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2 years ago

10

Repetitive movements at work can lead to injuries. True or False

1 answer:

OverLord2011 [107]2 years ago

8 0

Answer

True

Explanation

RSI can occur when you do repetitive movements. Those movements can cause your muscles and tendons to become damaged over time. Some activities that can increase your risk for RSI are: stressing the same muscles through repetition.

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The depletion in the Ozone layer is caused by:

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Read 2 more answers

Convert 850 nm wavelength into frequency, eV, wavenumber, joules and ergs.

Sholpan [36]

Answer:

Frequency = $3.5294\times 10^{14}s^{-1}$

Wavenumber = $1.1765\times 10^6m^{-1}$

Energy = $2.3365\times 10^{-19}J$

Energy = 1.4579 eV

Energy = $2.3365\times 10^{-12}erg$

Explanation:

As we are given the wavelength = 850 nm

conversion used : $(1nm=10^{-9}m)$

So, wavelength is $850\times 10^{-9}m$

The relation between frequency and wavelength is shown below as:

$Frequency=\frac{c}{Wavelength}$

Where, c is the speed of light having value = $3\times 10^8m/s$

So, Frequency is:

$Frequency=\frac{3\times 10^8m/s}{850\times 10^{-9}m}$

$Frequency=3.5294\times 10^{14}s^{-1}$

Wavenumber is the reciprocal of wavelength.

So,

$Wavenumber=\frac{1}{Wavelength}=\frac{1}{850\times 10^{-9}m}$

$Wavenumber=1.1765\times 10^6m^{-1}$

Also,

$Energy=h\times frequency$

where, h is Plank's constant having value as $6.62\times 10^{-34}J.s$

So,

$Energy=(6.62\times 10^{-34}J.s)\times (3.5294\times 10^{14}s^{-1})$

$Energy=2.3365\times 10^{-19}J$

Also,

$1J=6.24\times 10^{18}eV$

So,

$Energy=(2.3365\times 10^{-19})\times (6.24\times 10^{18}eV)$

$Energy=1.4579eV$

Also,

$1J=10^7erg$

So,

$Energy=(2.3365\times 10^{-19})\times 10^7erg$

$Energy=2.3365\times 10^{-12}erg$

5 0

3 years ago

Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters a

Arada [10]

Answer:

a) $h_c = 0.1599 W/m^2-K$

b) $H_{loss} = 5.02 W$

c) $T_s = 302 K$

d) $\dot{Q} = 25.125 W$

Explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

Entry temperature, T₁ = 304 + 273 = 577 K

Exit temperature, T₂ = 284 + 273 = 557 K

Ambient temperature, $T_a = 25^0 C = 298 K$

Pipe length, L = 10 m

Area, A = 2πrL

A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

$\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600} = 1.25 kg/sec$

Rate of heat transfer,

$\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W$

a) To calculate the convection coefficient relationship for heat transfer by convection:

$\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K$

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe, $k_c = 0.8$

$\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K$

b) Heat loss from the pipe to the environment:

$H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W$

d) The required fan control power is 25.125 W as calculated earlier above

5 0

3 years ago