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zysi [14]
3 years ago
9

Many adventures like to go rafting on the Colorado River through Grand Canyon National Park. There are many locations where the

river becomes more narrow, both the distance between the canyon walls as well as the depth changes due to debris like boulders on the bottom of the river; this leads to changes in the water speed. In the park, the Colorado River has an average width of 100m and an average depth of 8m, and an average speed of 3 m/s. At the Lava Falls Rapids, the river has an average width of about 25m and an average depth of about 15m. What is the approximate speed of the water in this location
Physics
1 answer:
GalinKa [24]3 years ago
6 0

Answer:

6.4 m/s

Explanation:

Given that :

The average width of the Colorado river = 100 m

Average depth of the river is = 8 m

Therefore, area = $A_1= 100 \ m \times 8 \ m$

Speed of the river, $v_1 = 3 \ m/s$

After the lava falls on the river,

Width of the river becomes = 25 m

Depth of the river became = 15 m

Therefore, area = $A_2= 25 \ m \times 15 \ m$

Now, since the volume flow rate of the Colorado river is same, then from the Continuity equation,

$Q_1=Q_2$

$A_1v_1=A_2v_2$

∴ $100 \times 8 \times3 = 25 \times 15 \times v_2$

$v_2=\frac{100 \times 8 \times 3}{25 \times 15}$

    = 6.4 m/s

Therefore, the speed of the river in this location is 6.4 m/s

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Answer:

Q1)  Time taking by particle to travel the 40 km wrt. earth = 1.34\times10^{-6} sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = 1.285\times10^{-5} sec.

Explanation:

Given :

Speed of particle wrt. earth v=0.99537c

Distance between where particle is created and earth surface = 40 km

we know that,

⇒       v = \frac{x}{t}

Where x = 40\times10^{3} m, v = 0.99537c, we know speed of light c = 3 \times10^{8}

∴      t = \frac{x}{v}

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According to the lorentz transformation,

⇒    l = l_{o} \sqrt{1-\frac{v^2}{c^2} }

Where l = improper length, l_{o} =proper length (distance measured wrt. rest frame) = 40 km

     l = 40 \sqrt{1-\frac{v^2}{c^2 }

     l = 40 \times 0.096

     l = 3.84 km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

   \Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }

Where \Delta t = improper time (wrt. earth frame time) =1.34\times10^{-6} sec ,  \Delta t _{o} = proper time (wrt. particle frame).

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