energy associate with position or shape
Answer:
Force that acted on the body was F = 13 N
Explanation:
If once accelerated, the body covers 60 meters in 6 seconds, then its velocity is 60/6 m/s = 10 m/s
When the force was acting (for 10 seconds) the object accelerated from rest (initial velocity vi = 0) to 10 m/s (its final velocity). therefore we can use the kinematic equation for the velocity in an accelerated motion given by:

which in our case becomes;

and we can solve for the acceleration as:
a = 10/10 m/s^2 = 1 m/s^2
Therefore the force acting on the body, based on Newton's 2nd Law expression: F = m * a is:
F = 13 kg * 1 m/s^2 = 13 N
The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.
2. This is immediate from (2.2.7).
3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =
A wind turbine turns wind energy into electricity using the aerodynamic force
Answer: 39.8 μC
Explanation:
The magnitude of the electric field generated by a capacitor is given by:

d is the distance between the plates.
For a capacitor, charge Q = CV where C is the capacitance and V is the voltage.

where A is the area of the plate and ε₀ is the absolute permittivity.
substituting, we get

It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.
radius of the plates of the capacitor, r = 69 cm = 0.69 m
Area of the plates, A = πr² = 1.5 m²
Thus, the maximum charge that can be placed on disks without a spark is:
Q = E×ε₀×A
⇒ Q = 3 × 10⁶ N/C × 8.85 × 10⁻¹² F/m × 1.5 m² = 39.8 × 10⁻⁶ C = 39.8 μC.