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sladkih [1.3K]
3 years ago
13

Air-conditioning principles are being discussed. Technician A states that heat always

Engineering
1 answer:
Sunny_sXe [5.5K]3 years ago
7 0

Answer: C.

Explanation:

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Sea B = 5.00 m a 60.0°. Sea C que tiene la misma magnitud que A y un ángulo de dirección mayor que el de A en 25.0°. Sea A ⦁ B =
uranmaximum [27]

Answer:

\| \vec A \| = 6.163\,m

Explanation:

Sean A, B y C vectores coplanares tal que:

\vec A = (\| \vec A \|\cdot \cos \theta_{A},\| \vec A \|\cdot \sin \theta_{A}), \vec B = (\| \vec B \|\cdot \cos \theta_{B},\| \vec B \|\cdot \sin \theta_{B}) y \vec C = (\| \vec C \|\cdot \cos \theta_{C},\| \vec C \|\cdot \sin \theta_{C})

Donde \| \vec A \|, \| \vec B \| y \| \vec C \| son las normas o magnitudes respectivas de los vectores A, B y C, mientras que \theta_{A}, \theta_{B} y \theta_{C} son las direcciones respectivas de aquellos vectores, medidas en grados sexagesimales.

Por definición de producto escalar, se encuentra que:

\vec A \,\bullet\, \vec B = \|\vec A \| \| \vec B \| \cos \theta_{B}\cdot \cos \theta_{A} + \|\vec A \| \| \vec B \| \sin \theta_{B}\cdot \sin \theta_{A}

\vec B \,\bullet\, \vec C = \|\vec B \| \| \vec C \| \cos \theta_{B}\cdot \cos \theta_{C} + \|\vec B \| \| \vec C \| \sin \theta_{B}\cdot \sin \theta_{C}

Asimismo, se sabe que \| \vec B \| = 5\,m, \theta_{B} = 60^{\circ}, \vec A \,\bullet \,\vec B = 30\,m^{2}, \vec B\, \bullet\, \vec C = 35\,m^{2}, \|\vec A \| = \| \vec C \| y \theta_{C} = \theta_{A} + 25^{\circ}. Entonces, las ecuaciones quedan simplificadas como siguen:

30\,m^{2} = 5\|\vec A \| \cdot (\cos 60^{\circ}\cdot \cos \theta_{A} + \sin 60^{\circ}\cdot \sin \theta_{A})

35\,m^{2} = 5\|\vec A \| \cdot [\cos 60^{\circ}\cdot \cos (\theta_{A}+25^{\circ}) + \sin 60^{\circ}\cdot \sin (\theta_{A}+25^{\circ})]

Es decir,

30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})

35\,m^{2} = \| \vec A \| \cdot [2.5\cdot \cos (\theta_{A}+25^{\circ})+4.330\cdot \sin (\theta_{A}+25^{\circ}})]

Luego, se aplica las siguientes identidades trigonométricas para sumas de ángulos:

\cos (\theta_{A}+25^{\circ}) = \cos \theta_{A}\cdot \cos 25^{\circ} - \sin \theta_{A}\cdot \sin 25^{\circ}

\sin (\theta_{A}+25^{\circ}) = \sin \theta_{A}\cdot \cos 25^{\circ} + \cos \theta_{A} \cdot \sin 25^{\circ}

Es decir,

\cos (\theta_{A}+25^{\circ}) = 0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A}

\sin (\theta_{A}+25^{\circ}) = 0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A}

Las nuevas expresiones son las siguientes:

30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})

35\,m^{2} = \| \vec A \| \cdot [2.5\cdot (0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A})+4.330\cdot (0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A})]

Ahora se simplifican las expresiones, se elimina la norma de \vec A y se desarrolla y simplifica la ecuación resultante:

30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})

35\,m^{2} = \| \vec A \| \cdot (4.097\cdot \cos \theta_{A} +2.865\cdot \sin \theta_{A})

\frac{30\,m^{2}}{2.5\cdot \cos \theta_{A}+ 4.330\cdot \sin \theta_{A}} = \frac{35\,m^{2}}{4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}}

30\cdot (4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}) = 35\cdot (2.5\cdot \cos \theta_{A}+4.330\cdot \sin \theta_{A})

122.91\cdot \cos \theta_{A} + 85.95\cdot \sin \theta_{A} = 87.5\cdot \cos \theta_{A} + 151.55\cdot \sin \theta_{A}

35.41\cdot \cos \theta_{A} = 65.6\cdot \sin \theta_{A}

\tan \theta_{A} = \frac{35.41}{65.6}

\tan \theta_{A} = 0.540

Ahora se determina el ángulo de \vec A:

\theta_{A} = \tan^{-1} \left(0.540\right)

La función tangente es positiva en el primer y tercer cuadrantes y tiene un periodicidad de 180 grados, entonces existen al menos dos soluciones del ángulo citado:

\theta_{A, 1} \approx 28.369^{\circ} y \theta_{A, 2} \approx 208.369^{\circ}

Ahora, la magnitud de \vec A es:

\| \vec A \| = \frac{35\,m^{2}}{4.097\cdot \cos 28.369^{\circ} + 2.865\cdot \sin 28.369^{\circ}}

\| \vec A \| = 6.163\,m

8 0
4 years ago
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
3 years ago
A nail that is rectangular in<br> cross section and has a blunt<br> point is called a
Travka [436]

Answer:

A nail that is rectangular in cross section and has a blunt point is called a Cut nail.

6 0
3 years ago
On some late model cars, climate control systems wil create trouble codes for trouble shooting purposes.
Advocard [28]
I think it’s true because I mean that where you have air conditioner and stuff but as though it’s probably not codes but in a car there probably will be codes
8 0
3 years ago
How's your day been so far? &lt;3
Rzqust [24]
My day as been going awesome
7 0
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