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) =
F=ma
F= 4x1.2
F= 4.8 N
F= 4gsin30 - Friction
Friction= 19.6 - 4.8 N
Friction= 14.8 N
Friction= u x 4gcos30
14.8 / 4gcos30 = u
u= 0.43596...
u= 0.44
coefficient is 0.44
Answer:
Magnetic field, B = 0.199 T
Explanation:
It is given that,
Radius of circular loop, r = 11.7 cm = 0.117 m
Magnetic flux through the loop, 
The magnetic flux linked through the loop is :


Here, 

or


B = 0.199 T
So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.
Answer:
Please see below as the answer is self-explanatory.
Explanation:
- We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
- Both movements are independent each other, due to they are perpendicular.
- In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
- Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

- Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

- In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
- Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

- In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

- Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

- Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.
Answer:
5.5N/m
Explanation:
Calculation for What is the spring constant
First step is to calculate the time period
T = 12 second/10
T = 1.2 second
Now let calculate the spring constant using this formula
k=4π²m/T²
Where,
m=0.2kg
T=1.2second
k represent spring constant=?
Let plug in the formula
k=4π²×0.2kg/(1.2)²
k=39.48×0.2kg/1.44
k=7.90/1.44
k=5.48N/m
k=5.5N/m ( Approximately)
Therefore the spring constant will be 5.5N/m