What is a diaganosis?

Un avión vuela a una velocidad de 900 km/h. Si tarda en viajar desde Canarias hasta la península 180 s ¿qué distancia recorre en

wel

Answer:

El avión recorrió 45 km en los 180 s.

Explanation:

La relación entre velocidad, distancia y tiempo se da de la siguiente manera;

$Velocidad= \dfrac{Distancia}{Hora}$

Por lo cual los parámetros dados son los siguientes;

Velocidad = 900 km/h = 250 m / s

Tiempo = 180 s

Estamos obligados a calcular la distancia recorrida

De la ecuación para la velocidad dada arriba, tenemos;

Distancia recorrida = Velocidad pf viaje × Tiempo de viaje

Distancia recorrida = 900 km/h × 180 s = 900

Distancia recorrida = 900 km/h × 1 h/60 min × 1 min/60 s × 180 s = 45 km

Por lo tanto, el avión viajó 45 km en 180 s.

8 0

4 years ago

okay so there is a wheel and it shows it when its high, in the middle, and below. my question is when is its potenial energy the

earnstyle [38]

At the highest point

cheers

4 0

4 years ago

A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w

Scilla [17]

Answer:

$v_f = 20 m/s$

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2$

now for pure rolling condition we will have

$v = R\omega$

also we have

$I = mR^2$

now we will have

$KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}$

$KE = mv^2$

now by work energy theorem we can say

$W = KE_f - KE_i$

$842 J = mv_f^2 - mv_i^2$

$842 = 3(v_f^2) - 3\times 11^2$

now solve for final speed

$v_f = 20 m/s$

3 0

3 years ago

What is meant by the term 'mass movement'?

vitfil [10]

Mass movement is the movement of surface materials caused by gravity. A great example would be a mud slide.

8 0

3 years ago

Read 2 more answers

If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu

ioda

Given :

Number of operations move through a pocket calculator during a full day's operation , $n=1.34 \times 10^{20}$ .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in one electron is :

$e^-=-1.6\times 10^{-19}\ coulombs$

So , charge on n electron is :

$C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C$

Therefore , -21.44 coulombs of charge is moved through it .

Hence , this is the required solution .

3 0

3 years ago