Answer:
The difference of head in the level of reservoir is 0.23 m.
Explanation:
For pipe 1

For pipe 2

Q=2.8 l/s
![Q=2.8\times 10^{-3]](https://tex.z-dn.net/?f=Q%3D2.8%5Ctimes%2010%5E%7B-3%5D)
We know that Q=AV




head loss (h)

Now putting the all values

So h=0.23 m
So the difference of head in the level of reservoir is 0.23 m.
Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂
Answer: a. Leave the lane closest to the emergency as soon as it is safe to do so, or slow down to a speed of 20 MPH below the posted speed limit.
Explanation:
Giving a way to the law enforcement vehicle and a medical emergency vehicle is necessary. If one approaches an emergency vehicle parked along the roadway one should change the lane as the vehicle may not move and the driver may also waste his or her time also one should also slow down his or her speed while approaching the vehicle as most of the emergency vehicle are in rush to reach the hospital so the driver should maintain some distance with the medical emergency vehicle.
Answer:
R = 31.9 x 10^(6) At/Wb
So option A is correct
Explanation:
Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A
Thus; R = L/μA,
Now from the question,
L = 4m
r_1 = 1.75cm = 0.0175m
r_2 = 2.2cm = 0.022m
So Area will be A_2 - A_1
Thus = π(r_2)² - π(r_1)²
A = π(0.0225)² - π(0.0175)²
A = π[0.0002]
A = 6.28 x 10^(-4) m²
We are given that;
L = 4m
μ_steel = 2 x 10^(-4) Wb/At - m
Thus, reluctance is calculated as;
R = 4/(2 x 10^(-4) x 6.28x 10^(-4))
R = 0.319 x 10^(8) At/Wb
R = 31.9 x 10^(6) At/Wb
If a clock frequency is applied to a cascaded counter, The lowest output frequency available will be
- The lowest output frequency will be =

<h3>
Cascade Counter</h3>
For a cascade counter,
Overall frequency = 
Overall frequency = 
<h3>Lowest F
requency</h3>
Therefore,
the lowest frequency

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