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stranger123
11.01.2020 •
Physics
Aball is thrown straight up with a speed of 3\frac{m}{s}, and it hits the ground with a speed of 15\frac{m}{s}. how tall is the tower? ignore air resistance, and answer in meters.
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Ответ:
Explanation:
We have equation of motion v² = u² + 2as
Initial velocity, u = 3 m/s upwards
Acceleration, a = -9.81 m/s² downwards
Final velocity, v = -15 m/s downwards
Substituting
v² = u² + 2as
(-15)² = 3² + 2 x -9.81 x s
s = -11 m that is downward
Form the initial position it travels 11 meter downward.
Height of tower is 11 meters
Ответ:
(a) La rapidez a la ida es 80
y la rapidez a la vuelta es 57.1428 ![\frac{km}{h}](/tpl/images/0722/5216/fa12f.png)
(b) La rapidez en todo el recorrido es 137.1428![\frac{km}{h}](/tpl/images/0722/5216/fa12f.png)
Explanation:
La rapidez hace referencia a la relación entre la distancia recorrida por un cuerpo y el tiempo empleado en cubrirla, es decir es distancia que recorre un objeto en un tiempo determinado. Entonces su cálculo se realiza mediante cociente entre la distancia y el tiempo:
Entonces, para ir de A y B conoces los siguientes datos:
distancia: 400 kmtiempo: 5 horasReemplazando:
y resolviendo se obtiene:
Para ir de B a A son conocidos los siguientes datos:
distancia: 400 kmtiempo: 7 horasReemplazando:
y resolviendo se obtiene:
(a) La rapidez a la ida es 80
y la rapidez a la vuelta es 57.1428 ![\frac{km}{h}](/tpl/images/0722/5216/fa12f.png)
La rapidez en todo el recorrido es calculada como la suma de la rapidez a la ida y a la vuelta:
rapidez total= 80
+ 57.1428
= 137.1428 ![\frac{km}{h}](/tpl/images/0722/5216/fa12f.png)
(b) La rapidez en todo el recorrido es 137.1428![\frac{km}{h}](/tpl/images/0722/5216/fa12f.png)