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3 years ago

Ughhh my cramps hurt sm

Engineering

2 answers:

kozerog [31]3 years ago

5 0

Answer:

Explanation:Come

tomate

vazorg [7]3 years ago

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Answer:

huh? wdym?

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You are given a rectangular piece of cloth with dimensions X by Y, where X and Y are positive integers, and a list of n products

Bond [772]

The proof that recursion is exponential and that dynamic programming is polynomial is given by the formula;

P(x,y) = max{

P(x,y)

max (1 <= h <= X) { P[h, Y] + P(X - h, Y) }

max (1 <= v <= Y) { P[X, v] + P[X, Y - v] }

}

To prove that the recursion is exponential and that dynamic programming is polynomial. we will do so as follows;

Let us first have the assumption that the cloth is in such a manner that either way, a product can be oriented. This implies that that after a cut, we will now have two pieces of cloth.

Now, we will make a list of the side lengths of the products that can fit in the piece after which we will consider a vertical cut for each of the side length as well as a horizontal cut for each of the side length, then we apply the same algorithm to each of the two resulting pieces.

Thus, after the point above, it is likely true that in some instances, there may be a place to cut that is not at a product side length. However, It might be better for us to make a list of lengths composed of one or more pieces side by side as long as the sum is less than the length of the side being considered.

Lastly, we would note that this recursive approach is not limited to just two -dimensional problems as It could also be applied to a single or more than two dimensions. A useful proof would be to prove it for one dimension, then assuming it is true for n dimensions, prove it is true for n + 1 dimensions.

Thus;

P(x,y) = max{

P(x,y)

max (1 <= h <= X) { P[h, Y] + P(X - h, Y) }

max (1 <= v <= Y) { P[X, v] + P[X, Y - v] }

}

Read more at; brainly.com/question/11665190

3 0

2 years ago

For some transformation having kinetics that obey the Avrami equation , the parameter n is known to have a value of 1.1. If, aft

ivanzaharov [21]

Answer:

total time = 304.21 s

Explanation:

given data

y = 50% = 0.5

n = 1.1

t = 114 s

y = 1 - exp(-kt^n)

solution

first we get here k value by given equation

y = 1 - $e^{(-kt^n)}$ ...........1

put here value and we get

0.5 = 1 - e^{(-k(114)^{1.1})}

solve it we get

k = 0.003786 = 37.86 × $10^{4}$

so here

y = 1 - $e^{(-kt^n)}$

1 - y = $e^{(-kt^n)}$

take ln both side

ln(1-y) = -k × $t^n$

t = $\sqrt[n]{-\frac{ln(1-y)}{k}}$ .............2

now we will put the value of y = 87% in equation with k and find out t

t = $\sqrt[1.1]{-\frac{ln(1-0.87)}{37.86*10^{-4}}}$

total time = 304.21 s

7 0

3 years ago

Why does voltage have so many names

BARSIC [14]

Answer:

Europe and most other countries in the world use a voltage which is twice that of the US. ... Originally Europe was 120 V too, just like Japan and the US today, but it was deemed necessary to increase voltage to get more power with fewer losses and less voltage drop from the same copper wire diameter

www.worldstandards.eu › electricity › why-no-standard-voltage

Explanation:

6 0

3 years ago

Read 2 more answers

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

A gasoline engine takes in air at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which th

Inessa [10]

Answer:

attached below

Explanation:

4 0

4 years ago