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3 years ago

6

Of the core elements of successful safety and health programs,management leadership,worker participation and what else directly

relate to individuals
roles

1 answer:

ch4aika [34]3 years ago

7 0

I think it would definitely be dinvi and the ghosts just because I really relate to that story

You might be interested in

timofeeve [1]

A. The factor by which a machine multiplies a force

6 0

3 years ago

The coefficient of static friction for both wedge surfaces is μw=0.4 and that between the 27-kg concrete block and the β=20° inc

balandron [24]

Assuming the wedge has an angle of 5°.The minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

<h3>Minimum value of force P</h3>

First step

Using this formula to find the weight of the block

W=mg

W=27×9.81

W=264.87 N

Second step

Angles of friction ∅A and ∅B

∅A=tan^-1(μA)

∅A=tan^-1(0.70)

∅A=34.99°

∅B=tan^-1(μB)

∅B=tan^-1(0.40)

∅B=21.80°

Third step

Equate the sum of forces in m-direction to 0 in order to find the reaction force at B.

∑fm=0

W sin (∅A+20°) + RB cos (∅B+∅A)=0

264.87 sin(34.99°+20°) + RB cos (21.80°+34.99°)=0

216.94+0.5477Rb=0

RB=216.94/0.5477

RB=396.09 N

Fourth step

Equate the sum of forces in x-direction to 0 in order to find force Rc.

∑fx=0

RB cos (∅B) - RC cos (∅B+ 5°)=0

396.09 cos(21.80°) - RC cos (21.80°+5°)=0

RC=396.09 cos(21.80°)/cos(26.80°)

RC=412.02 N

Last step

Equate the sum of forces in y-direction to 0 in order to find force P required to move the block up the incline.

∑fy=0

RB sin (∅B) + RC sin (∅B)-P=0

P=Rb sin (∅B) + RC sin (5°+∅B)

P=396.09 sin(21.80°) +412.02sin (5°+21.80°)

P=322.84 N

Inconclusion the minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

Learn more about Minimum value of force P here:brainly.com/question/20522149

7 0

2 years ago

A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s

Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

$\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})$

where

'H' is the initial depth of the layer

$C_c$ is the Compression index

$e_o$ is the inital void ratio

$\bar{\sigma_o}$ is the initial effective stress at the depth

$\Delta \bar{\sigma_o}$ is the change in the effective stress at the given depth

Applying the given values we get

$\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04$

3 0

3 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

3 years ago

Engineering Careers Scavenger Hunt

emmasim [6.3K]

Answer:

c

Explanation:

it's the only engineering career

6 0

3 years ago

Read 2 more answers