Answer:
The answer to this question is 1273885.3 ∅
Explanation:
<em>The first step is to determine the required hydraulic flow rate liquid if working pressure and if a cylinder with a piston diameter of 100 mm is available.</em>
<em>Given that,</em>
<em>The distance = 50mm</em>
<em>The time t =10 seconds</em>
<em>The force F = 10kN</em>
<em>The piston diameter is = 100mm</em>
<em>The pressure = F/A</em>
<em> 10 * 10^3/Δ/Δ </em>
<em> P = 1273885.3503 pa</em>
<em>Then</em>
<em>Power = work/time = Force * distance /time</em>
<em> = 10 * 1000 * 0.050/10</em>
<em>which is =50 watt</em>
<em>Power =∅ΔP</em>
<em>50 = 1273885.3 ∅</em>
Answer:
Z = 3 + 0.23t
The water level is rising
Explanation:
Please see attachment for the equation
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:
×

Explanation:
Please kindly find the attached document for the answer.