Se deja caer un objeto desde la cornisa de una casa la cual tarda en llegar al suelo 645/500 segundos. ¿Cúal es la altura de la
1 answer:
8.15m. La altura de la casa de la cual se deja caer un objeto que tarda en caer es de 8.15m.
La clave para resolver este problema es mediante caída libre, la velocidad inicial del objeto es 0 por lo que podemos calcular la altura h de la casa mediante la relación .
Conocemos la gravedad y el tiempo en que tarda en caer el objeto
, sustituyendo los valores tenemos que:
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Answer:
a) The centripetal acceleration of the car is 0.68 m/s²
b) The force that maintains circular motion is 940.03 N.
c) The minimum coefficient of static friction between the tires and the road is 0.069.
Explanation:
a) The centripetal acceleration of the car can be found using the following equation:
Where:
v: is the velocity of the car = 51.1 km/h
r: is the radius = 2.95x10² m
Hence, the centripetal acceleration of the car is 0.68 m/s².
b) The force that maintains circular motion is the centripetal force:
Where:
m: is the mass of the car
The mass is given by:
Where P is the weight of the car = 13561 N
Now, the centripetal force is:
Then, the force that maintains circular motion is 940.03 N.
c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:
Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.
I hope it helps you!