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3 years ago

7 to 1 inch above the stock

1 answer:

6 0

B. Pass a written test on safety and operating procedures of the table saw with a ... 1. Blade guard. 8. Fence lock. 2. Table insert. 9. Fence carriage. 3. Table. 10. ... Never raise the saw blade more than ¼ inch above the material being cut. ... should extend no more than ______ inches above the stock being cut. a. ¼ b. 1/8 c.

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1. Looking at the case study provided under the Companion Material section, what is the main problem that is addressed in this c

Fiesta28 [93]

Incomplete question. However, I provided information that could assist you in identifying the main problem or issue addressed in any case study.

Explanation:

First, note that a case study is simply a learning aid that allows one to learn from a real-life scenario.

To determine the main problems of a case study one needs to:

Read the case as many times as possible to become familiar with the message been expressed. For example,<em> by highlighting or underlining the most important facts </em>it can help you to discover the main problem or issue.
Check for any facts provided in the case study, by so doing you can identify the most important problems.

Thus, by taking these few steps you may be able to determine the main problem in that case study.

6 0

3 years ago

A cylinder with a piston restrained by a linear spring contains 2 kg of carbon dioxide at 500 kPa and 400°C. It is cooled to 40°

leonid [27]

Answer:

heat transfer for the process is - 643.3 kJ

Explanation:

given data

mass m = 2 kg

pressure p1 = 500 kPa

temperature t1 = 400°C = 673.15 K

temperature t2 = 40°C = 313.15 K

pressure p2 = 300 kPa

to find out

heat transfer for the process

solution

we know here mass is constant so

m1 = m2

so by energy equation

m ( u2 - u1 ) = Q - W

Q is heat transfer

and in process P = A+ N that is linear spring

W = ∫PdV

= 0.5 ( P1+P2) ( V1 - V2)

so for case 1

P1V1 = mRT

put here value

500 V1 = 2 (0.18892) (673.15)

V1 = 0.5087 m³

and

for case 2

P2V2 = nRT

300 V2 = 2 (0.18892) (313.15)

V2 = 0.3944 m³

and

here W will be

W = 0.5 ( 500 + 300 ) ( 0.3944 - 0.5087 )

W = -45.72 kJ

and

Q is here for Cv = 0.83 from ideal gas table

Q = mCv ( T2-T1 ) + W

Q = 2 × 0.83 ( 40 - 400 ) - 45.72

Q = - 643.3 kJ

heat transfer for the process is - 643.3 kJ

7 0

3 years ago

A cross beam in a highway bridge experiences a stress of 14 ksi due to the dead weight of the bridge structure. When a fully loa

zlopas [31]

Answer:

a) 2.452

b) 1.256

Explanation:

Stress due to dead weight. = 14 Ksi

Stress due to fully loaded tractor-trailer = 45Ksi

ultimate tensile strength of beam = 76 Ksi

yield strength = 50 Ksi

endurance limit = 38 Ksi

Determine the safety factor for an infinite fatigue life

a) If mean stress on fatigue strength is ignored

β = ( 45 - 14 ) / 2

= 15.5 Ksi

hence FOS ( factor of safety ) = endurance limit / β

= 38 / 15.5 = 2.452

b) When mean stress on fatigue strength is considered

β2 = 45 + 14 / 2

= 29.5 Ksi

Ratio = β / β2 = 15.5 / 29.5 = 0.5254

Next step: applying Goodman method

Sa = [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]

= 19.47 Ksi

hence the FOS ( factor of safety ) = Sa / β

= 19.47 / 15.5 = 1.256

8 0

2 years ago

The car travels around the portion of a circular track having a radius of r = 500 ft such that when it is at point A it has a ve

stellarik [79]

Answer:

Explanation:

Given

velocity at A is $v=4\ ft/s$

For $r=500\ ft$

velocity is increasing at $\dot{v}=0.004t\ ft/s^2$

Tangential acceleration is given by

$a_t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a_t=0.004t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int 0.004tdt=\int dv$

$\int dv=\int 0.004tdt$

$v=0.002t^2+c$

at $t=0\ v=4\ ft/s$

$4=0.002\cdot 0+c$

$c=4\ ft/s$

thus $v=0.002t^2+4$

Velocity in terms of Displacement is given by

$v=\frac{\mathrm{d} s}{\mathrm{d} t}$

$\Rightarrow \int ds=\int \left ( 0.002t^2+4\right )dt$

$\Rightarrow s=\frac{0.002t^3}{3}+4t$

When car has traveled $\frac{3}{4}$ th of distance i.e.

$s=\frac{3}{4}\times (2\pi r)=\frac{3\pi r}{2}$

$s=750\pi$

$750\pi =\frac{0.002t^3}{3}+4t$

$\Rightarrow \frac{0.002t^3}{3}+4t-2356.5=0$

on solving we get $t=139.23\ s$

Thus velocity at $t=139.23\ s$

$v=42.76\ s$

(b)Acceleration when car has traveled three-fourth the way of track

normal acceleration $a_n=\frac{v^2}{r}=\frac{(42.76)^2}{500}$

$a_n=3.658\ m/s^2$

Tangential acceleration $a_t at t=139.23\ s$

$a_t=0.556\ m/s^2$

Net acceleration $a_t=\sqrt{(a_n)^2+(a_t)^2}$

$a_n=\sqrt{(3.658)^2+(0.556)^2}$

$a_n=3.7\ m/s^2$

8 0

3 years ago

. In order to prevent injury from inflating air bags, it is recommended that vehicle occupants sit at least __________ inches aw

scZoUnD [109]

10 inches is the answer

8 0

2 years ago

Read 2 more answers