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3 years ago

From the hazards blisted below, determine which could require foot protection

1 answer:

8 0

Foot protection could be required with Sharps, Slippery areas, Hazardous liquids,and Falling objects.

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A 10-m long steel linkage is to be designed so that it can transmit 2 kN of force without stretching more than 5 mm nor having a

Fed [463]

Answer:

minimum required diameter of the steel linkage is 3.57 mm

Explanation:

original length of linkage l = 10 m

force to be transmitted f = 2 kN = 2000 N

extension e = 5 mm= 0.005 m

maximum stress σ = 200 N/mm^2 = $2*10^{8} N/m^{2}$

maximum stress allowed on material σ = force/area

imputing values,

200 = 2000/area

area = 2000/( $2*10^{8}$ ) = $10^{-5}$ m^2

recall that area = $\pi d^{2} /4$

$10^{-5}$ = $\frac{3.142*d^{2} }{4}$ = $0.7855d^{2}$

$d^{2} = \frac{10^{-5} }{0.7855}$ = $1.273*10^{-5}$

$d = \sqrt{1.273*10^{-5} }$ = $3.57*10^{-3}$ m = 3.57 mm

maximum diameter of the steel linkage d = 3.57 mm

4 0

3 years ago

Nate needs to replace the cable to his lamp. He is stripping it to connect it to the termils. What should he remember to do with

pshichka [43]

Answer: i got you its d

Explanation:had the smae question as you

5 0

3 years ago

A solid steel shaft has to transmit 100 kW at 160 RPM. Taking allowable shear stress at 70 Mpa, find the suitable diameter of th

MA_775_DIABLO [31]

Answer:

The diameter of the shaft is 80.5 mm.

Explanation:

Torsion equation is applied for the diameter of the solid shaft.

Step1

Given:

Power of the shaft is 100 kw.

Revolution per minute is 160 RPM.

Allowable shear stress is 70 Mpa.

Maximum torque is 20% more than the mean torque.

Step2

Mean torque is calculated as follows:

$P=T\omega$

$P=T(\frac{2\pi N}{60})$

$100\times 1000=T(\frac{2\pi 160}{60})$

T=5968.31 N-m

Step3

Maximum torque is calculated as follows:

$T_{max}=(1+\frac{20}{100})T$

$T_{max}=1.2T$

$T_{max}=1.2\times 5968.31$

T_{max}=7161.97 N-m

Step4

Apply torsional equation for diameter of shaft as follows:

$\tau _{max}=\frac{T_{max}}{\frac{\pi d^{3}}{16}}$

$70\times 10^{6}=\frac{7161.97}{\frac{\pi d^{3}}{16}}$

$d^{3}=5.211\times 10^{-4}$

d=0.0805 m

or,

d=80.5 mm

Thus, the diameter of the shaft is 80.5 mm.

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4 years ago

What type of treatment is Dr. Wayne experimenting on?

shepuryov [24]

Answer:

Vaccine

Explanation:

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3 years ago

A 2-m-long and 3-m-wide horizontal rectangular plate is submerged in water. The distance of the top surface from the free surfac

UkoKoshka [18]

Answer:

864 KN

Explanation:

Atmospheric pressure is defined as the force per unit area exerted against a surface by the weight of the air above that surface.

Please kindly check attachment for the step by step solution of the given problem.

3 0

3 years ago

From the hazards blisted below, determine which could require foot protection​

From the hazards blisted below, determine which could require foot protection