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Rufina [12.5K]
3 years ago
13

A barometer reads 780 mm Hg. Mercury has a density of 1.36 x 10^4 kg /m^3.

Physics
2 answers:
iris [78.8K]3 years ago
7 0

The pressure of the atmosphere, when a barometer reads 780 mm Hg.    Mercury which a density of 1.36 x 10^4 kg /m^3 is B 1.1 x 10^5 N/m^2

This problem can be solved using the formula below

P = dgh................. Equation 1

Where P = Pressure of the atmosphere, d = density of the mercury, h = height of the mercury, g = acceleration due to gravity.

From the question,

Given: d = 1.36×10⁴ kg/m³, h = 780 mm = 0.78 m,

Constant: g = 10 m/s²

Substitute these values into equation 1

P = (1.36×10⁴)(10)(0.78)

P = 10.608×10⁴ N/m²

P ≈ 1.1×10⁵ N/m²

Hence the right answer is B. 1.1×10⁵ N/m²

Learn more about Pressure here: brainly.com/question/23603188

frutty [35]3 years ago
5 0

Answer:

B

Explanation:

Pressure = density × g × height

p \:  = (1.36 \times  {10}^{4} )(10)(780 \times  {10}^{ - 3}  ) \\ p = 1.1 \times  {10}^{5}

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System diagram for given situation is shown in attached Fig. 1

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<h3>PROOF:</h3>

rotational inertia for cylinders  is given as:

                                  I=\frac{1}{2}MR^{2} -----(1)

Newton's second law for angular motion is:

                                             ∑τ = Iα ------(2)

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                                               τ  = RF --------(4)        

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                                       RF=\frac{1}{2}MR^{2}\alpha

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

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As angular acceleration is related to linear by:

                                          a= R\alpha

Eq (5) becomes

                                    f_{s}=\frac{1}{2}Ma_{x}---- (6)

aₓ shows displacement in horizontal direction

From (3)

                                              ∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

                                  \sum F_{x} = f_{s} - kx ----(7)

Put (7) in (3)

                                  f_{s} - kx  = Ma_{x}[/tex] -----(8)

Using (6) in (8)

                               \frac{1}{2}Ma_{x} - kx =Ma_{x}

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                                  \omega^{2} = \frac{2k}{3M}

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Time period is related to angular frequency as:

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                                  T = 2\pi \sqrt{\frac{3M}{2k}

                           

 

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