Answer:
V=20cm/s
Explanation:
The average speed is the distance total divided the time total:
First stage:
T1=5s
But, 
(decelerates to rest)
then: 
on the other hand:
X1=75cm
Second stage:
T2=5s
X2=125cm
Finally:
X=X1+X2=200cm
T=T1+T2=10s
V=X/T=20cm/s
Answer:
Yes, the capacitor's Q load varies inversely proportional to the distance between plates.
Explanation:
In the attached files you see the inverse relationship between capacity and distance between plates "d".
In the following formula we see its relationship with the "Q" load
Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?
Answer:
225 N
Explanation:
From Pascal's principle,
F/A = f/a ...................... Equation 1
Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.
Making f the subject of the equation,
f = F(a)/A ..................... Equation 2
Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².
Substituting into equation 2
f = 15000(3/200)
f = 225 N.
Hence the downward force that must be applied to small piston = 225 N
Shear stress = 1.0 N/m² (Pa)
For water, the dynamic viscosity = 10⁻³ Pa-s at 20°C.
The velocity gradient required = (Shear stress)/(Dynamic viscosity)
= (1.0 Pa)/( 10⁻³ Pa-s)
= 10³ 1/s
Answer: 10³ s⁻¹
