NEED HELPASAP what is the moisture content of a board if a test sample that originally weighed 11. 5 oz was found to weigh 10 oz after oven drying?

Answers

Answer 1

Answer:

11.5-10/10= 0.15x100 =15%

Explanation:

Answer 2

The moisture content of the board is 13.04% as per the given data.

What is moisture content?

Simply said, a product's moisture content is its water content. It affects a substance's weight, density, viscosity, conductivity, and other physical characteristics.

The original weight of the test sample was 11.5 oz, and the weight of the sample after oven-drying was 10 oz.

The amount of moisture lost during the oven-drying process is equal to the original weight of the sample minus the weight of the sample after drying, or:

11.5 oz - 10 oz = 1.5 oz

To calculate the moisture content, you then divide the amount of moisture lost by the original weight of the sample and multiply by 100 to express the result as a percentage:

(1.5 oz / 11.5 oz) x 100% = 13.04%

Thus, the moisture content of the board is 13.04%.

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Select the correct answer. Which of the following devices is a simple machine? A.
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B.
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Answer:

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Explanation:

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letter B.

What is the main purpose of the alternater

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Answer:

to keep the battery charged when the vehicle is running, it also helps the battery keep the electrical components in your vehicle charged

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8.
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Please explain the term ‘causal link’. What is the importance of the causal link
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Answers

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A plant has ten machines and currently operates two 8-hr shift per day, 5 days per week, 50 weeks per year. the ten machines produce the same part each at a rate of 30 pc/hr. (a) determine annual production capacity of this plant. (b) if the plant were to operate three 8-hr shifts per day, 6 days per week, 51 weeks per year, determine the annual percentage increase in plant capacity

Answers

Answer:

  83.6%

Explanation:

(a)

On its current schedule, the plant can theoretically produce ...

  (30 pc/h/machine)(10 machine)(8 h/shift)(2 shift/day)(5 day/wk)(50 wk/yr)

  = (30)(10)(8)(2)(5)(50) pc/yr = 1,200,000 pc/yr

__

(b)

On the proposed schedule, this production becomes ...

  (30 pc/mach)(10 mach)(8 h/sh)(3 sh/da)(6 da/wk)(51 wk/yr)

  = (30)(10)(8)(3)(6)(51) pc/yr = 2,203,200 pc/yr

The increase in capacity is ...

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_____

Additional comment

The number of parts per shift did not change. Only the number of shifts per year changed. It went up by a factor of (3/2)(6/5)(51/50) = 1.836. Hence the 83.6% increase in capacity.

We have to assume that maintenance and repair are done as effectvely as before in the reduced down time that each machine has.

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Answer:

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Explanation:

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Answers

The theoretical density of diamond given that the C-C distance and bond angle are 0.154 nm and 109.5° respectively is; 3.54 g/cm³

What is the Theoretical Density?

The first thing we will do is to get the unit cell edge length from the given C - C distance.

From the structure of diamond;

Φ = ¹/₂ * bond angle

Φ = ¹/₂ * 109.5°

Φ = 54.75°

Thus;

θ = 90° - 54.75°

θ = 35.25°

Let the height from a point to the base of the cube of the structure be x. Thus;

x = a/4 = y sin θ

where a is unit cell edge length and y is C -C length. Thus;

a = 4y sin θ

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a = 3.56 * 10⁻⁸ cm

The unit cell volume is;

V_c = a³

Thus; V_c = (3.56 * 10⁻⁸)³

V_c = 4.51 * 10⁻²³ cm³

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ρ = (n * A_c)/(V_c * N_a)

A_c is atomic weight of diamond = 12.01 g/mol

N_a is avogadro's number = 6.022 * 10²² atoms/g

Thus;

ρ = (8 * 12.01)/(4.51 * 10⁻²³ * 6.022 * 10²²)

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Answers

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How to calculate the average temperature of the water.

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[tex]h_{out}= \frac{q}{w} + h_{in}\\\\h_{out}= \frac{1.635 \times 10^9}{34 \times 10^6} + 483.02\\\\h_{out}= 48.09 + 483.02\\\\h_{out}= 531.11\;Btu/lb[/tex]

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[tex]q_{av}=\frac{q}{W} =\frac{q}{\pi A^2H \times 23142} \\\\q_{av}=\frac{485 \times 10^3}{3.142 \times 0.0698^2 \times 91.9 \times 0.01639 \times 23142} \\\\q_{av}=199.5\;kW/liter.[/tex]

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In this scenario, we would assume the reactor core is bare. Thus, the heat is given by:

[tex]q(0) = \frac{2.32PE_1}{HE_R} \\\\q(0) =\frac{2.32 \times 485 \times 180}{23142 \times 200} \\\\q(0) =0.044 \;MW = 1.5 \times 10^8 \;Btu/hr^2[/tex]

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[tex]q_{max}=\frac{q(0)}{2HA^2} \\\\q_{max}=\frac{1.5 \times 10^8}{2 \times (\frac{75.4}{2} \times \frac{1}{12}) \times (\frac{0.298}{2} \times \frac{1}{12})} \\\\q_{max}=\frac{1.5 \times 10^8}{2 \times 3.1417 \times 0.00124} \\\\q_{max}=\frac{1.5 \times 10^5}{0.007791416}\\\\q_{max}=1.55 \times 10^8 \;Btu/hr.ft^3[/tex]

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6.
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A blue and a yellow cubes are rolled- What is the probability that a yellow cube is a multiple of 3 and the product is 6?

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Answer:

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Answer:

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Answers

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