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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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A 3-ft-diameter vertical cylindrical tank open to the atmosphere contains 1-ft-high water. The tank is now rotated about the cen

arlik [135]

Answer:

The angular velocity is 7.56 rad/s

the maximum water height is 2 ft

Explanation:

The z-position as a function of r is equal to

$z_{s(r)} =h_{0} -\frac{w^{2}(R^{2}-2r^{2} }{4g}$ (eq. 1)

where

h0 = initial height = 1 ft

w = angular velocity

R = radius of the cylinder = 1.5 ft

zs(r) = 0 when the free surface is lowest at the centre

Replacing and clearing w

$w=\sqrt{\frac{4gh_{0} }{R^{2} } } =\sqrt{\frac{4*32.17*1}{1.5^{2} } } =7.56rad/s$

if you consider the equation 1 for the free surface at the edge is equal to

$z_{s(R)} =h_{0} +\frac{w^{2}R^{2} }{4g} =1+\frac{(7.56^{2})*(1.5^{2} ) }{4*32.17} =1.99ft=2ft$

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3 years ago

den301095 [7]

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3 years ago

What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a

Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18 )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

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2 years ago

What type of foundation do engineers use for a small and light building and when the load of the building is borne by columns? A

ikadub [295]

Answer:

Explanation:

Individual footings are the commonest, and they are often used if the load of the building is borne by columns. Typically, every column will have an own footing. The footing is usually only a rectangular or square pad of concrete on which the column is erected

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Should aircraft wings have infinite stiffness?

Colt1911 [192]

Answer:

No, they need to be somewhat flexible so that forces such as turbulance don't shear the wing off.

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