All categories

2 years ago

What are the four types of physical hazards?

2 answers:

7 0

Answer:Physical hazards include ergonomic hazards, radiation, heat and cold stress, vibration hazards, and noise hazards. Engineering controls are often used to mitigate physical hazards.

Explanation:

Vedmedyk [2.9K]2 years ago

6 0

Physical hazards include ergonomic hazards, radiation, heat and cold stress, vibration hazards, and noise hazards.

You might be interested in

What are the controlling LRFD load combinations for dead and floor live load?

yuradex [85]

Answer:

1) 1.4(D + F)

2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)

4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)

5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S

6) 0.9D + 1.6W + 1.6H

7) 0.9D + 1.0E + 1.6H

Explanation:

Load and Resistance Factor Design

there are 7 basic load combination of LRFD that is

1) 1.4(D + F)

2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)

4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)

5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S

6) 0.9D + 1.6W + 1.6H

7) 0.9D + 1.0E + 1.6H

and

here load factor for L given ( * ) mean it is permitted = 0.5 for occupancies when live load is less than or equal to 100 psf

here

D is dead load and L is live load

E is earth quake load and S is snow load

W is wind load and R is rain load

Lr is roof live load

3 0

3 years ago

A motor is mounted on a platform that is observed to vibrate excessively at an operating speed of 6000 rpm producing a 250-N for

vichka [17]

Answer:

The amplitude of the absorbed mass can be found

for ka:

$X_{a} =0.002m=\frac{F_{0} }{K_{a} } =\frac{250}{K_{a} } =125000N/m$

now

$w^2=\frac{K_{a} }{m_{a} } \\m_{a} =\frac{K_{a} }{w^2} =\frac{125000}{[6000*2\pi /60]^2} =0.317kg$

4 0

3 years ago

Express the following quantities to the nearest standard prefix using no more than three digits.(a) 20,000,000 Hz(b) 1025 W(c) 0

bija089 [108]

Answer:

(a) 20 MHz

(b) 1.025 KW

(d) 33 pF

Explanation:

(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = 20 MHz

(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = 1.025 KW

(d) 33 x10^(-12)F = 33 pico F = 33 pF

8 0

3 years ago

The minimum requirements for engineering documents are enumerated in

vladimir1956 [14]

Answer:

The answer will be Rule 61G15-23 F.A.C, relating to Seals.

Explanation:

According to the description given by: Florida administrative code&Florida administrative register the Minimum requirements for engineering documents are in the section 'Final 61G15-23' from 11/3/2015. This document provides specifications of materials required for the safe operation of the system that is the result of engineering calculations, knowledge and experience.

8 0

3 years ago

A 5-cm-diameter shaft rotates at 4500 rpm in a 15-cmlong, 8-cm-outer-diameter cast iron bearing (k = 70 W/m·K) with a uniform cl

-BARSIC- [3]

Answer:

(a) the rate of heat transfer to the coolant is Q = 139.71W

(b) the surface temperature of the shaft T = 40.97°C

Explanation:

See explanation in the attached files

5 0

3 years ago