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3 years ago

6

Of the core elements of successful safety and health programs,management leadership,worker participation and what else directly

relate to individuals
roles

1 answer:

ch4aika [34]3 years ago

7 0

I think it would definitely be dinvi and the ghosts just because I really relate to that story

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Manufacturers frequently make choices about their suppliers of raw materials based on their impact on society and the environmen

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3 years ago

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:

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

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

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4 years ago

What is a thermal reservoir?

Mekhanik [1.2K]

Answer:

Explanation:

Thermal reservoir:

It is the body which have infinite amount of heat capacity and store large amount of heat.The temperature of thermal reservoir is constant and does not change with time.The temperature of thermal reservoir is not change even heat is going out from reservoir or heat going inside the reservoir.The temperature of thermal reservoir remains constant and does not depend on the surroundings.

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4 years ago

Explain biometric senser.

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Biometric sensors are used to collect measurable biological characteristics from a human being, which can then be used in conjunction with biometric recognition algorithms to perform automated person identification.

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3 years ago

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3 years ago