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2 years ago

Sí un auto viaja a 8m/s determine Tiempo en llegar a 200km de distancia Distancia que recorre en 40 minutos

Physics

1 answer:

GrogVix [38]2 years ago

7 0

Answer:

a. Tiempo = 25000 segundos

b. Distancia = 19200 metros

Explanation:

Dados los siguientes datos;

Velocidad = 8 m/s

Distancia = 200 km a metros = 200 * 1000 = 200,000

Para encontrar el tiempo para cubrir la distancia anterior;

Tiempo = distancia/velocidad

Tiempo = 200000/8

Tiempo = 25000 segundos

b. Para encontrar la distancia recorrida en 40 minutos;

Tiempo = 40 minutos a segundos = 40 * 60 = 2400 segundos

Distancia = velocidad * tiempo

Distancia = 8 * 2400

Distancia = 19200 metros

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Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

Gennadij [26K]

Answer:

a. The station is rotating at $1.496 \frac{rev}{min}$

b. the rotation needed is $2.8502 \frac{rev}{min}$

Explanation:

We know that the centripetal acceleration is

$a_{c}= \omega ^2 r$

where $\omega$ is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

The rotational speed is :

$2.7 \frac{m}{s^2} = \omega ^2 * 110 \ m$

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$\omega = 0.1567 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.1567 \frac{rad}{s} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 1.496 \frac{rev}{min}$

The rotational speed needed is :

$9.8 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.08909 \frac{rad^2}{s^2} }$

$\omega = 0.2985 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

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$\omega = 2.8502 \frac{rev}{min}$

3 0

3 years ago

Read 2 more answers

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

gayaneshka [121]

Force applied on the car due to engine is given as

$F_1 = 300 N$ towards right

Also there is a force on the car towards left due to air drag

$F_2 = 150 N$ towards left

now the net force on the car will be given as

$\vec F_{net} = \vec F_1 + \vec F_2$

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

$F_{net} = F_1 - F_2$