Answer:
Option B: An MSDS
Explanation:
A dictionary is used to check up the meaning of general words and not for checking if a substance being used is hazardous. Option A is wrong.
MSDS means "Material Safety Data Sheet" and it contains documents with information that relates to occupational health & safety for checking various substances and products. Thus, option B is correct.
SAE stands for Society of Automotive Engineering and their standards pertain to mainly Automobiles. Thus option C is wrong.
EPA guidelines are mainly for checking facility and environmental health and safety compliance. Thus, option D is wrong.
Answer:
<em>hi</em><em> </em><em>sa</em><em>m</em><em>e</em><em> </em><em>here</em><em>.</em><em> </em><em>hru</em><em> </em><em>btw</em>
<em>안녕하세요 모두가 잘되기를 바랍니다. 안전 유지</em>
Answer:
it goes low to high off and on
Explanation:
because i wrote it
Answer:
Z = 29.938Ω ∠22.04°
I = 2.494A
Explanation:
Impedance Z is defined as the total opposition to the flow of current in an AC circuit. In an R-L-C AC circuit, Impedance is expressed as shown:
Z² = R²+(Xl-Xc)²
Z = √R²+(Xl-Xc)²
R is the resistance = 4Ω
Xl is the inductive reactance = ωL
Xc is the capacitive reactance =
1/ωc
Given C = 12 μF, L = 6 mH and ω = 2000 rad/sec
Xl = 2000×6×10^-3
Xl = 12Ω
Xc = 1/2000×12×10^-6
Xc = 1/24000×10^-6
Xc = 1/0.024
Xc = 41.67Ω
Z = √4²+(12-41.67)²
Z = √16+880.31
Z = √896.31
Z = 29.938Ω (to 3dp)
θ = tan^-1(Xl-Xc)/R
θ = tan^-1(12-41.67)/12
θ = tan^-1(-29.67)/12
θ = tan^-1 -2.47
θ = -67.96°
θ = 90-67.96
θ = 22.04° (to 2dp)
To determine the current, we will use the relationship
V = IZ
I =V/Z
Given V = 12V
I = 29.93/12
I = 2.494A (3dp)
Answer:
a) the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb
Explanation:
Given the data in the question;
Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in
Diameter d = 3/16 in
Unit weight w = 490 pcf
First we determine the area of the wire;
A = π/4 × d²
we substitute
A = π/4 × (3/16)²
A = 0.0276 in²
Next we get the Volume
V = Area × Length of wire
we substitute
V = 0.0276 × 4800
V = 132.48 in³
Weight of the steel wire will be;
W = Unit weight × Volume
we substitute
W = 490 × ( 132.48 / 12³ )
W = 490 × 0.076666
W = 37.57 lb
a) the maximum tensile stress due to the weight of the wire;
σ
= W / A
we substitute
σ = 37.57 / 0.0276
= 1361.23 psi
Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi
Maximum load P that the wire can safely support its lower end will be;
P = ( σ
- σ )A
we substitute
P = ( 24000 - 1361.23 )0.0276
P = 22638.77 × 0.0276
P = 624.83 lb
Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb
