Answer:
The pressure is 2.167 psi.
Explanation:
Given that,
Diameter = 1.5 feet
Height = 10 feet
We need to calculate the psi at 5 feet
Using formula of pressure at a depth in a fluid
Suppose the fluid is water.
Then, the pressure is

Where, P = pressure
= density
h = height
Put the value into the formula


Pressure in psi is


Hence, The pressure is 2.167 psi.
<span>4.5 m/s
This is an exercise in centripetal force. The formula is
F = mv^2/r
where
m = mass
v = velocity
r = radius
Now to add a little extra twist to the fun, we're swinging in a vertical plane so gravity comes into effect. At the bottom of the swing, the force experienced is the F above plus the acceleration due to gravity, and at the top of the swing, the force experienced is the F above minus the acceleration due to gravity. I will assume you're capable of changing the velocity of the ball quickly so you don't break the string at the bottom of the loop.
Let's determine the force we get from gravity.
0.34 kg * 9.8 m/s^2 = 3.332 kg m/s^2 = 3.332 N
Since we're getting some help from gravity, the force that will break the string is 9.9 N + 3.332 N = 13.232 N
Plug known values into formula.
F = mv^2/r
13.232 kg m/s^2 = 0.34 kg V^2 / 0.52 m
6.88064 kg m^2/s^2 = 0.34 kg V^2
20.23717647 m^2/s^2 = V^2
4.498574938 m/s = V
Rounding to 2 significant figures gives 4.5 m/s
The actual obtainable velocity is likely to be much lower. You may handle 13.232 N at the top of the swing where gravity is helping to keep you from breaking the string, but at the bottom of the swing, you can only handle 6.568 N where gravity is working against you, making the string easier to break.</span>
Answer:
The capacitance of your capacitor is 5.476 x 10⁻⁵ μF
Explanation:
Given;
diameter of the aluminum pie plates = 16 cm = 0.16 m
separation distance, d = 3.25 mm = 0.00325 m
voltage across the parallel plates = 6 V

where;
C is the capacitance of your capacitor
ε is the permittivity of free space = 8.85 x 10⁻¹² (F/m)
d is separation distance
A is the area of the plate = ¹/₄ (πd²) = 0.25 (π x 0.16²) = 0.02011 m²

Therefore, the capacitance of your capacitor is 5.476 x 10⁻⁵ μF
Answer:
The position of stable equilibrium is -a
And the period of small oscillations must be: c/(ma^3)
Explanation:
Since the potential is:

We first look for a position of stable equilibrium. This posiiton must satisfy two considtions, that the first derivative of the potential must vanish at this point and the second derivative must be positive.

Which vanish for
x = a ; x =-a
The second derivative of V(x) is:

And:

Therefore:
a)
The position of stable equilibrium is -a
And the period of small oscillations must be:

(c/(ma^3))^1/2
b)
Let's find the maximum amplitude if the particle starts at this point with velocity v
If this is the case, the total energy will be:
(mv^2)/2
And the maximum amplitude will be
x = a^3/c mv^2 = (m v^2 a^3)/ c