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KatRina [158]
3 years ago
12

Suppose an individual is lying on his stomach with sheets of paper stacked on his back. If each sheet of paper has a mass of 0.0

0350 kg and is the standard letter size of 8.5 in by 11 in ( 0.216 m by 0.279 m ), how many sheets must be stacked to produce a pressure on his back equal to atmospheric pressure (roughly 101325 Pa )
Physics
1 answer:
AURORKA [14]3 years ago
5 0

Answer:

N = 177843 sheets

Explanation:

We are given;

Mass;m = 0.0035 kg

Pressure; p = 101325 pa = 101325 N/m²

L = 0.279m

W = 0.216m

The weight of N sheets is N(mg)

Where;

m is the mass of one sheet

N is number of sheets

g is the acceleration due to gravity.

The pressure equals weight divided by the area on which the weight presses:

Thus,

p= F/A = Nmg/(L•W)

Therefore, making N the subject;

N = pLW/(mg)

N = 101325 x 0.279 x 0.216/ (0.0035 x 9.81)

N = 177843

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  • The range of motion is: A = 4.44 cm

<h3>Oscillatory movement.</h3>

The oscillatory periodic motion of a system occurs when there is a recovered force, in the special case that this force is proportional to the displacement is called simple harmonic motion, which is described by the expression.

            x = A cos (wt + Ф)

            w² = k/m

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Let's find the angular velocity/

            w= \sqrt{ \frac{200}{0.300} }

            w = 25.8 rad/s

Let's look for the constant Ф, as the system is released from rest its initial velocity is zero, for zero time. The definition of speed is:

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             0 = -A w sin Ф

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They indicate that at a given instant of the time the velocity is v= 50.0 cm/s and it is in a position x= 4.00 cm, let us write the equations for this time

Position.

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Speed.

              50.0 = - At 25.8 sin 25.8t

To solve the system, ;et's square and add.

              Cos² 25.8t = \frac{16}{A^2}

              sin² 25.8t = \frac{3.756}{A^2 }

              1 = \frac{1}{A^2} \ (16 + 3.756)

               A = \sqrt{19.756}

               A= 4.44 cm

In conclusion using the solution for the oscillatory movement we can find the result for the amplitude of the initial displacement is:

The range of motion is: A = 444 cm

Learn more about oscillatory motion here:  brainly.com/question/14311816

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