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:
15.8
0.0944
Explanation:
L = 1.5
B = 1.0
Speed of water = 15cm
Temperature = 20⁰C
At 20⁰C
Specific weight = 9790
Kinematic viscosity v = 1.00x10^-4m²/s
Dynamic viscosity u = 1.00x10^-3
Density p = 998kg/m²
Reynolds number
= 0.15x1.5/1.00x10^-4
= 225000
S = 5
5x1.5/225000^1/2
= 0.0158
= 15.8mm
Resistance on one side of plate
F = 0.664x1x1.0x10^-3x0.15x225000^1/2
= 0.04724N
Total resistance
= 2N
= 2x0.04724
= 0.0944N
The exciter provides fully coherent receiver local oscillator signals at radar frequency band as well as requisite, auxiliary high frequency clock signals. The exciter function is divided into an internal frequency synthesizer and an upconverter.
Hope this helps :)))
Answer:
Option C (practice by..............technical) would be the appropriate choice.
Explanation:
- A prerogative seems to be the discipline or nature of the design process to ensure.
- All certificates must, therefore, throughout compliance with current requirements including professional competence, express one's right and privilege of providing offers perhaps throughout the areas of qualified professionals.
Some other alternatives that are given aren't connected to a particular scenario. But that's the best solution above.
