Answer:
“An OSHA 300 log is where companies record the injuries that occur at the workplace,” said Luna. “By law, they have to report all the injuries to OSHA.” The OSHA law gives workers and their unions the right to have access to injury logs.
Answer: a) 0.948 b) 117.5µf
Explanation:
Given the load, a total of 2.4kw and 0.8pf
V= 120V, 60 Hz
P= 2.4 kw, cos θ= 80
P= S sin θ - (p/cos θ) sin θ
= P tan θ(cos^-1 (0.8)
=2.4 tan(36.87)= 1.8KVAR
S= 2.4 + j1. 8KVA
1 load absorbs 1.5 kW at 0.707 pf lagging
P= 1.5 kW, cos θ= 0.707 and θ=45 degree
Q= Ptan θ= tan 45°
Q=P=1.5kw
S1= 1.5 +1.5j KVA
S1 + S2= S
2.4+j1.8= 1.5+1.5j + S2
S2= 0.9 + 0.3j KVA
S2= 0.949= 18.43 °
Pf= cos(18.43°) = 0.948
b.) pf to 0.9, a capacitor is needed.
Pf = 0.9
Cos θ= 0.9
θ= 25.84 °
(WC) V^2= P (tan θ1 - tan θ2)
C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2
f=60, π=22/7
C= 117.5µf
Answer:
Elastic modulus of steel = 202.27 GPa
Explanation:
given data
long = 110 mm = 0.11 m
cross section 22 mm = 0.022 m
load = 89,000 N
elongation = 0.10 mm = 1 ×
m
solution
we know that Elastic modulus is express as
Elastic modulus =
................1
here stress is
Stress =
.................2
Area = (0.022)²
and
Strain =
.............3
so here put value in equation 1 we get
Elastic modulus =
Elastic modulus of steel = 202.27 ×
Pa
Elastic modulus of steel = 202.27 GPa