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

1 answer:

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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A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. immediat

koban [17]

<span>We'll use the momentum-impulse theorem. The x-component of the total momentum in that direction is given by p_(f) = p_(1) + p_(2) + p_(3) = 0.
So p_(1x) = m1v1 = 0.2 * 2 = 0.4 Also p_(2x) = m2v2 = 0 and p_(3x) = m3v3 = 0.1 *v3 where v3 is unknown speed and m3 is the mass of the third particle with the unknown speed
Similarly, the 235g particle, y-component of the total momentum in that direction is given by p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
So p_(1y) = 0, p_(2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_(3y) = m3v3 = 0.1 * v3 where m3 is third particle mass.
So p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0.4 + 0.1v3; v3 = 0.4/-0.1 = - 4
Also p_(fy) = 0.3525 + 0.1v3; v3 = - 0.3525/0.1 = -3.525
So v_3x = -4 and v_3y = 3.525.
The speed is their resultant = âš (-4)^2 + (-3.525)^2 = 5.335</span>

4 0

1 year ago

Read 2 more answers

A 4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb)1 = 6 ft>s. if the attached cord is pulled

Leya [2.2K]

Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s

By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:
Radius, r₂ = 2 ft

By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²

Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
= (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.

The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb

Answer: 44.25 ft-lb