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In-s [12.5K]
2 years ago
9

¿Por que una una tapa pequeña de agua purificada se mantiene quieta por más tiempo sobre una rampa, en comparación al tiempo red

ucido que permanece un carrito del supermercado sobre la misma superficie?
Physics
1 answer:
shusha [124]2 years ago
5 0

Answer:

Esto sucede principalmente por el movimiento rotacional de las ruedas del carrito.

Como sabemos, existen dos tipos de coeficientes de fricción, el estacionario y el cinético.

En principio, la tapa se mantendrá quieta por que no tendrá la suficiente fuerza como para vencer al coeficiente de fricción estático, por lo que no podrá moverse hasta que esta reciba un pequeño impulso, como puede ser una pequeña corriente de aire. (asumo que la superficie curva de la tapita es la que esta en contacto con la rampa)

(Voy a simplificar el movimiento rotacional, pero creo que es suficiente para explicar la situación)

Ahora, en el carrito, las ruedas están conectadas a un eje.

La fuerza gravitatoria podemos pensarla que esta arriba de este eje, mas o menos en el centro del carrito, y esta fuerza intentara que el carrito baje de la rampa (lo mismo pasa para la tapita), pero a diferencia de la tapita, esta fuerza es aplicada a una distancia dada del eje (o del centro de las rueditas) lo que podemos pensar que funciona como una palanca, amplificando así la fuerza, por lo que esta vez el rozamiento estático podrá detener la superficie de la rueda que esta en contacto con el suelo, pero cuanto el centro de masa se mueva un poco, el punto de contacto entre la rueda y el suelo se moverá, de esta manera aparece lo que llamamos movimiento rotacional (si no hubiera fricción, las ruedas simplemente deslizarían sobre la rampa) que le permite al carrito ignorar en parte al coeficiente de fricción estático.

Otro factor importante, es que las tapitas suelen tener una superficie rugosa y un centro de masa desfazado, lo que no las hace buenas rotadoras.

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You are a visitor aboard the New International Space Station, which is in a circular orbit around the Earth with an orbital spee
Alchen [17]

Answer:

The minimum total speed is 11.2km/s

Explanation:

We are been asked to find the escape velocity.

Escape velocity is defined as the minimum initial velocity that will take a body(projectile)away above the surface of a planet(earth) when it's projected vertically upwards.

The formula to calculate the escape velocity is Ve = √2gR

For the earth g = 9.8m/s2 , R = 6.4*10^6

Substituting into the equation Ve = √2*9.8*6.4*10^6 = 11.2*10^3m/s

=11.2km/s

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By considering action-reacton forces,identify why water rises up a thin capillary tube
netineya [11]

The forces of attraction between water molecules and the glass walls and within the molecules of water themselves are what enable the water to rise in a thin tube immersed in water.

<h3>What is force?</h3>

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

Surface or interfacial forces lead to capillarity. The forces of attraction between the water molecules and the glass walls and among the water molecules themselves are what causes the water in a thin tube submerged in water to rise.

Hence, the water rises up a thin capillary tube can be explained by Newton's third law.

To learn more about the force refer to the link;

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When a cracker or bread dissolves in your mouth, is that a physical or chemical change?
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a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if
klio [65]

Answer:

Velocity of the two balls after collision: 60\; \rm m \cdot s^{-1}.

100\; \rm J of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

  • Mass of the first ball: 100\; \rm g = 0.1\; \rm kg.
  • Mass of the second ball: 400\; \rm g = 0.4 \; \rm kg.

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass m and velocity v is: p = m \cdot v.

Momentum of the two balls before collision:

  • First ball: p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}.
  • Second ball: p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}.
  • Sum: 10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1} given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be 30\; \rm kg \cdot m \cdot s^{-1}. The mass of the two balls, combined, is 0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg. Let the velocity of the two balls after the collision v\; \rm m \cdot s^{-1}. (There's only one velocity because the collision had sticked the two balls to each other.)

  • Momentum after the collision from p = m \cdot v: (0.5\, v)\; \rm kg \cdot m \cdot s^{-1.
  • Momentum after the collision from the conservation of momentum: 30\; \rm kg \cdot m \cdot s^{-1}.

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about v:

0.5\, v = 30.

v = 60.

In other words, the velocity of the two balls right after the collision should be 60\; \rm m \cdot s^{-1}.

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass m and velocity v is \displaystyle \frac{1}{2}\, m \cdot v^{2}.

Kinetic energy before the collision:

  • First ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Second ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Sum: 500\; \rm J + 500\; \rm J = 1000\; \rm J.

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

  • Mass of the two balls, combined: 0.5\; \rm kg.
  • Velocity of the two balls right after the collision: 60\; \rm m\cdot s^{-1}.

Sum of the kinetic energies of the two balls right after the collision:

\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J.

Therefore, 1000\; \rm J - 900\; \rm J = 100\; \rm J of kinetic energy would be lost during this collision.

7 0
3 years ago
A child of mass 25 kg is skating fast, +10 m/s, and tries to get revenge by colliding with the 60 kg adult who is sitting still.
tankabanditka [31]

Answer: -4.4 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:

p_{o}=p_{f} (1)

Where:

p_{o}=m_{1}V_{o}+m_{2}U_{o} (2)

p_{f}=m_{1}V_{f}+m_{2}U_{f} (3)

m_{1}=25 kg is the mass of the child

V_{o}=10 m/s is the initial velocity of the child

m_{2}=60 kg is the mass of the adult

U_{o}=0 m/s is the initial velocity of the adult (it is sitting still)

V_{f} is the final velocity of the child

U_{f}=6 m/s is the final velocity of the adult

Substituting (2) and (3) in (1):

m_{1}V_{o}+m_{2}U_{o}=m_{1}V_{f}+m_{2}U_{f} (4)

Isolating V_{f}:

V_{f}=\frac{m_{1}V_{o}-m_{2}U_{f}}{m_{1}} (5)

V_{f}=\frac{(25 kg)(10 m/s)-(60 kg)(6 m/s)}{25 kg} (6)

Finally:

V_{f}=-4.4 m/s This means the velocity of the child is in the opposite direction

4 0
3 years ago
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