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DaniilM [7]
4 years ago
10

A 1.6-cm-thick wood floor covers a 4.0m×5.5m room. Part A The subfloor on which the flooring sits is at a temperature of 16.1 ∘C

, while the air in the room is at 19.5 ∘C. What is the rate of heat conduction through the floor? Use 0.200 W/m⋅K for the thermal conductivity of wood.
Physics
1 answer:
Bogdan [553]4 years ago
3 0

Answer:

Rate of heat conduction will be 935 Watt

Explanation:

We have given thickness of the floor d = 1.6 cm = 0.016 m

Area A=4\times 5.5=22m^2

Temperature difference \Delta T=19.5-16.1=3.4^{\circ}C

Thermal conductivity of the wood k = 0.200 w/M-K

We have to find heat conduction

Heat conduction is given by \frac{Q}{t}=\frac{kA\Delta T}{d}=\frac{0.200\times 22\times 3.4}{0.016}=935Watt

Rate of heat conduction will be 935 Watt

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Kyle lays a mirror flat on the floor and aims a laser at the mirror. The laser beam reflects off the mirror and strikes an adjac
fgiga [73]

Answer:

θr = 55.2 °

Explanation:

The law of reflection states that the angle of incidence and the angles of reflection is the same.

Let's use trigonometry to find the angle between the reflected ray, where the horizontal distance x = 55.9 cm and the vertical distance is y = 38.9 cm

    tan θ = y / x

    tan θ = 38.9 / 55.9

    θ = tan⁻¹ (0.6959)

    θ = 34.8°

This angle is measured with respect to the x-axis (horizontal), but in general the angles in optics are measured from the y-axis so that the angle is

    θ = 90 - 34.8

    θr = 55.2 °

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The air around a pool and the water in the pool receive equal amounts of energy from the sun. Why does the air experience a grea
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Solve using correct significant figures and indicating maximum absolute uncertainty.
Vera_Pavlovna [14]

We use the criterion of significant figures to find the result with reliable figures

          X = 9.2 10-5

now with the propagation of errors we obtain the result with its uncertainty

         X ± ΔX = (9.2 ± 0.5) 10⁻⁵

given Parameter

     * expression values ​​with their absolute errors

to find

     * the result with the correct significant figures

     * the absolute error of the expression

Significant figures are defined with the number of decimals that give information, the number of figures in a quantity gives information about the uncertainty of this quantity.

There are two criteria for applying significant figures:

     * Add and subtract the result of going with the number of decimal places of the figure that has the least

    * Product and division as a result of going with the least number of significant figures than the value that has the least.

Remember that the zero to the left do not form a pair of the significant figures

Let's apply this belief to the case presented, let's write the precaution

 

              x = \frac{a-b}{c}

where in this case they are worth

         a = 0.0336 ± 0.0002

         b = 0.010 ± 0.001

         c = 255.4 ± 0.4

We see that the significant figures of each parameterize (a, b, c) and their absolute errors are correct.

Let's apply the criteria to the operation

          a-b = 0.0336 - 0.010

          a- b = 0.0236

we apply the criterion of significant figures for the subtraction, the result must be with 3 decimal places

        a - b = 0.024

let's do the other operation

         X = \frac{a-b}{c}

         X = 0.024 / 255.4

         X = 9.24 10⁻⁵

We apply the criterion of significant figures for the division, in this case the result is left with two significant figures

         X = 9.2 10⁻⁵

The uncertainty or error of the measurements is of most importance as it determines how many significative figures are reliable at a given magnitude.

If the magnitudes are measured with some type of instrument, the absolute error is given by the appreciation of the instrument, if the magnitude is calculated using some equation, the errors must be propagated using the variations of each parameter in the worst case.

           

the uncertainty of the calculated quantity (X) is

        \Delta X = | \frac{dX}{da}| \Delta a + | \frac{dX}{db} | \Delta b + | \frac{dX}{dc}| \Delta c

let's perform the derivatives

        \frac{dX}{da} = \frac{1}{c}

        \frac{dX}{db} = - \frac{1}{c}

        \frac{dX}{dc} = - \frac{a-b}{c^2}

we substitute

remember that the bulk value guarantees that we tune the worst case. So all the mistakes add up

          ΔX = \frac{1}{c}  Δa + \frac{1}{c} Δb + \frac{a-b}{c^2}  Δc

          ΔX = \frac{1}{c} (Δa + Δb) + \frac{a-b}{c^2} Δc

we substitute

         ΔX = \frac{1}{255.4}  (0.0002 + 0.001) + \frac{0.0336-0.010}{255.4^2}  0.4

         ΔX = 4.698 10⁻⁶ + 1.45 10⁻⁷

         DX = 4.8 10-6

Absolute errors must be given with a single significant figure

         ΔX = 5 10⁻⁶

The result of the requested quantity using the criterion of significant figures and propagation of errors is

          X ± ΔX = (9.2 ± 0.5) 10⁻⁵

learn more about   significative figure here:

brainly.com/question/18955573

8 0
3 years ago
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