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sergeinik [125]
2 years ago
11

A house has a roof (colored gray) with the dimensions shown in the drawing. Determine the magnitude of the net force that the

Physics
1 answer:
Sav [38]2 years ago
5 0

The magnitude of the net force that the atmosphere applies to the roof when the outside pressure drops suddenly is  1.9 x 10⁵ N.

<h3>What is pressure?</h3>

The pressure is the amount of force applied per unit area.

Pressure p = Force/Area

Given is a house has a roof (colored gray) with the dimensions shown in the drawing. The outside pressure drops suddenly by 13.2 mm of mercury, before the pressure in the attic can adjust.

The pressure difference ΔP = 13.2 mm of Hg

The length of the roof l = 14.5m

the breadth of the roof h = 4.21m

The force exerted by pressure is

Force, F = P x A

         = (13.2 mm of Hg) [(133 N/m²) /1 mm of Hg ](14.5 x 4.21)

         = 107,170.6 N

Then the net vertical force

    Fnet = 2F cos30

    Fnet = 2 ( 107,170.6) cos30

    Fnet = 185625 N

    Fnet =1.9 x 10⁵ N

The direction of the force is downwards, since the horizontal components of the forces cancel each other.

Hence,  magnitude of the net force is 1.9 x 10⁵ N

Learn more about pressure.

brainly.com/question/12971272

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<h3>With a uniform acceleration of 2 m/s ²</h3>
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3 years ago
Please answer this question sqdancefan​
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Answer:

  (D)  4

Explanation:

The percentage error in each of the contributors to the calculation is 1%. The maximum error in the calculation is approximately the sum of the errors of each contributor, multiplied by the number of times it is a factor in the calculation.

  density = mass/volume

  density = mass/(π(radius^2)(length))

So, mass and length are each a factor once, and radius is a factor twice. Then the total percentage error is approximately 1% +1% +2×1% = 4%.

_____

If you look at the maximum and minimum density, you find they are ...

  {0.0611718, 0.0662668} g/(mm²·cm)

The ratio of the maximum value to the mean of these values is about 1.03998. So, the maximum is 3.998% higher than the "nominal" density.

The error is about 4%.

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<em>Additional comment</em>

If you work through the details of the math, you will see that the above-described sum of error percentages is <em>just an approximation</em>. If you need a more exact error estimate, it is best to work with the ranges of the numbers involved, and/or their distributions.

Using numbers with uniformly distributed errors will give different results than with normally distributed errors. When such distributions are involved, you need to carefully define what you mean by a maximum error. (By definition, normal distributions extend to infinity in both directions.) While the central limit theorem tends to apply, the actual shape of the error distribution may not be precisely normal.

4 0
2 years ago
Read 2 more answers
A 0.3423 g sample of pentane, C5H12, was burned in a bomb (constant-volume) calorimeter. The temperature of the calorimeter and
Yakvenalex [24]

Answer:

16.82 kJ/g

Explanation:

The Heat energy generated by the pentance combusion caused the temperature increasing of the caloriemeter and the water inside it.

Heat energy of a substace can be calculated using

Q = mcθ

C= mc

where Q = Heat energy

          m =mass

          c = specific heat capacity

          C = heat capacity

          θ =temperature difference

(Temperature difference = final temperature - initial temperature.)

heat absorbed by water = mcθ

                                         = 1  * 4.184 * (22.82-20.22)

                                         = 10.88 J --------------(1)

heat absorbed by calorimeter = Cθ

                                                  = 2210 *(22.82-20.22)

                                                  = 5746 J ---------------(2)

Combustion energy = heat absorbed by water + heat absorbed by                                                  calorimeter

Combustion energy =  10.88+ 5746 = 5756.88 J

                                 = 5.757 kJ

heat of combustion of pentane = Combustion energy / mass

                                                    = 5.757 / 0.3423

                                                    = 16.82 kJ/g

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Answer:  Compression

Explanation:

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