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MissTica
2 years ago
7

Some students are investigating how resonance works using two tuning forks. Rearrange the steps of the procedure they should fol

low in the correct order. Place the first step at the top and the last step at the bottom.
The second tuning fork will begin to vibrate at the same rate as the first one.
Hold the vibrating tuning fork right beside the other one.
Strike one tuning fork on the edge of the table to make it vibrate.
Physics
1 answer:
Vanyuwa [196]2 years ago
4 0

Answer:

1. strike one tuning fork on the edge of the table to make it vibrate.

2. hold the vibrating tuning fork right beside the other one.

3. the second tuning fork will begin to vibrate at the same rate as the first one.

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The results of a recent television survey of American TV households revealed that 86 out of every 100 TV households have at leas
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q = \dfrac{14}{100}

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given,

86 out of every 100 TV households have at least one remote control

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Probability = \dfrac{favourable\ outcome}{Total\ outcome}

P(at\ least\ one\ remote\ control)=\dfrac{86}{100}

P(at\ least\ one\ remote\ control)=0.86

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3 years ago
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
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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
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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
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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
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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'
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At the instant of peak acceleration, the weighing machine thinks that
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The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
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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
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That result fills me with an overwhelming sense of no-confidence.
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In any case, my conscience is clear ... I do feel that I've put in a good
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