Metering valves. These valves should be initially adjusted to provide adequate lubrication to each location
Answer:
0.064 mg/kg/day
6.25% from water, 93.75% from fish
Explanation:
Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.
The BCF = 10³, so the concentration of the chemical in the fish is:
10³ = x / (0.1 mg/kg)
x = 100 mg/kg
For 2 L of water and 30 g of fish:
2 kg × 0.1 mg/kg = 0.2 mg
0.030 kg × 100 mg/kg = 3 mg
The total daily intake is 3.2 mg. Divided by the woman's mass of 50 kg, the dosage is:
(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day
b) The percent from the water is:
0.2 mg / 3.2 mg = 6.25%
And the percent from the fish is:
3 mg / 3.2 mg = 93.75%
Here's mine!!!
If I were to write a book I would write my steps down as in figuring out what I what to talk about. Have an Introduction- Using an introduction is needed because you need to Describe your main idea, or what the essay is about, in one sentence. ... Develop a thesis statement, or what you want to say about the main idea. ... List three points or arguments that support your thesis in order of importance (one sentence for each). I would also use a Rising action- This is needed because, To construct the plot, nearly any story can be said to use increasing action. It serves the following purposes: It generates suspense and enhances the sense of urgency around the story's central conflict or problem. Then I would do a Climax- The object of a climax is not to provide the most possible conflict or action. It's also not just about producing the character's greatest reversal of fortunes. Too often we mistake the fortunes of our character with the story's arc, and although fortune is involved, it's not our story arc's key criterion. After that, I would just add finish touches and finish it off before it needs to be shipped off to make copies.
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
A is the answer for the sentence
