Question

X Densily Practice Problems (Up X Denty Practice Problems.pdf G Sucos G Suppose a albic freshwater agix mnos/u-180596bfb796c36287ad6dbf800242932befees/MyFiles/Downloads/Density%20Practioek20Problems%20(1).pdf Density and Weight Practice Problems 1) Suppose a cubic freshwater aquarium holds 150 liters of water. Assume the tank has only water in it.) What is the mass of the water in the tank? Give your answer in kilograms. What are the dimensions of the tank? What useful observation can you take away from this problem? 2) Compute the weight of the water in the tank in #1. 3) Suppose 2.50 10 grams of water are poured into a cylindrical container with a radius of 8.00 cm. What will the depth of the water be after it is poured into the container? Recall the volume formula for a cylinder, V = wrh. 4) Most ice has a density of about 0.92 What is the mass of a chunk of ice if its volume is 58.5 cm? What is the weight of the ice in Newtons? 5) What is the mass (g) of a 1.75 liter chunk of ice? What is the weight (N)of the ice? Would the mass and weight change if the ice melted? 6) If a substance is less dense than water, then it will float in water. The density of such a substance gives you the decimal form of a percent. The number tells you what percent of the object will be under water when it is placed in a body of water. Convert the densities below to a percent a. Pine: 0.77

Answer 1

1) Given:

V = volume of water = 150 lt = $150*10^{-3}m^{3}$

the container is cubical.

density of water = $10^{3}kg/m^{3}$ .

Let the side length of the cubical container is L.

a) mass of water taken = m = volume * density of water = .

b) The volume of container is $L^{3}$ .

therefore, $L^{3}=150*10^{-3}$

$\Rightarrow L=\left (150*10^{-3} \right )^{\frac{1}{3}}=0.53m$

therefore, the side length of the cubical container is 0.53 m. [answer]

c) The useful observations are, the amount of water in litre is equal to amount of water in kilogram.

2) weight of water in the tank is equal to

mg = 150*9.81 = 1471.5 N [answer]

3) Given:

r = radius of cylindrical container = 8 cm = 0.08 m

Let the height of water column be 'h'.

Volume of cylindrical shape is given by $V=\pi r^{2}h$ [where r = radius, h = height]

Now, volume of water taken : $V=\frac{2.5}{10^{3}}=0.0025m^{3}$

Now, equating the volume of water to the volume of cylindrical shape

$\pi r^{2}h=0.0025$

$\Rightarrow h=\frac{0.0025}{\pi r^{2}} = \frac{0.0025}{\pi*0.08^{2}}=0.1243m = 12.34cm$ [answer]

4) Given:

density of ice = $0.92\frac{g}{cm^{3}}=0.92*\frac{10^{-3}}{10^{-6}}kg/m^{3}=0.92*10^{3}kg/m^{3}$

volume of ice taken = $58.5cm^{3}=58.5*10^{-6}m^{3}$

mass of ice = m = volume * density = [answer]

weight of ice = mg = [answer]