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3 years ago

Explain jack plane. please dont copy from go ogle

2 answers:

5 0

Jack planes are frequently used to flatten rough stock and reduce it to final size. However, depending on the size of the workpiece, they can be beneficial for smoothing and jointing.

Ksju [112]3 years ago

4 0

Jack Plane is designed to take off heavy shavings and squares up rough timber to correct size and quickly removes waste wood.

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The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content

MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

$Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf$

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

$\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}$

Therefore, $V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}$

$V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy$

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since $1 yd^{3}= 27 ft^{3}$

The embankment requires water of 6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

$Unit_{weight}=\frac {17451720}{498594}=35.00186 lb$

1 gallon is approximately $8.35 yd^{3}$ hence

$\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}$

That's approximately 4.2 gallons

7 0

3 years ago

The Texas Sure program is designed to:

Romashka [77]

TexasSure is designed to reduce the number of uninsured drivers and cut costs for responsible Texans, who now pay almost $900 million a year to protect themselves against those with no coverage. Currently, an estimated 20 percent of Texas drivers are uninsured.

TexasSure is the financial responsibility verification program created by the 79th Texas Legislature, Regular Session, in Senate Bill 1670.

4 0

2 years ago

A 360 kg/min stream of steam enters a turbine at 40 bar pressure and 100 degrees of superheat. The steam exits the turbine as a

Brrunno [24]

Answer:
skskkdkdkfkgkgkgkkgkgkgigooigigi lol
Explanation:
Oof

7 0

2 years ago

A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel. The paddle wheel does 240 kJ of work on the

Sergeu [11.5K]

To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.

By definition the entropy change would be defined as

$\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{P_2}{P_1})$

Using the Boyle equation we have

$\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{v_1T_2}{v_2T_1})$

Where,

$C_p$ = Specific heat at constant pressure

$T_1$ = Initial temperature of gas

$T_2$ = Final temprature of gas

R = Universal gas constant

$v_1$ = Initial specific Volume of gas

$v_2$ = Final specific volume of gas

According to the statement, it is an isothermal process and the tank is therefore rigid

$T_1 = T_2, v_2=v_1$

The equation would turn out as

$\Delta S = C_p ln1-ln1$

<em>Therefore the entropy change of the ideal gas is 0</em>

Into the surroundings we have that

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat Exchange

T = Temperature in the surrounding

Replacing with our values we have that

$\Delta S = \frac{230kJ}{(30+273)K}$

$\Delta S = 0.76 kJ/K$

<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>

8 0

3 years ago

6. Driving with parking lights only (in place of headlights) is against the law. A. True B. False

trasher [3.6K]

Answer:

B false it is illegal to only have got fog lights on though and bright headlights because it can distract other drivers going last and if the y are distracted then that will cause a collision

Hope this helps :)

Explanation:

4 0

3 years ago

Read 2 more answers

Explain jack plane. please dont copy from go ogle​

Explain jack plane. please dont copy from go ogle