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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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Volume of a block can be found by: length × width × height. So:

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I hope it's helpful!

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a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?
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The distance between the two charges is 0.3 m

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

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In this problem, we are given the following:

q_1 = 8\cdot 10^{-5} C

q_2 = 3\cdot 10^{-5} C

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Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

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Brownian motion is due to:
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Answer:

b

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Answer: Option B

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1.5 × 10⁴ m.

Explanation:

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1.5 × 10⁴ m.

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