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kap26 [50]
2 years ago
10

>En cual de las siguientes situaciones la fuerza neta sobre el cuerpo es cero?

Physics
1 answer:
ExtremeBDS [4]2 years ago
3 0
La respuesta es "Un avion que vuela al norte con rapidez constante y altitud constante".

Para que la fuerza neta sea 0, la aceleracion debe ser 0, para esto la velocidad debe ser constante.  

Para que la velocidad sea constante el objecto debe estar moviendo con rapidez (magnitud de la velocidad) constante y sin cambiar direccion; ya que la velocidad es un vector asi es que depende en magnitud y direccion.

En las demas opciones la magnitud de la velocidad (rapidez) cambia y/o la direccion.
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Small blocks, each with mass m, are clamped at the ends and at the center of a rod of length L and negligible mass. Compute the
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Answer:

Answered

Explanation:

a) Two balls are at a distance of L/2 from the axis of rotation and one block at the center. ( center of rod).

therefore,

I=2\times m\frac{L}{2}^2

I= \frac{1}{2}mL^2

b) two balls at a distance L/4 at the from the axis and 1 ball at a distance 3L/4 from the from the axis.

I= 2\times m\times(L/4)^2 + m(\frac{3L}{4})^2

= \frac{1}{16} mL^2(2+9)= \frac{11}{16}mL^2

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A series of waves with decreasing wavelength labeled radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamm
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Answer:

label A= radio waves, label C= infrared, Label D= visible Light, Label G= gamma rays.

Explanation:

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Sara and Saba are identical twins who are the same in every way, including their weights. One day, Sara and Saba decided to go f
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A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by
aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

x(t)=\alpha t^2 - \beta t^3

where

\alpha = 1.50 m/s^2

\beta = 0.05 m/s^3

The average velocity is given by the ratio between the displacement and the time taken:

v=\frac{\Delta x}{\Delta t}

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

The position at t = 2.00 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

So the displacement is

\Delta x = x(2)-x(0)=5.6-0=5.6 m

The time interval is

\Delta t = 2.0 s - 0 s = 2.0 s

And so, the average velocity in this interval is

v=\frac{5.6 m}{2.0 s}=2.8 m/s

b)

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

While the position at t = 4.00 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = x(4)-x(0)=20.8-0=20.8 m

The time interval is

\Delta t = 4.0 - 0 = 4.0 s

So the average velocity here is

v=\frac{20.8}{4.0}=5.2 m/s

c)

The position at t = 2 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

While the position at t = 4 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = 20.8 - 5.6 = 15.2 m

While the time interval is

\Delta t = 4.0 - 2.0 = 2.0 s

So the average velocity is

v=\frac{15.2}{2.0}=7.6 m/s

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

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