Killed in accident or injured
By V=IR
A: 24=I*20
I = 1.2A
B: 220 = I*250
I = 0.88A
C: 6= I*3
I = 2 A
C,A,B
I'm going to assume that this gripping drama takes place on planet Earth, where the acceleration of gravity is 9.8 m/s². The solutions would be completely different if the same scenario were to play out in other places.
A ball is thrown upward with a speed of 40 m/s. Gravity decreases its upward speed (increases its downward speed) by 9.8 m/s every second.
So, the ball reaches its highest point after (40 m/s)/(9.8 m/s²) = <em>4.08 seconds</em>. At that point, it runs out of upward gas, and begins falling.
Just like so many other aspects of life, the downward fall is an exact "mirror image" of the upward trip. After another 4.08 seconds, the ball has returned to the height of the hand which flung it. In total, the ball is in the air for <em>8.16 seconds</em> up and down.
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:
La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h
Explanation:
La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.
Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?
distancia= 0.025 km
Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?
distancia= 25 metros
<u><em>La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h</em></u>
