You should use a hammer to remove a guard while a tool is in use.
Answer: 3002.86kg
Explanation:
Hydraulic cylinder diameter =125mm
Ambient pressure =1bar
Pressure =2500kpa
Piston Mass (MP) =?
F|(when it moves upward )=PA=F|(when it moves downward) =PoA+Mpg
Po=1 bar=100kpa
A=(π/4)D^2=(π/4)*0.125^2=0.01227m^2
Mp=(P-Po) A/g=(2500-100)*1000*0.01227/9.80665
Mp=3002.86kg.
Answer:
True
Explanation:
Engineering controls are those techniques used to reduce or eliminate hazards of any condition, thereby protecting the workers.
These are mostly products that act as barriers between the worker and the hazard. This may include machinery or equipment. The common engineering controls used are glovebox, biosafety cabinet, fume hood, vented balance safety enclosure, HVAC system, lockout-tagout, sticky mat and rupture disc.
Answer:
Power consume by compressor=113,726.87 KW
Explanation:
Given:
Actually compressor is an open system, so here we will use first law of thermodynamics for open system .
We know that first law of thermodynamics for steady flow
We know thatand we take the air as ideal gas.
System is in steady state then mass flow rate in =mass flow rate out
Mass flow rate=
So mass flow rate = ,
=1.23×200×2 Kg/s
=541.17 Kg/s
,
=80.07 m/s
Enthalpy of ideal gas h=
So
Now by putting the values
Here Q=0 because heat transfer is zero here.
W= -210.15 KJ/kg
So power consume by compressor=541.17×210.15
=113,726.87 KW
Scissors to avoid any rough edges