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mojhsa [17]
3 years ago
9

Hard woods and tight curves should be cut using what blade speeds.

Engineering
1 answer:
erica [24]3 years ago
4 0

Answer:

slower

Explanation:

When it comes to using scroll saws, it is important to consider the <u>type of wood</u> in order to choose the correct speed. The blades of the saw tends to break easily, thus it is essential to choose a <em>"slower speed"</em> when shaping hard woods. Such speed is also used for<em> "tight curves."</em> Furthermore, it is important not to force the wood to the blade because it will also cause breakage.

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Please answer fast. With full step by step solution.​
lina2011 [118]

Let <em>f(z)</em> = (4<em>z </em>² + 2<em>z</em>) / (2<em>z </em>² - 3<em>z</em> + 1).

First, carry out the division:

<em>f(z)</em> = 2 + (8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1)

Observe that

2<em>z </em>² - 3<em>z</em> + 1 = (2<em>z</em> - 1) (<em>z</em> - 1)

so you can separate the rational part of <em>f(z)</em> into partial fractions. We have

(8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1) = <em>a</em> / (2<em>z</em> - 1) + <em>b</em> / (<em>z</em> - 1)

8<em>z</em> - 2 = <em>a</em> (<em>z</em> - 1) + <em>b</em> (2<em>z</em> - 1)

8<em>z</em> - 2 = (<em>a</em> + 2<em>b</em>) <em>z</em> - (<em>a</em> + <em>b</em>)

so that <em>a</em> + 2<em>b</em> = 8 and <em>a</em> + <em>b</em> = 2, yielding <em>a</em> = -4 and <em>b</em> = 6.

So we have

<em>f(z)</em> = 2 - 4 / (2<em>z</em> - 1) + 6 / (<em>z</em> - 1)

or

<em>f(z)</em> = 2 - (2/<em>z</em>) (1 / (1 - 1/(2<em>z</em>))) + (6/<em>z</em>) (1 / (1 - 1/<em>z</em>))

Recall that for |<em>z</em>| < 1, we have

\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n

Replace <em>z</em> with 1/<em>z</em> to get

\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}

so that by substitution, we can write

\displaystyle f(z) = 2 - \frac2z \sum_{n=0}^\infty (2z)^{-n} + \frac6z \sum_{n=0}^\infty z^{-n}

Now condense <em>f(z)</em> into one series:

\displaystyle f(z) = 2 - \sum_{n=0}^\infty 2^{-n+1} z^{-(n+1)} + 6 \sum_{n=0}^\infty z^{-n-1}

\displaystyle f(z) = 2 - \sum_{n=0}^\infty \left(6+2^{-n+1}\right) z^{-(n+1)}

\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{-(n-1)+1}\right) z^{-n}

\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{2-n}\right) z^{-n}

So, the inverse <em>Z</em> transform of <em>f(z)</em> is \boxed{6+2^{2-n}}.

4 0
3 years ago
Sea A una matriz 3x3 con la propiedad de que la transformada lineal x → Ax mapea R³ sobre R³.
skelet666 [1.2K]

Answer:

ax

Explanation:

7 0
3 years ago
Quadrature encoders operate with two photodetectors offset by _______ degrees. (a) 45 (b) 90 (c) 180 (d) 270
Lemur [1.5K]

Answer:

i think it should be 90 degree

5 0
2 years ago
An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75
Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process            T_{3} = 2042.56 K

(b). The value of pressure at the end of heat addition process                    P_{3} = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle   E_{otto} = 0.4478

(d). The value of mean effective pressure of the cycle P_{m} = 1506.41 \frac{k pa}{kg}

Explanation:

Compression ratio r_{p} = 8

Initial pressure P_{1} = 95 k pa

Initial temperature T_{1} = 278 °c = 551 K

Final pressure P_{2} = 8 × P_{1} = 8 × 95 = 760 k pa

Final temperature T_{2} = T_{1} × r_{p} ^{\frac{\gamma - 1}{\gamma} }

Final temperature T_{2} = 551 × 8 ^{\frac{1.4 - 1}{1.4} }

Final temperature T_{2} = 998 K

Heat transferred at constant volume Q = 750 \frac{KJ}{kg}

(a). We know that Heat transferred at constant volume Q_{S} = m C_{v} ( T_{3} - T_{2}  )

⇒ 1 × 0.718 × ( T_{3} - 998 ) = 750

⇒ T_{3} = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

⇒ \frac{P_{3} }{P_{2} } = \frac{T_{3} }{T_{2} }

⇒ P_{3} = \frac{2042.56}{998} × 760

⇒ P_{3} = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle Q_{R} = m C_{v} ( T_{4} - T_{1}  )

For the compression and expansion process,

⇒ \frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }

⇒ \frac{2042.56}{998} = \frac{T_{4} }{551}

⇒ T_{4} = 1127.7 K

Heat rejected Q_{R} = 1 × 0.718 × ( 1127.7 - 551)

⇒ Q_{R} = 414.07 \frac{KJ}{kg}

Net heat interaction from the cycle Q_{net} = Q_{S} - Q_{R}

Put the values of Q_{S} & Q_{R}  we get,

⇒ Q_{net} = 750 - 414.07

⇒ Q_{net} = 335.93 \frac{KJ}{kg}

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ Q_{net} = W_{net}

⇒ W_{net} = 335.93 \frac{KJ}{kg}

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

E_{otto} = 1- \frac{T_{1} }{T_{2} }

Put the values of T_{1} & T_{2} in the above formula we get,

E_{otto} = 1- \frac{551 }{998 }

⇒ E_{otto} = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure P_{m} :-

We know that mean effective pressure of  the Otto cycle is  given by

P_{m} = \frac{W_{net} }{V_{s} } ---------- (1)

where V_{s} is the swept volume.

V_{s} = V_{1}  - V_{2} ---------- ( 2 )

From ideal gas equation P_{1} V_{1} = m × R × T_{1}

Put all the values in above formula we get,

⇒ 95 × V_{1} = 1 × 0.287 × 551

⇒ V_{1} = 0.6 m^{3}

From the same ideal gas equation

P_{2} V_{2} = m × R × T_{2}

⇒ 760 × V_{2} = 1 × 0.287 × 998

⇒ V_{2} = 0.377 m^{3}

Thus swept volume V_{s} = 0.6 - 0.377

⇒ V_{s} = 0.223 m^{3}

Thus from equation 1 the mean effective pressure

⇒ P_{m} = \frac{335.93}{0.223}

⇒ P_{m} = 1506.41 \frac{k pa}{kg}

This is the value of mean effective pressure of the cycle.

4 0
3 years ago
In a residence, the control of a room's baseboard electric resistance heating system would be accomplished automatically using a
Alex73 [517]

Answer:

I think D is the correct answer

Explanation:

hope it is

3 0
3 years ago
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