Part 1
If water does not spill at the top point of the circular motion then for the minimum speed condition we can say normal force will be zero at the top position
given that
R = 1 m
g = 9.8 m/s^2
now from above equation we have
Part b)
for minimum value of angular speed we will have
Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0
Explanation:
- a) First of all find the distance between the two charges;
- x = 0, y = 0.30 and x = 0.40 m, y = 0
hence, the force F = 2Kq1q2cosθ /r²...............equation 1
but cosθ = y/r = 0.3/0.5
cosθ = 0.6
plugging back to equation 1;
F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2
F = 540 x 10^-3
Magnitude of Force = 0.54N
b) Direction is at angle 90
Answer: 4.8 s
Explanation:
We have the following data:
the mass of the raft
the force applied by Sawyer
the raft's final speed
the raft's initial speed (assuming it starts from rest)
We have to find the time 
Well, according to Newton's second law of motion we have:
(1)
Where 
is the acceleration, which can be expressed as:
(2)
Substituting (2) in (1):
(3)
Where 
Isolating from (3):
(4)
Finally:
Gravitational potential energy can be described as m*g*h (mass times gravity times height).
Originally,
15kg * 9.8m/s^2 *0.3 m = 44.1 kg*m^2/s^2 = 44.1 Joules.
After it is moved to a 1m shelf:
15kg * 9.8m/s * 1 = 147 kg*m^2/s^2= 147 Joules.
To find how much energy was added, we subtract final energy from initial energy:
147 J - 44.1 J = 102.9 Joules.
Answer:
The average magnetic flux through each turn of the inner solenoid is 
Explanation:
Given that,
Number of turns = 22 turns
Number of turns another coil = 330 turns
Length of solenoid = 21.0 cm
Diameter = 2.30 cm
Current in inner solenoid = 0.140 A
Rate = 1800 A/s
Suppose For this time, calculate the average magnetic flux through each turn of the inner solenoid
We need to calculate the magnetic flux
Using formula of magnetic flux
Put the value into the formula
Hence, The average magnetic flux through each turn of the inner solenoid is
