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mihalych1998 [28]
3 years ago
13

Cody's family went hiking along a stream in Georgia and found a waterfall. At the bottom of the waterfall, they noticed many sma

ll pebbles with smooth edges on the bottom of the stream. If Cody's family added more rocks to the stream at the bottom of the waterfall, what should they expect to happen to the size of the rocks?
Physics
1 answer:
Vitek1552 [10]3 years ago
4 0

Answer:

They can expect the size of the rocks to get smaller as the time passes.

Explanation:

They can expect the size of the rocks to get smaller as the time passes.

  • This is so because the water falling from certain height under waterfall will exert mechanical impact on the rock due to the mass of the water. This will act as water hammer <em>causing wear and tear</em> to the rock particles and flowing them away with the stream of water.
  • This hydraulic shock is generated when there is sudden stoppage in the motion of water. In sync with Newton's second law of motion, when the momentum of water is changed within a short course of time it creates a hammer like impact which has potential  of wear and tear when an object is continuously subjected to it.
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A circular window of 30 cm diameter in a submarine can withstand a maximum force of 5.20 × 105 N. The maximum depth in a lake to
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Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

         F = 5.20 \times 10^{5} N

         g = 9.8 m/s

         radius = \frac{diameter}{2}

                    = \frac{30 cm}{2} = 15 cm = 0.15 m   (as 1 m = 100 cm)

Formula to calculate depth is as follows.

        F = \rho \times g \times h \times A

or,      h = \frac{F}{\rho \times g \times A}        

       h = \frac{5.2 \times 10^{5}}{1000 \times 9.8 \times (3.1416 \times (0.15 m^{2})}

           = 751 m

Thus, we can conclude that the maximum depth in a lake to which the submarine can go without damaging the window is closest 750 m.

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3 years ago
*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude
kvv77 [185]

Answer:

Approximately 3.86\; {\rm N} (given that the magnitude of this charge is -7.35\; {\rm \mu C}.)

Explanation:

If a charge of magnitude q is placed in an electric field of magnitude E, the magnitude of the electrostatic force on that charge would be F = E\, q.

The magnitude of this charge is q = 7.35\; {\rm \mu C}. Apply the unit conversion 1\; {\rm \mu C} = 10^{-6}\; {\rm C}:

\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}.

An electric field of magnitude E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}} would exert on this charge a force with a magnitude of:

\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}.

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

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