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luda_lava [24]
3 years ago
11

As a consultant to the soft-drink industry, Dr. J is given the task of conducting the ultimate Pepsi taste test. This is Dr. J's

tenth taste test, which puts him seven up on his nearest consultant, who had only done three. Of course Dr. J is very qualified, having been hooked on soft drinks (especially orange soda) since he was Nehi to a pop bottle. Dr. J mounts a rather large container of Pepsi on a ledge some 3 meters above the ground. A bullet of mass 5 grams is then fired into the container, thus killing the taste. Not only that, but the Pepsi falls through the bullet hole onto the ground below (causing the taste to go flat). The wall of the container is 2 cm thick. The velocity of the bullet changes from an initial value of 500 m/sec just before striking the container wall to 5 m/sec upon leaving the container wall and entering the Pepsi. It finally fizzles out at a point 25 cm from the container wall.A. How much work does the container wall do on the bullet?
B. How much work does the Pepsi do on the bullet?
C. At what velocity does the Pepsi hit the floor?
Physics
1 answer:
dem82 [27]3 years ago
3 0

Answer:

A) How much work does the Pepsi do on the bullet = 0.0625J

B) At what velocity does the Pepsi hit the floor = 7.67m/s

Explanation:

  • Given mass of bullet = 5g
  • initial velocity = 500 m/sec
  • final velocity = 5 m/sec

From work done = Force X Distance

  • WD = Mad
  • a = v - u /t

Workdone ; m(v - u )/t

A) The work done by the Pepsi is equal to the change in kinetic energy while in the Pepsi:

  • but The work done by the container wall on the bullet is equal to the change in kinetic energy on either side of the wall: W = change in kinetic energy = 1/2 mv²

= 1/2 X 0.005 x 500² - 1/2 X 0.055 x 5²

= 624.94J

  • Therefore ; The work done by the Pepsi is equal to the change in kinetic energy while in the Pepsi = W = change in KE = 0.0625 - 0 = 0.0625J

B) At what velocity does the Pepsi hit the floor?

  • From conservation of energy principle; PE = KE
  • mgh = 1/2mv²
  • 0.98 X 3 = 1/2v²
  • velocity = 7.67m/s

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When you apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed ().When you apply a f
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Answer:

Explanation:

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It implies that a force ( frictional ) acts on it which is equal to 76 N in opposite direction ( friction )

When we apply  a greater force on it it starts moving with acceleration .

This time kinetic friction acts on it due to rough ground equal to 76 N .This is limiting friction ( maximum friction )

Net force on the body in later case

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un avión aterriza en la superficie de un portaaviones a 50 m/s y se detiene por completo en 120 metros, ¿cuál es la aceleración
sergejj [24]

Answer:

La aceleración necesaria para detener el avión es - 10.42 m/s².

Explanation:

Un movimiento uniformemente acelerado (M.U.A) es aquél cuya aceleración es constante y la velocidad de un objeto cambia a medida que el movimiento evoluciona.

Siendo la aceleración "a" el cambio de velocidad al tiempo transcurrido en un punto A a B, la velocidad inicial la velocidad que tiene un cuerpo al iniciar su movimiento en un período de tiempo y la velocidad final la velocidad que tiene un cuerpo al finalizar su movimiento en un período de tiempo, entonces en  M.U.A se cumple:

Vf² - Vo² = 2*a*d

donde:

  • Vf: Velocidad final
  • Vo: Velocidad inicial
  • a: Aceleración
  • d: Distancia recorrida

En este  caso:

  • Vf: 0 m/s, porque el avión se detiene
  • Vo: 50 m/s
  • a: ?
  • d: 120 m

Reemplazando:

(0 m/s)² - (50 m/s)² = 2*a*120 m

Resolviendo:

a=\frac{(0 m/s)^{2} -(50 m/s)^{2} }{2*120 m}

a= - 10.42 m/s²

<u><em>La aceleración necesaria para detener el avión es - 10.42 m/s².</em></u>

5 0
3 years ago
Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
3 years ago
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