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4 years ago

2. Can you make adjustments on a saw while the trigger is in the on position.

Engineering

2 answers:

Alchen [17]4 years ago

4 0

No is the answer I guess

Liula [17]4 years ago

3 0

Umm... Heck no, that is very dangerous and could cause you to loose a finger... OR A HAND. To be safe I would always turn it off but if the saw has an on and off switch and a rev switch I think it would be ok.

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Determine the design angle ϕ (0∘≤ϕ≤90 ∘) between struts AB and AC so that the 400 lb horizontal force has a component of 600 lb

Svetllana [295]

Answer:

design angle ∅ = 4.9968 ≈ 5⁰

Explanation:

First calculate the force Fac :

Fac = $\sqrt{400^2 + 650^2 - 2(400)(650)cos30}$

= $\sqrt{160000 + 422500 - 80210}$

= 708.72 Ib

using the sine law to determine the design angle

$\frac{sin}{400} = \frac{sin 30}{Fac}$

hence ∅ = $sin^{-1} (\frac{sin 30 *400}{708.72} )$

= $sin^{-1} 0.0871$ = 4.9968 ≈ 5⁰

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3 years ago

Three single-phase, 10 kVA, 2400/280 V, 60-Hz transformers are connected to form a three-phase, 2400/480 V transformer The equiv

Dominik [7]

The answer to this question is letter A

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4 years ago

Where the velocity is highest in the radial direction? Why?

posledela

Answer:

In the center and directed away from it.

Explanation:

The direction along the radius and directed away from the center is known as radial direction.

The velocity is highest in the radial direction pointing away from the center, this is because of the reason that when the particle executes its motion in the direction that is radial, then it is not acted upon by any force that opposes the motion of the particle and thus there is no obstruction to the velocity of the particle and it is therefore, the highest in the radial direction.

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3 years ago

What are some "vital signs" that we consider to tell us about the economy?

Wittaler [7]

Explanation:

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if the budget is good

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3 years ago

Read 2 more answers

Miller Indices:

svetlana [45]

Answer:

A) The sketches for the required planes were drawn in the first attachment [1 2 1] and the second attachment [1 2 -4].

B) The closest distance between planes are d₁₂₁=a/√6 and d₁₂₋₄=a/√21 with lattice constant a.

C) Five posible directions that electrons can move on the surface of a [1 0 0] silicon crystal are: |0 0 1|, |0 1 3|, |0 1 1|, |0 3 1| and |0 0 1|.

Compleated question:

1. Miller Indices:

a. Sketch (on separate plots) the (121) and (12-4) planes for a face centered cubic crystal structure.

b. What are the closest distances between planes (called d₁₂₁ and d₁₂₋₄)?

c. List five possible directions (using the Miller Indices) the electron can move on the surface of a (100) silicon crystal.

Explanation:

A)To draw a plane in a face centered cubic lattice, you have to follow these instructions:

1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment) and the planes has 3 main coeficients shown as [l m n]

2- The coordinates of that plane are written as: π:[1/a₀ 1/b₀ 1/c₀] (if one of the coordinates is 0, for example [1 1 0], c₀ is ∞, therefore that plane never cross the direction c).

3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.

4- Join the points.

In this case, for [1 2 1]:

$l=1=1/a_0 \rightarrow a_0=1$

$m=2=2/b_0 \rightarrow b_0=0.5$

$n=1=1/c_0 \rightarrow c_0=1$

for $[1 2 \overline{4}]$ :

$l=1=1/a_0 \rightarrow a_0=1$

$m=2=2/b_0 \rightarrow b_0=0.5$

$n=\overline{4}=-4/c_0 \rightarrow c_0=-0.25$

B) The closest distance between planes with the same Miller indices can be calculated as:

With $\pi:[l m n]$ ,the distance is $d_{lmn}= \displaystyle \frac{a}{\sqrt{l^2+m^2+n^2}}$ with lattice constant a.

In this case, for [1 2 1]:

$d_{121}= \displaystyle \frac{a}{\sqrt{1^2+2^2+1^2}}=\frac{a}{\sqrt{6}}=0.41a$

for $[1 2 \overline{4}]$ :

$d_{12\overline{4}}= \displaystyle \frac{a}{\sqrt{1^2+2^2+(-4)^2}}=\frac{a}{\sqrt{21}}=0.22a$

C) The possible directions that electrons can move on a surface of a crystallographic plane are the directions contain in that plane that point in the direction between nuclei. In a silicon crystal, an fcc structure, in the plane [1 0 0], we can point in the directions between the nuclei in the vertex (0 0 0) and e nuclei in each other vertex. Also, we can point in the direction between the nuclei in the vertex (0 0 0) and e nuclei in the center of the face of the adjacent crystals above and sideways. Therefore:

dir₁=|0 0 1|

dir₂=|0 0.5 1.5|≡|0 1 3|

dir₃=|0 1 1|

dir₄=|0 1.5 0.5|≡|0 3 1|

dir₅=|0 0 1|

5 0

3 years ago