Ask question

Ask question

All categories

3 years ago

3

Ergonomics- an applied science devoted to incorporating comfort, efficiency, and safety into the design of items in the work pla

ce.
True/False

1 answer:

jasenka [17]3 years ago

7 0

Answer:

True

Explanation:

Ergonomics is the study of people's efficiency in their working environment.

When the science of Ergonomics is applied in the workplace, it is found that the efficiency of the overall system in the workplace is increased.

You might be interested in

A 3-phase stepping motor is to be used to drive a linear axis for a robot. The motor output shaft will be connected to a screw t

Nikolay [14]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

6 0

3 years ago

I can’t find the code

Ymorist [56]

Answer:

what code for

Explanation:

8 0

4 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:

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

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

8 0

4 years ago

The disk of radius 0.4 m is originally rotating at ωo=4 rad/sec. If it is subjected to a constant angular acceleration of α=5 ra

Lina20 [59]

Answer:32.4m/ $s^2$

Explanation:

Given data

$radius\left ( r\right )$ =0.4m

Intial angular velocity $\left ( \omega_0\right )$ =4rad/s

angular acceleration $\left ( \alpha\right )$ =5rad/ $s^2$

angular velocity after 1 sec

$\omega$ = $\omega_0$ + $\alpha\times\t$

$\omega$ =4+5 $\left ( 1\right )$

$\omega$ =9rad/s

Velocity of point on the outer surface of disc $\left ( v\right )$ = $\omega_0\timesr$

v= $9\times0.4$ m/s=3.6m/s

Normal component of acceleration $\left ( a_c\right )$ = $\frac{v^2}{r}$

$a_c$ = $\frac{3.6\times3.6}{0.4}$ =32.4m/ $s^2$

3 0

4 years ago

An AM radio transmitter radiates 550 kW at a frequency of 740 kHz. How many photons per second does the emitter emit?

kifflom [539]

Answer:

1121.7 × 10³⁰ photons per second

Explanation:

Data provided in the question:

Power transmitted by the AM radio,P = 550 kW = 550 × 10³ W

Frequency of AM radio, f = 740 kHz = 740 × 10³ Hz

Now,

P = $\frac{NE}{t}$

here,

N is the number of photons

t is the time

E = energy = hf

h = plank's constant = 6.626 × 10⁻³⁴ m² kg / s

Thus,

P = $\frac{NE}{t}$ = $\frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}$ [t = 1 s for per second]

or

550 × 10³ = $\frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}$

or

550 = N × 4903.24 × 10⁻³⁴

or

N = 0.11217 × 10³⁴ = 1121.7 × 10³⁰ photons per second

7 0

3 years ago