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3 years ago

Which of the following statements best describes ergonomics?

Physics

2 answers:

Romashka-Z-Leto [24]3 years ago

6 0

Ergonomics involves the study of the human body.

Ergonomics is the science of adjusting environments, tasks, or procedures to fit you, the healthcare worker.

Ergonomics is the study of human psychology.

Ergonomics is the science of adjusting you, the healthcare worker, to fit the environment, tasks, or procedures.

emmainna [20.7K]3 years ago

3 0

Can you provide the multiple choices ?

You might be interested in

I really need help with this. please! thank you.

zhannawk [14.2K]

Solo estoy aquí por la libra, lo siento, no pude ayudar

7 0

3 years ago

A quantity that is fully described by magnitude alone is a ___________ quantity. A quantity that is fully described by both magn

deff fn [24]

Answer:

Scalar quantity

Vector quantity

Explanation:

A scalar quantity is a quantity that is fully described by magnitude alone. Examples include; mass, temperature etc

A vector quantity is described by both magnitude and direction. E.g force, weight etc

7 0

3 years ago

anyone know where I can find stuff (answer key, tables, etc.) for my Newton's Law of Motion lab report on edge2020? need answers

Zolol [24]

Usually the full tables are found on quizlet or quizzes

7 0

3 years ago

After an earthquake, which type of seismic wave arrives last at a seismometer?

amm1812

I believe L waves arrive last at a seismometer.
hope this helps!

8 0

3 years ago

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y_n= \frac{n \lambda D}{a}$ (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
$y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m$ (2)

If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

7 0

4 years ago