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alina1380
3 months ago
5

A ball collides elastically with an immovable wall fixed to the earth’s surface. Which statement is false? 1. The ball's speed i

s unchanged by the collision.2. The ball does no net work on the wall.3. The ball's kinetic energy is unchanged by the collision.4. the wall doos no net work on the ball.5. the ball's size and shape are unchanged by the collision.6. All of these statements are true7. The ball's momentum is unchanged by the collision
Physics
1 answer:
Maru [3.3K]3 months ago
4 0

Answer:

Statements 4, 6 & 7 are incorrect.

Explanation:

In any elastic collision, the overall momentum vector sum of the system remains zero.

In this scenario, an elastic collision occurs between the ball and a stationary wall. The ball's velocity will consistently revert after the impact, leading to a change in direction of momentum.

The initial momentum of the ball is represented as:

p=m.v

where:

m = mass of the ball

v = initial velocity of the body

post-collision for the elastic interaction:

p=m.(-v)

  • Here, the momentum changes solely in direction, thus contradicting statement 7.
  • During the impact, both the ball and the wall exert forces on each other that are equal and opposite. The wall remains motionless, while the ball is influenced by the wall's reaction force, performing work on it, which contradicts statement 4.
  • Given that this collision is elastic, the ball's form and dimensions do not alter.
  • The previous points clearly indicate that not all provided statements hold true, thus violating statement 6.
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Answer:

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Explanation:

initially given information

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diameter = 100 mm = 0.1 m

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dynamic viscosity = 1.75 ×10^{-5} Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ \frac{du}{dy}

therefore

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substituting the values

= 1.75 ×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

the area between the air and puck is given by

Area = \frac{\pi }{4} d^{2}

area = \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

thus, the force on the puck can be represented as

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force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v

now applying Newton's second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

solving for t = \frac{0.1 v * 0.03}{1.37*10^{-3} v}

the time needed after impact for the puck is 2.18 seconds

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3 months ago
No official standard exists for extinguisher-cylinder colors. that said, a red cylinder is typically associated with which type
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ValentinkaMS [3465]

Answer:

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Explanation:

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