Hi pupil here's your answer ::
➡➡➡➡➡➡➡➡➡➡➡➡➡
Action and Reaction do not act on the same body !! If they acted on the same body, the resultant force will be zero and their could be never accelerated motion.
If both the forces acted on the same body, then if they are equal to opposite direction the object will remain stationary. If on of the forces is greater than other the object will move in the direction of greater force.
If both acted in the same direction there would be an accelrated motion.
⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅
Hope this helps . . . . .
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) =
Answer:
y = 4 Sin (2πt)
Explanation:
Amplitude, A = 4
frequency, f = 1
Wave function is given by
y = A sinωt
where, ω is angular frequency
ω = 2 π f = 2 π x 1 = 2π
So, the desired wave function
y = 4 Sin (2πt)
What effect man there are a million effects the only one that arent true are fusion or fission and magnetism unless that bowling ball is iron
Answer:
You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.
Explanation:
Given Values:
L = 50 cm = 0.5 m
H = 170 j/s
To find the diameter of the rod, we have to find the area of the rod using the following formula.
Here Tc = 100.0° C
k = 50.2
H = k × A × ![\frac{[T_{H -}T_{C} ] }{L}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BT_%7BH%20-%7DT_%7BC%7D%20%5D%20%7D%7BL%7D)
Solving for A
A = ![\frac{H * L }{k * [ T_{H}- T_{C} ] }](https://tex.z-dn.net/?f=%5Cfrac%7BH%20%2A%20L%20%7D%7Bk%20%2A%20%5B%20T_%7BH%7D-%20T_%7BC%7D%20%5D%20%7D)
A = ![\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}](https://tex.z-dn.net/?f=%5Cfrac%7B170%20%2A%200.5%7D%7B50.2%20%2A%20%5B%20350%20-%20100%20%5D%7D)
A = 
= 6.77 ×
m²
Now Area of cylinder is :
A = 
d²
solving for d:
d = 
d = 9.28 cm
