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2 years ago

What is the importance of the causal link in work accidents?

1 answer:

3 0

Causal link is very important to know the correlation between a cause and an outcome of a workplace accident which is useful for social insurance cover.

<h3>Understanding work accidents?</h3>

In work accidents, there is the need for evaluation criteria and injury determination. Now, this is where causal link comes in because we'll need to know whether the damage is a permanent disability or not.

Now, causal link is important because it helps us to know the correlation between a cause and an outcome of a workplace accident and whether it was intentional or caused by other factors. This is very important because of social insurance cover.

Read more about Work Accidents at; brainly.com/question/17583177

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Alla [95]

C.nicotine
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3 years ago

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Using the tables for water determine the specified property data at the indicated state. For H2O at T = 140 °C and v = 0.2 m3/kg

Dennis_Churaev [7]

Answer:

$h = 1429.74\,\frac{kJ}{kg}$

Explanation:

The determination of any further properties requires the knowledge of two independent properties. (Temperature and specific volume in this case). The specific volumes for saturated liquid and vapor at 140 °C are, respectively:

$\nu_{f} = 0.001080\,\frac{m^{3}}{kg}$

$\nu_{g} = 0.50850\,\frac{m^{3}}{kg}$

Since $\nu_{f} < \nu < \nu_{g}$ , it is a liquid-vapor mixture. The quality of the mixture is:

$x = \frac{\nu-\nu_{f}}{\nu_{g}-\nu_{f}}$

$x = \frac{0.2\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }{0.50850\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }$

$x = 0.392$

The specific enthalpies for saturated liquid and vapor at 140 °C are, respectively:

$h_{f} = 589.16\,\frac{kJ}{kg}$

$h_{g} = 2733.5\,\frac{kJ}{kg}$

The specific enthalpy is:

$h = h_{f}+x\cdot (h_{g}-h_{f})$

$h = 589.16\,\frac{kJ}{kg}+0.392\cdot \left( 2733.5\,\frac{kJ}{kg} - 589.16\,\frac{kJ}{kg} \right)$

$h = 1429.74\,\frac{kJ}{kg}$

6 0

4 years ago

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A viscometer may be created by using a pair of closely fitting concentric cylinders and filling the gap between the cylinder wit

kari74 [83]

Answer:

The viscosity of the fluid is 6.05 Pa-s

Explanation:

The viscosity obtained from a rotating type viscometer is given by the equation

$\mu =\frac{3Tt}{2\pi \omega R_o(R_o^3-R_i^3)}$

where

T is the torque applied in the viscometer

't' is the thickness of gap

$\omega$ is the angular speed of rotation

$R_o,R_i$ are the outer and the inner raddi respectively

Since it is given that

$\omega =100rpm=\frac{100\times 2\pi }{60}=10.47rad/sec$

Applying the values in the above relation we get

$\mu =\frac{3\times 0.021\times 0.02\times 10^{-3}}{2\pi \times 10.47\times 37.5\times 10^{-3}(37.52^3-37.5^3)\times 10^{-9}}=6.04Pa-s$

5 0

3 years ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Assoli18 [71]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

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4 years ago

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3 years ago