C. Tripling the length and reducing the radius by a factor of 2 is the change to a pipe would increase the conductance by a factor of 12.
<u>Explanation:</u>
As we know that the resistance is directly proportional to the length of the pipe and it is inversely proportional to the cross sectional area of the pipe.
So it is represented as,
R∝ l /A [ area is radius square]
So k is the proportionality constant used.
R = kl/A
Conductance is the inverse of resistance, so it is given as,
C= 1/R.
R₁ = kl₁ / A₁
R₂ = kl₂/A₂
R₂/R₁ = 1/12 [∵ conductance is the inverse of resistance]
= l₂A₁ / l₁A₂
If we chose l₁/l₂= 3 and A₂/A₁= 4 So R₂/R₁= 1/3×4 = 1/12
So tripling the length and reducing the radius by a factor of 2 would increase the conductance by a factor of 12.
Answer:
b i think because i think so
Answer:
3.84 Ω
Explanation:
From the question given above, the following data were obtained:
Electrical power (P) = 150 W
Voltage (V) = 24 V
Resistance (R) =?
P = IV
Recall:
V = IR
Divide both side by R
I = V/R
P = V/R × V
P = V² / R
Where:
P => Electrical power
V => Voltage
I => Current
R => Resistance
With the above formula (i.e P = V²/R), we can calculate resistance as illustrated below:
Electrical power (P) = 150 W
Voltage (V) = 24 V
Resistance (R) =?
P = V²/R
150 = 24² / R
150 = 576 / R
Cross multiply
150 × R = 576
Divide both side by 150
R = 576 / 150
R = 3.84 Ω
Thus, the resistance is 3.84 Ω
Answer:
The second column on the periodic table of the chemical elements is collectively called the alkaline earth metal group: beryllium, magnesium, calcium, strontium, barium, and radium. Because the outer electron structure in all of these elements is similar, they all have somewhat similar chemical and physical properties.
Explanation:
