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lawyer [7]
3 years ago
15

A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass

uming the motion is vertical calculate the amplitude of the ship's motion, if the scale reading of the machine varies between limits of 55.0kg and 65.0kg.
Physics
1 answer:
STALIN [3.7K]3 years ago
8 0

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 .

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6 points + Brainliest asap!
Artist 52 [7]

Answer:

  The acceleration would double

Explanation:

Assuming the same box and spring, the maximum acceleration is proportional to amplitude. When the amplitude doubles, the acceleration would double.

__

A spring is generally considered to have a linear force vs. distance characteristic. Hence, doubling the distance doubles the force. The acceleration is proportional to the force.

3 0
3 years ago
Course hero N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches
aleksandrvk [35]

Answer:

Therefore it is save to carry a 62kg adult

Explanation:

From the question we are told that:

Mass m=380kg

Height of supporting Rock X=85cm

Length of BoardL_r=4.5m

Mass of board M_b=22kg

Mass of adult M_a=62  

Generally the moment of balance about wedge part about  is mathematically given by

N -Q + R = Mg + mg

0.85*N - Mg*2.25 - mg*(2.25 + x) = 0

0.85*N  = + Mg*2.25 + mg*(2.25 + x)

where

N+R=4547

therefore

N = 570.70588 + 1608.3529 + 714.823 x

if N=0 at fallen person

x=3.04m

Therefore it is save to carry a 62kg adult

8 0
2 years ago
An open diving chamber rests on the ocean floor at a water depth of 60 meter. Find the air pressure (gage pressure relative to t
Fofino [41]

Answer:

Gauge Pressure required = 606.258 kPa

Explanation:

Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.

Thus at a depth of 60m we have pressure of water equals

P(z)=P_{0}+\rho _wgh

Now the gauge pressure is given by

P(z)-P_{0}=\rho _wgh

Applying values we get

P(z)-P_{0}=\rho _wgh\\\\P_{gauge}=1.03\times 1000\times 9.81\times 60Pa\\\\P_{gauge}=606258Pascals\\\\P_{gauge}=606.258kPa

8 0
3 years ago
Answer this please. thanks in advance!! please tel me                                                                 a christma
saw5 [17]
If 50 identical light bulbs are connected in series across
a single power source, then the voltage across each bulb
is ( 1/50 ) of the voltage delivered by the power source.
6 0
3 years ago
Read 2 more answers
An airplane travels 4000m in 16 seconds on a heading of 35 whats is its velocity
Marizza181 [45]

In the question, you just gave a complete and detailed
description of the plane's velocity vector:

       4,000/16  meters/second , heading 35 degrees .

You might want to simplify the speed and make it a unit rate,
but otherwise, it's perfect.

         250 meters/second, heading 35 degrees .

6 0
3 years ago
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