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Nataly_w [17]
3 years ago
13

1. Si tengo medio kilo de fruta y te doy un cuarto y tú me das tres cuartos de kilo, ¿cuánto tengo? 2. Si en una carrera te qued

a por recorrer la mitad de la mitad de 1 km, ¿cuánto te falta? 3. ¿Qué pesa mas, un kilo y medio de hierro o tres medios kilos de paja? porfavor es urgente.
Physics
1 answer:
Kipish [7]3 years ago
6 0

Answer:

1. Tienes 1 kg de fruta.

2. Queda por recorrer 1/4 km.

3. Ambos pesan lo mismo.

Explanation:

1. Tienes 1/2 kg y cuando te doy 1/4 te queda:

m = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}

Ahora cuando te doy 3/4 kg te queda en total:

m_{T} = \frac{1}{4} + \frac{3}{4} = 1 kg

Por lo tanto, tienes 1 kg de fruta al final.

2. Si falta por recorrer la mitad de la mitad, tenemos:

d = \frac{1/2}{2} = \frac{1}{4}

Entonces, queda por recorrer 1/4 km.

3. El peso (P) del hierro es:

P = m*g    

P = (1 + 1/2)kg*9.81 m/s^{2} = 14.72 N

Y el peso de la paja es:

P = 3/2 kg*9.81 m/s^{2} = 14.72 N

Por lo tanto, ambos pesan lo mismo.

Espero que te sea de utilidad!

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There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = \frac{v}{2}

Part C: v_1 = \frac{v}{3}; v_2 = \frac{4v}{3}

Part D: v_1 = v_2 = \frac{v}{4}

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = \frac{1}{2}m.v^{2}

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

m_{1}.v_{1i} + m_{2}.v_{2i} = m_{1}.v_{1f} + m_{2}.v_{2f}

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f}  )

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i}  + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}

v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}

For all the collisions, object 2 is static, i.e. v_{2i} = 0

<u>Part A</u>: Both objects have the same mass (m), v_{1i} = v and collision is elastic:

v_1 = \frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}

v_1 = 0

v_2 = \frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}

v_2 = \frac{2.m}{m+m}.v

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

<u>Part B</u>: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

m_{1}.v_{1i} + m_{2}.v_{2i} = m_{1}.v_{1f} + m_{2}.v_{2f}

m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}

v_{f} =  \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }

v_1 = v_2 = \frac{m.v}{m+m}

v_1 = v_2 = \frac{v}{2}

<u>Part C:</u> Object 1 is 2m, object 2 is m and elastic collision:

v_1 = \frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}

v_1 = \frac{2m - m}{2m + m } . v

v_1 = \frac{v}{3}

v_2 = \frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}

v_2 = \frac{2.2m}{2m+m}.v

v_2 = \frac{4v}{3}

<u>Part D</u>: Object 1 is m, object is 3m and collision is inelastic:

v_1 = v_2 = v_{f} =  \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }

v_1 = v_2 = \frac{m}{m+3m}.v

v_1 = v_2 = \frac{v}{4}

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Answer:

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(b) The stationary time of the wildebeest is 675 s

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speed of the wildebeest, v_w = 16 m/s

speed of the chicken, v_c = 2.5 m/s

Time for wildebeest to finish the race without stopping, 2000 / 16 = 125 s

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Thus, the distance of the chicken from the finish line = 2000 m -  1937.5 m

the distance of the chicken from the finish line = 62.5 m

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