(8%) Problem 3: Sound in water travels at a velocity governed by the relation v = √(B/rho) where B is the bulk modulus and rho is the density. For salt water, take B = 2.38 × 109 Pa and rho = 1046 kg/m3. A whale sends out a high frequency (9 kHz) song to another whale 1.0 km away. How much time, in seconds, does it take for the sound to travel between the whales, t?

the time required for the sound to travel between the whales 0.66 S.

Explanation:

As given in the problem, the velocity of sound wave () is governed by the equation

Given, and

So for salt water, the velocity of sound wave () can be written as

As the whales are d = 1 Km or 1000 m apart from each other, so the time () required for the sound wave to travel this distance is given by

depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation:

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz