Answer
given,
flow rate = p = 660 kg/m³
outer radius = 2.8 cm
P₁ - P₂ = 1.20 k Pa
inlet radius = 1.40 cm
using continuity equation
A₁ v₁ = A₂ v₂
π r₁² v₁ = π r₁² v₂
Applying Bernoulli's equation
v₂ = 1.97 m/s
b) fluid flow rate
Q = A₂ V₂
Q = π (0.014)² x 1.97
Q = 1.21 x 10⁻³ m³/s
Answer:
Explanation:
Force between charge is given by the following expression
F = k Q₁ Q₂ / R² , k = 9 x 10⁹ , Q₁ and Q₂ are charges , R is distance between charges .
Putting the given values ,
.85 = 9 x 10⁹ x 7.1 x 10⁻⁶ x Q₂ / 5.4²
Q₂ = .85 x 5.4² / (9 x 10⁹ x 7.1 x 10⁻⁶ )
= .38788 x 10⁻³ C .
= 387.88 x 10⁻⁶ C .
Magic, Nah im just kidding. A battery has two parts, the anode and the cathode. Which anode is positive and cathode is negative, which they are connected to the electrolyte. Once they are connected to a device they once start working from separate ends. Which is the flow of energy
Answer:
1.56 J
Explanation:
The potential energy only depends on the vertical height from the ground level.
We consider the ground level to have zero P.E.
So when it is 2 m above the ground level,
P.E. = mgh
= 0.078×10×2
= 1.56 J
The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as
- t=0.476v
- t=1.967v
- V2=4.323v
<h3>What is the potential across the capacitor?</h3>
Question Parameters:
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
at
- t = 1.0 seconds
- 5.0 seconds
- 20.0 seconds.
Generally, the equation for the Voltage is mathematically given as
v(t)=Vmax=(i-e^{-t/t})
Therefore
For t=1
V=5(i-e^{-1/10})
t=0.476v
For t=5s
V2=5(i-e^{-5/10})
t=1.967
For t=20s
V2=5(i-e^{-20/10})
V2=4.323v
Therefore, the values of voltages at the various times are
- t=0.476v
- t=1.967v
- V2=4.323v
Read more about Voltage
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Complete Question
A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.
Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.
