining pigs is 51.5 dB.
Explanation:
Decibel (dB) is a unit of measure of the intensity of a given sound.
Number of pigs = 199, noise level = 74.3 dB.
Given that the intensity (I) of the sound from the pen is proportional to the number of pigs (N), thus:
I 
N
I = kN
where k is the constant of proportionality.
⇒ k = 
= 
k = 0.3734
When 61 numbers of pigs were removed, the number of remaining pigs (N) squealing at their original level is 138.
Thus, the becibel level (I) of the remaining pigs can be determined by:
I = kN
= 0.3734 × 138
= 51.53 dB
The becibel level (I) of the remaining pigs is 51.53 dB.
