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Colt1911 [192]
3 years ago
13

What are the causes of kickback on a table-saw?

Engineering
1 answer:
love history [14]3 years ago
4 0
From the spinning blade pushing (away) on the material.
It should be held down tightly
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Experimental Design Application Production engineers wish to find the optimal process for etching circuit boards quickly. They c
Veseljchak [2.6K]

Answer:

Hello your question is incomplete attached below is the missing part and answer

options :

Effect A

Effect B

Effect C

Effect D

Effect AB

Effect AC

Effect AD

Effect BC

Effect BD

Effect CD

Answer :

A  = significant

 B  = significant

C  = Non-significant

D  = Non-significant

AB  = Non-significant

AC  = significant

AD  = Non-significant

BC  = Non-significant

 BD  = Non-significant

 CD = Non-significant

Explanation:

The dependent variable here is Time

Effect of A  = significant

Effect of B  = significant

Effect of C  = Non-significant

Effect of D  = Non-significant

Effect of AB  = Non-significant

Effect of AC  = significant

Effect of AD  = Non-significant

Effect of BC  = Non-significant

Effect of BD  = Non-significant

Effect of CD = Non-significant

8 0
3 years ago
What do you mean by searching?​
nikitadnepr [17]

Answer:

thoroughly scrutinizing, especially in a disconcerting way.

Explanation:

8 0
3 years ago
Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of
earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant

Since the losses are neglected thus applying this theorm between upper and lower porion we have

\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}

Now by continuity equation we have

A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s

Applying the values in the Bernoulli's equation we get

\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars

6 0
3 years ago
It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 630 MPa. Calc
ratelena [41]

Answer:

The answer is below

Explanation:

Given that:

Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength (\sigma_{cd})=630\ MPa = 630*10^6\ Pa, Fracture strength

(\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa

a) The critical length (L_c) is given by:

L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm

The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.

b) The volume fraction (Vf) is gotten from the formula:

\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}}  \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456

6 0
3 years ago
In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water
Orlov [11]

Answer:

The amount of energy transferred to the water is 4.214 J

Explanation:

The given parameters are;

The mass of the object that drops = 5 kg

The height from which it drops = 86 mm (0.086 m)

The potential energy P.E. is given by the following formula

P.E = m·g·h

Where;

m = The mass of the object = 5 kg

g = The acceleration de to gravity = 9.8 m/s²

h = The height from which the object is dropped = 0.086 m

Therefore;

P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J

Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;

The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.

6 0
2 years ago
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