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scZoUnD [109]
2 years ago
5

A fisherman and his young son are in a boat on a small pond. Both are wearing life jackets. The son is holding a large helium fi

lled balloon by a string, holding a significant part of his weight. Consider each action below independently and indicate whether the level of the water in the pond, Rises, Falls, is Unchanged or Cannot tell.
The son pops the helium balloon.
The fisherman knocks the tackle box overboard and it sinks to the bottom.
The son finds a cup and bails some water out of the bottom of the boat
The fisherman lowers himself in the water and floats on his back.
The fisherman lowers the anchor and it hangs one foot above the bottom of the pond. (the anchor is initially inside the boat)
The son gets in the water, looses his grip on the string, letting the balloon escape upwards.
Physics
1 answer:
Flura [38]2 years ago
5 0

(a) The son pops the helium balloon (unchanged)

(b) The fisherman knocks the tackle box overboard and it sinks to the bottom (rises)

(c) The son finds a cup and bails some water out of the bottom of the boat (falls)

(d) The fisherman lowers himself in the water and floats on his back (unchanged)

(e) The fisherman lowers the anchor and it hangs one foot above the bottom of the pond (rises)

(f) The son gets in the water, looses his grip on the string, letting the balloon escape upwards (rises).

<h3>Displaced volume of water in the pond</h3>

According to Archimedes principle, when a body is partially or completely immersed in a fluid, it experiences an upthrust which is equal to weight of to fluid displaced.

F = ρVg

V = F/ρg

where;

  • V is volume of water displaced
  • ρ is density of the immersed object
  • g is acceleration due to gravity

When an object with a significant weight is dropped into the pond, the level of water in the pond will rise and vice versa.

<h3> The son pops the helium balloon</h3>

The water level in the pond will be Unchanged, since the weight of the balloon is almost insignificant.

<h3>Fisherman knocks the tackle box overboard and it sinks to the bottom</h3>

The water level will rise because the weight of the tackle box is significant.

<h3>Son finds a cup and bails some water out of the bottom of the boat</h3>

The water level will fall since some volume of water is being removed.

<h3>Fisherman lowers himself in the water and floats on his back</h3>

The water level will be unchanged, since nothing new is added into the pond.

<h3>Fisherman lowers the anchor and it hangs one foot above the bottom of the pond</h3>

The water level will rise because the anchor will displace some volume of water upwards when it is dropped inside the pond.

<h3>Son gets in the water, looses his grip on the string, letting the balloon escape upwards.</h3>

The water level will rise, because the balloon will displace the water upwards when it escape upwards.

Learn more about upthrust here: brainly.com/question/23931817

#SPJ1

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A 10 kg frictionless cart is pushed at a constant force of 5.0 N for a distance of 10 m. The work done on the cart is 50J.
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1) Final kinetic energy of the cart: B) 50 J

2) Final speed of the cart: C) 3.2 m/s

3) Height reached along the ramp: A) 0.5 m

Explanation:

1)

We can solve this part of the problem by using the work-energy theorem, which states that the work done on an object is equal to the kinetic energy gained by the object itself. Mathematically:

W=K_f - K_i

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K_i is the initial kinetic energy

In this problem, the work done on the cart is

W = 50 J

And assuming it starts from rest, its initial kinetic energy is zero:

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Therefore, the final kinetic energy is:

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2)

The kinetic energy of an object is the energy possessed by an object due to its motion; it is calculated as

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For the cart in this problem, we have:

K = 50 J is its final kinetic energy

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v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(50)}{10}}=3.2 m/s

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We can think this problem in terms of conservation of energy. In fact, as the cart rolls up the ramp, its kinetic energy is converted into gravitational potential energy, which is given by

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h is the heigth of the cart

When the cart reaches the maximum height, all the kinetic energy has been converted into potential energy, so we can write:

K=PE\\\frac{1}{2}mv^2=mgh

Re-arranging,

h=\frac{v^2}{2g}

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we can find the maximum height reached along the ramp:

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