Answer:
0.023 Ohms
Explanation:
Given data
Length= 2.8m
radius= 1.03mm
current I= 1.35 A
voltage V= 0.032V
We know that from Ohm's law
V= IR
Now R= V/I
Substitute
R= 0.032/1.35
R= 0.023 Ohms
Hence the resistance is 0.023 Ohms
Answer:
1) λ < 2d, 2) nfrared imaging technique, 3) each color there is a different index of refraction
Explanation:
We are going to answer the three questions
1) When x-rays pass through matter in order to be dispersed, their wavelength must be of the order of the length of separation in the atoms and molecules of the body, in solid bones this length is similar and they scatter and reflect the x-rays therefore they can be observed, the fat and the soft tissue have a much greater separation therefore the x-rays cannot be reflected and consequently it is not observable by this technique.
2) At airports they use the infrared imaging technique, where the image is taken for the infrared wavelength, which is the heat part of the electromagnetic spectrum; consequently, when the image is viewed, the hottest areas appear brighter and, since when a person has a virus, his temperature rises, his temperature rises, it is possible to observe people with a higher temperature.
3) when white light hits a prism it is refracted with the equation
n₁ sin θ₁ = n₂ sin θ₂
where the incidence of refraction depends on the wavelength, therefore for each color there is a different index of refraction and consequently the light is separated in its different colors.
The longer you continue to listen, the more beats will be heard.
They'll occur at the rate of (260Hz - 254Hz) = 6 Hz .
Answer:
208
Explanation:
add it together for the answer
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) =
