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Blizzard [7]
4 years ago
15

you measure the depth to the bottom with a lead line and find out the bottom is 6000 meters deep. How many seconds would it take

for a seconds would it take for a sound signal to leave the ship, bounce off the bottom and return to ship? Show your work
Chemistry
1 answer:
son4ous [18]4 years ago
7 0
<h3>Answer:</h3>

7.838 seconds

<h3>Explanation:</h3>
  • Assuming the sound of sound in seawater is 1531 m/s we can calculate the time taken by the signal to leave the ship and back.
  • The depth to the bottom is 6000 meters
  • The sound signal will cover the distance twice, to and fro the bottom to the ship.
  • Time is calculated by dividing the distance by the velocity or speed.

In this case;

Time = (2× distance) ÷ velocity

        = (2×6000) m ÷ 1531 m/s

        = 12000 m ÷ 1531 m/s

        = 7.838 s

Therefore, it will take 7.838 s for the sound signal to leave the ship and bounce back off and return to the ship.

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If a concentration of 10 fluorescent molecules per μm2of cell membrane is needed to visualize a cell under the microscope, how m
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Answer : Total molecules that will be needed to visualize a single egg will be 78500 molecules of dye.

Explanation : As a single egg cell has an approximately diameter of 100 μm.

We can use this formula to calculate area of the cell membrane;

A = π (100)^{2} / 4;  

We can take π as 3.14 and we get;

A = 3.14 X (100)^{2} / 4  

Soving we get;

A =  7850 μm^{2}  

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here 1 μm^{2} = 7850 μm^{2} dye molecules.

Therefore, 10 fluorescent molecules will need;  

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4 years ago
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n = ???

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x = 111 * 0.622
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