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Blizzard [7]
3 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]3 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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2 years ago
Suppose you wanted to dissolve 40.0 g NaOH in enough H2O to make 6.00 dm3 of solution
dezoksy [38]

Molarity of solution = 1.6 M

<h3>Further explanation</h3>

Given

40 g NaOH

6 L solution

Required

Steps to solve the problem of molarity

Solution

No additional information about the question.

If you want to make the solution above, then we just need to put the existing NaOH (40 g) into 6 L of water, then do the stirring (in a warm temperature above the hot plate will speed up the NaOH dissolving process)

But if you want to know the molarity of a solution, then

  • 1. we calculate the moles of NaOH

\tt mol=\dfrac{mass}{MW}

MW(molecular weight) of NaOH=

Ar Na+ Ar O + Ar H

23 + 16 + 1 = 40 g/mol

so mol NaOH :

\tt mol=\dfrac{40~g}{40~g/mol}=1~mol

  • 2. Molarity(M)

\tt M=\dfrac{n}{V}\\\\M=\dfrac{1}{6}\\\\M=0.16

5 0
3 years ago
How much salt (NaCl) is carried by a river flowing at 30.0 m3/s and containing 50.0 mg/L of salt
denis23 [38]

Answer:

129,600kg/day

Explanation:

The river is flowing at 30.0

1 m³ is equivalent to 1000L

flowrate of river = 30*1000 =30,000L/s

Convert L/s to litre per day by multiplying by 24*60*60

flowrate of river = 30,000 * 24*60*60 L/day

                    = 2,592,000,000L/day

if the river contains 50mg of salt  in 1L of solution

lets find how many mg of salt Y is contained in 2,592,000,000L/day

by cross multiplying we have

Y=\frac{2592000000*50}{1}

Y= 129,600,000,000 mg/day

convert this value to kg/day by dividing by 1 million

Y= 129,600,000,000/1000000

Y= 129,600kg/day

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