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ollegr [7]
3 years ago
15

Jeremy accidently dropped his toy stuffed animal from the balcony of his apartment on the fourth floor. The toy hit the ground a

t a velocity of 16.0 meters/second. At impact, it took 2.0 seconds for the toy’s velocity to reach 0 meters/second. If the toy has a mass of 0.25 kilograms, what's the force of the impact?
Physics
2 answers:
GalinKa [24]3 years ago
6 0

Ayy the answer is 2 newtons well for plato anyways

uysha [10]3 years ago
3 0

F = (mass)(acceleration) = ma

m = 0.25 kg

Vi = 16 m/s

t = 2 s

Vf = 0 m/s (since it was put to stop)

a=(Vf-Vi)/t

a=(0-16)/2

a = 8 m/s^2 (decelerating)

F = ma = (0.25 kg)(8 m/s^2)

F = 2 N

<span>Hope this answer will be a good h<span>elp for you.</span></span>

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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
STALIN [3.7K]

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,
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If the mass of the man standing on the weighing machine is 60kg, then
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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
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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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m = (0.000249 kg) (2.205 lbm/kg)

m = 0.000549 lbm

m = 5.49×10⁻⁴ lbm

3 0
3 years ago
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