Answer:
A safety manager is a person who designs and maintains the safety elements at workplace. A balance should be required for production and the job in providing work environment. As a safety officer in a medium sized manufacturing facility the following organizational system can be designed and maintained:
- Maintaining a workplace as per the guidelines by Occupational safety and health association. The rules and regulation should be such that maintains the manufacturing facilities.
- For warning to workers proper labelling, floor mapping, signs, posters should be used.
- Procurement and usage of safe tools.
- A guideline that describes safety standard and precautionary measures should be available to the workers. They should be aware about all the steps that needs to be taken in crisis.
- Ensuring that the workers have enough training safety and health or accident prevention.
- Identify and eliminate the hazardous elements from the workplace.
- A strict action should be taken against the worker in case of violation of rules and not adhering with guidelines.
Answer:
A wheelbarrow, a bottle opener, and an oar are examples of second class levers
Answer:
work done = 48.88 × 
J
Explanation:
given data
mass = 100 kN
velocity = 310 m/s
time = 30 min = 1800 s
drag force = 12 kN
descends = 2200 m
to find out
work done by the shuttle engine
solution
we know that work done here is
work done = accelerating work - drag work - descending work
put here all value
work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J
work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J
work done = 48.88 × J
Answer:
(a) 561.12 W/ m² (b) 196.39 MW
Explanation:
Solution
(a) Determine the energy and power of the wave per unit area
The energy per unit are of the wave is defined as:
E = 1 /16ρgH²
= 1/16 * 1025 kg/ m3* 9.81 m/s² * (2.5 m )²
=3927. 83 J/m²
Thus,
The power of the wave per unit area is,
P = E/ t
= 3927. 83 J/m² / 7 s = 561.12 W/ m²
(b) The average and work power output of a wave power plant
W = E * л * A
= 3927. 83 J/m² * 0.35 * 1 *10^6 m²
= 1374.74 MJ
Then,
The power produced by the wave for one km²
P = P * л * A
= 5612.12 W/m² * 0.35 * 1* 10^6 m²
=196.39 MW
Given:
diameter of sphere, d = 6 inches
radius of sphere, r = 
= 3 inches
density, 
= 493 lbm/ 
S.G = 1.0027
g = 9.8 m/ 
= 386.22 inch/ 
Solution:
Using the formula for terminal velocity,
= 
(1)
where,
V = volume of sphere
= drag coefficient
Now,
Surface area of sphere, A = 
Volume of sphere, V = 
Using the above formulae in eqn (1):
= 
= 
= 
Therefore, terminal velcity is given by:
= 
inch/sec
