Hmmm... I’d say that’s pretty accurate. Good job! Well wishes on whatever it is you’re doing.
Answer:
So the repulsive force between the pith ball will be 
Explanation:
We have given that the pith ball have the equal charge q = -40 nC =
Distance between the charges = 10 cm =0.1 m
According to coulombs law 
So the repulsive force between the pith ball will be
Answer:
a) Fi = 85.76 N
b) Fi = 87.8 N
Explanation:
Given:-
- Density of hydraulic oil, ρ = 804 kg/m^3
- The radius of input piston, ri = 0.00861 m
- The radius of output piston, ro = 0.141 m
Find:-
What input force F is needed to support the 23000-N combined weight of a car and the output plunger, when
(a) the bottom surfaces of the piston and plunger are at the same level
(b) the bottom surface of the output plunger is 1.10 m above that of the input plunger?
Solution:-
For part a.
- We see that both plungers are equal levels or there is no pressure due to elevation head. So we are only dealing with static pressure exerted by the hydraulic oil on both plungers to be equal. This part is an application of Pascal's Law:
Pi = Po
Fi / Ai = Fo / Ao
Fi = Ai / Ao * Fo
Fi = (ri/ro)^2 * Fo
Fi = ( 0.00861 / 0.141 )^2 * 23000
Fi = 85.76 N
For part b.
- We see that both plungers are at different levels so there is pressure due to elevation head. So we are only dealing with static pressure exerted by the hydraulic oil on both plungers plus the differential in heads. This part is an application of Bernoulli's Equation:
Pi = Po + ρ*g*h
Where, h = Elevation head = 1.10m
Fi = (ri/ro)^2 * Fo + ρ*g*h*π*ri^2
Fi = 85.76 + (804*9.81*1.1*3.142*0.00861^2)
Fi = 87.8 N
Answer:
Current, I = 2.45 T
Explanation:
It is given that,
Magnetic field, B = 0.92 T
Length of wire, l = 2.6 m
Mass, m = 0.6 kg
We need to find the minimum current needed to levitate the wire. It is given by balancing its weight to the magnetic force i.e.
I = 2.45 A
So, the minimum current to levitate the wire is 2.45 T. Hence, this is the required solution.
