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3 years ago

Which type of finish is absorbed into the wood?

1 answer:

7 0

Penetrating finishes are absorbed by the wood and dry inside the wood. Interior wood stain is used to accentuate the grain in natural woods or plywood while still adding some color. While wood stains offer some level of protection on wood, it's recommended to use stain in combination with a surface finish.

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Whatever they said bbbbbbbbbbbbbbbbbb

solong [7]

Answer:

yh and I said aaaaaaaaaaaa

Explanation:

7 0

3 years ago

Read 2 more answers

What steps would you take to design an improved toothpaste container?

algol [13]

Answer:

A. Identify the need, recognize limitations of current toothpaste containers, and then brainstorm ideas on how to improve the existing

Explanation:

To design an improved toothpaste container, we must identify the needs of the customer, one of the major need is to make the container attractive to the sight. This is the first thing that will prompt a customer to wanting to buy the product (The reflectance/appearance).

Then recognize the limitation of the current design, what needed change. This will help in determining what is needed to be included and what should be removed based on identified customers need.

The last step is to brainstorm ideas on how to improve the existing designs. Get ideas from other colleagues because there is a saying that two heads are better than one. This will help in coming to a reasonable conclusion on the new design after taking careful consideration of people's opinion.

7 0

3 years ago

When an object is moving, we use the following coefficient for friction calculations a)-μk b)-μs c)-γk d)- γs

Reika [66]

Answer: $\mu_{k}$

Explanation:

We use kinetic friction when a body is moving i.e. $\mu_{k}$ for calculations.

Static friction is used when a body is in rest while kinetic friction is used when a body is moving and its value is quite low as compared to static friction .

Static friction value increases as we apply more force while kinetic friction occurs when there is relative motion between bodies.

3 0

4 years ago

A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

Bad White [126]

Answer:

the net work per cycle $\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume $V_2 = 0.16 V_1$

the crankshaft N rotates at a speed of 2400 RPM.

At the beginning of the compression , temperature $T_1$ = 60 F = 519.67 R

and;

Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

$V_1-V_2 = \dfrac{\pi}{4}D^2L$

$V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)$

$V_1-0.16V_1= 36.55714291$

$0.84 V_1 =36.55714291$

$V_1 =\dfrac{36.55714291}{0.84 }$

$V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3$

$V_1 = 0.02518 \ ft^3$

the mass in air ( lb) can be determined by using the formula:

$m = \dfrac{P_1V_1}{RT}$

where;

R = 53.3533 ft.lbf/lb.R°

$m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}$

m = 0.0018962 lb

From the tables of ideal gas properties at Temperature 519.67 R

$v_{r1} =158.58$

$u_1 = 88.62 Btu/lb$

At state of volume 2; the relative volume can be determined as:

$v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}$

$v_{r2} = 158.58 \times 0.16$

$v_{r2} = 25.3728$

The specific energy $u_2$ at $v_{r2} = 25.3728$ is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

$v_{r3} = 0.1828$

$u_3 = 1098 \ Btu/lb$

To determine the relative volume at state 4; we have:

$v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}$

$v_{r4} =0.1828 \times \dfrac{1}{0.16}$

$v_{r4} =1.1425$

The specific energy $u_4$ at $v_{r4} =1.1425$ is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

$W_{net} = Heat \ supplied - Heat \ rejected$

$W_{net} = m(u_3-u_2)-m(u_4 - u_1)$

$W_{net} = m(u_3-u_2- u_4 + u_1)$

$W_{net} = m(1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (410.08)$

$\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

$W = 4 \times N' \times W_{net$

where ;

$N' = \dfrac{2400}{2}$

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec × 0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0

3 years ago

Explain why the following scenario fails to meet the definition of a project description.

s344n2d4d5 [400]

Answer:

The youth hockey training facility

Explanation:

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3 years ago