What is the depth dm of the mercury? Use pw = 1000 kg/m³ as the density of water and pm = 13,600 kg/m³ as the density of mercury. Express your answer in centimeters. ▸ View Available Hint(s) ΑΣΦ ? dm = 73.5 Submit Previous Answers × Incorrect; Try Again; 29 attempts remaining cm

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You have an 18-carat gold coin (an alloy of gold and copper). Its mass is 6 grams. Using the fact that the number of carats is equal to (24*mass of gold/total mass), calculate the volume of gold in the coin. (density of pure gold = 19300 kg/m3, density of copper = 8960 kg/m3.) a.) 1.68*10-7 m3 b) 2.33*10-7 m3                c) 4*10-7 m3.   In the problem above, what is the density of the coin? a.) 14963 kg/m3 b) 12333 kg/m3                c) 19300 kg/m3

Water is being poured into a cylindrical can that is 20 inches in height and has a radius of 8 inches as shown in Figure 2. The water is being poured at a rate of 3 inch/ sec. Calculate the height of the water in the can changing when the height is 8 feet deep. ( HINT: Volume of cylinder, V = r²h). r= 8 20 Figure 2

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Items 13 - 16, refer to the problem inside the box. Then, answer the questions that follow. The length of a tarpaulin is 4 more than twice the width. Its area is no less than 96 meters. 13.) Which of the following represents the length of the tarpaulin? А. х В. х + 4 С. 2х + 4 D. 2x 14.) Which of the following expressions represents the width of the tarpaulin? A. x В. х + 4 С. 2х + 4 D. 2x 15.) Which inequality represents the situation? A. x(x) 2 96 B. x(2x + 4) 2 96 C. x(x + 4) 2 96 D. x(2x) 2 96 16.) What is the least length of the tarpaulin? A. 16-meters C. 6-meters B. 10-meters D. 4-meters

6. Convert the following measurements mg.um kg.m а. 68 ms? hr2 b. 49mg/cm³ to ng/mm³ 7. Using Dimensional Analysis, check which of the following equations is or are correct. а. рgh3 — та b. avt = vh?

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