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Ipatiy [6.2K]
3 years ago
15

1. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott

le, and the bottle flies forward at 25 m/s. How fast is the ball traveling after hitting the bottle?
Physics
1 answer:
Kobotan [32]3 years ago
7 0
 0.4 x 18 = 7.2 kg m/s

The momentum of the bottle after being hit is 0.2 x 25 = 5 kg m/s

7.2 - 5 = 2.2 kg m/s is the motmentum of the ball now 

 the velocity  is 2.2/0.4 = 5.5 m/s
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An object weighs 15N in air and 13N when completely
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Answer:

The volume of water displaced = 0.204 m³

Explanation:

<em>From Archimedes' principle,</em>

<em>Upthrust = lost in weight = weight of water displaced.</em>

<em>U = W₁ - W₂..................... Equation 1</em>

<em>Where U = upthrust, W₁ = weight in air, W₂ = weight when completely submerged in water.</em>

<em>Given: W₁ = 15 N, W₂ = 13 N</em>

<em>Substituting these values into equation 1</em>

<em>U = 15 - 13 </em>

<em>U = 2 N</em>

<em>But,</em>

<em>Weight of water displaced = mass of water displaced. × acceleration due to gravity.</em>

<em>Mass of water displaced = weight of water displaced/ acceleration due to gravity.................... Equation 2</em>

Where: acceleration due to gravity = 9.8 m/s², weight of water displaced = 2 N

Substituting these values into equation 2

Mass of water displaced = 2/9.8

Mass of water displaced = 0.204 kg.

Also,

Volume of water displaced = mass of water displaced/ Density of water.............. Equation 3

Where Density of water = 1.0 kg/m³, mass of water displaced = 0.204 kg

Substituting these values into equation 3 above,

Volume of water displaced = 0.204/1

volume of water displaced = 0.204 m³

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Answer:

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Si el coeficiente de fricción cinética entre los neumáticos y el pavimento seco es de 0.80. ¿Cuál es la distancia mínima para de
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Answer: 52.9 metros.

Explanation:

Podemos escribir la fuerza de fricción cinética como

F = μ*N

donde N es la fuerza normal entre el coche y el suelo, cuya magnitud es igual al peso en esta situación.

F = μ*m*g

donde m es la masa del coche y g es 9.8m/s^2

y sabemos que μ = 0.8

Por la segunda ley de Newton, sabemos que:

F = m*a

fuerza es igual a masa por aceleración.

a = F/m

entonces la aceleración causada por la fuerza de rozamiento es:

F = 0.8*m*g

a = F/m = (0.8*m*g)/m = 0.8*g.

Entonces ya encontramos la aceleración, hay que recordar que esta aceleración es en sentido opuesto a la sentido de movimiento, entonces podemos escribir la aceleración como:

a(t) = -0.8*g

Para la velocidad, podemos integrar sobre el tiempo para obtener.

v(t) = -0.8*g*t + v0

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v(t) = -0.8*g*t + 28.8m/s

Ahora podemos encontrar el tiempo necesario para que la velocidad del coche sea cero, en ese momento, como deja de moverse, ya no tendremos rozamiento cinético, entonces no habrá aceleración y el coche se detendrá completamente.

v(t) = 0m/s = -0.8*9.8m/s^2*t + 28.8m/s

7.84m/s^2*t = 28.8m/s

                 t   = (28.8m/s)/(7.84m/s^2) = 3.63 segundos.

Ahora vamos a la ecuación de movimiento, donde asumimos que la posición inicial del coche es 0m, así que no tendremos constante de integración.

p(t) = -(1/2)*(0.8*9.8m/s^2)*t^2 + 28.8m/s*t

Ahora podemos evaluar la posición en t = 3.63 segundos, y esto nos dara la distancia que el coche se movio mientras frenaba.

p(3.63s) = -(1/2)*(0.8*9.8m/s^2)*(3.63s)^2 + 28.8m/s*(3.63s) = 52.9 metros.

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