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crimeas [40]
3 years ago
15

a passenger on cruise between San Juan, Puerto Rico and Miami, Florida accidentally drops a souvenir metal cube over the side of

the boat, into the water. Each side of the metal cube measures 1 meter. The cube promptly sinks to the deepest part of the Puero Rico Trench. Once at the bottom, what pressure does the cube experience? Neglect Atmospheric Pressure. Use wikipedia to see depth of Trench!
Physics
1 answer:
Viefleur [7K]3 years ago
6 0

Answer:

P = 84.1 MPa

Explanation:

  • The pressure at the bottom of column of of salt water of height h, is given by the following expression:

       P = \rho * g * h  (1)

       where ρ = density of salt water (in Kg/m³),

       g = acceleration due to  gravity  (in m/s²)

       h = height of the column of water.

  • Replacing by their values in (1):

      P = \rho * g * h  = 1023.6kg/m3*9.8m/s2*8380m = 84.1 MPa  (2)

  • Neglecting the atmospheric pressure, the pressure on the cube at the bottom of Puerto Rico Trench is given by (2):
  • P = 84.1 MPa.
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A crude model of the human throat is that of a pipe open at both ends with a vibrating source to introduce the sound into the pi
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Cannonball a is launched across level ground with speed v₀ at an angle of 45. Cannonball b is launched from the same location wi
postnew [5]

Answer:

Cannonball b spends more time in the air than cannonball a.

Explanation:

Starting with the definition of acceleration, we have that:

a=\frac{\Delta v}{\Delta t}\\\\\Delta t= \frac{\Delta v}{a}

Since both cannonballs will stop in their maximum height, their final velocity is zero. And since the acceleration in the y-axis is g, we have:

\Delta t= -\frac{v_{oy}}{g}

Now, this time interval is from the moment the cannonballs are launched to the moment of their maximum height, exactly the half of their time in the air. So their flying time t_f is (the minus sign is ignored since we are interested in the magnitudes only):

t_f=2\frac{v_{oy}}{g}

Then, we can see that the time the cannonballs spend in the air is proportional to the vertical component of the initial velocity. And we know that:

v_{oy}=v_o\sin\theta\\\\\implies t_f=2\frac{v_o\sin\theta}{g}

Finally, since \sin60\°=\frac{\sqrt{3} }{2} and \sin45\°=\frac{\sqrt{2} }{2}, we can conclude that:

t_{fa}=\sqrt{2}\frac{v_o }{g} \\\\t_{fb}=\sqrt{3}\frac{v_o }{g}\\\\\implies t_{fb}>t_{fa}

In words, the cannonball b spends more time in the air than cannonball a.

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3 years ago
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