The length of the arm is the main part of natur
Answer:
correct option is c. a testable prediction leading to design of an experiment
Explanation:
The results of raising tadpoles were estimated to be water with an atrazine level of 0.1 ppb compared to those grown in pure water. So, this is not the question. If this assumption can now be tested, an experiment can be made in which a certain number of tadpoles can be raised in pure water and the same number of tadpoles can be raised in water with a 0.1ppb atrazine level can. The difference between the two populations can be estimated or compared.
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:
Time =t2=58.4 h
Explanation:
Since temperature is the same hence using condition
x^2/Dt=constant
where t is the time as temperature so D also remains constant
hence
x^2/t=constant
2.3^2/11=5.3^2/t2
time=t^2=58.4 h
Answer:
50421.6 m³
Explanation:
The river has an average rate of water flow of 59.6 m³/s.
Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:
Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s
The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken
time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds
The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³
