Explanation:
LD₁ = 10⁵ mm⁻²
LD₂ = 10⁴mm⁻²
V = 1000 mm³
Distance = (LD)(V)
Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m
Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m
Conversion to miles:
Distance₁ = 10×10⁴ m / 1609m = 62 miles
Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.
Answer:
659.01W
Explanation:
The cab has a mass of 1250 kg, the weight of the cab represented by Wc will be
Wc = mass of the cab × acceleration due to gravity in m/s²
Wc = 1250 × 9.81 = 12262.5 N
but the counter weight of the elevator represented by We = mass × acceleration due to gravity = 995 × 9.81 = 9760.95 N
Net weight = weight of the cab - counter weight of the elevator = Wc - We = 12262.5 - 9760.95 = 2501.55 N
the motor of the elevator will have to provide this in form of work
work done by the elevator to lift the cab to height of 49 m = net weight × distance (height) = 2501.55 × 49m
power provided by the motor of the elevator = workdone by the motor / time in seconds
Power = (2501.55 × 49) ÷ ( 3.1 × 60 seconds) = 659.01 W
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) =
Well, the rings surrounding a planet are made out of rock. A ring surrounding the sun would be impossible since the sun can reach more than 27 million degrees Fahrenheit (15 million degrees Celsius.)
Hope this helped.
<span>speed = wavelength x frequency
speed = 0.4m X 10 Hz
speed = 4 m/s</span>
