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2 years ago

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Manual handling of materials accounts for the primary source of injury in the workplace

1 answer:

irinina [24]2 years ago

7 0

I think the statement is true. Manual handling of materials accounts for the primary source of injury in the workplace. This can be found in OSHA. <span>Manual materials handling is the </span>principal source of compensable injuries in the American work force<span>, and four out of five of </span>these injuries<span> will affect the lower back.</span>

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The diagram below shows a longitudinal wave at one instant in its motion. Each oval represents a particle acted on by the ener t

Anton [14]

Answer:

the first line

Explanation:

A rarefaction is a region in a longitudinal wave where the particles are furthest apart. ... The region where the medium is compressed is known as a compression and the region where the medium is spread out is known as a rarefaction.

use this picture as a reference:

8 0

2 years ago

Read 2 more answers

What is the molarity of a 9.13 L soda that contains 13.83 of sugar?

choli [55]

Answer:

1.52 M

Explanation:

Molarity of a solution is calculated as follows:

Molarity = number of moles (n) ÷ volume (V)

Based on the information given in this question,

Volume of soda (V) = 9.13 L

number of moles = 13.83 mol

Molarity = 13.83 ÷ 9.13

Molarity = 1.52 M

7 0

2 years ago

How many carbon atoms are in 27.0g of carbon (C)?

Rama09 [41]

The answer is 12g of Carbon

8 0

2 years ago

calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?

inessss [21]

<u>Answer:</u>

$\Delta E=E_{final}-E_{initial}$

$\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}$

$\Delta E=-1312[0.111-1]KJ mol^{-1}$

$\Delta E=1166 KJ mol^{-1}$

$\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}$

$=193623 \times 10^{-23} \frac {J}{photon}$

$\Delta E=1.93623 \times 10^{-18} \frac {J}{photon}$

$\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}$

h is planck's constant

c is the speed of light

λ is the wavelength of light

$\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18} J/photon)}$

Wavelength

$\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m} =103 nm (Answer)$

<em>Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is </em><u><em>103 nm.</em></u>

7 0

2 years ago

71 points Chemistry tritiation please help me

Rudiy27

Answer:

Answer in the picture

Explanation:

3 0

2 years ago