To find the density of the block of lead, first calculate its volume using the formula ( \text{Volume} = \text{length} \times \text{width} \times \text{height} ). For the given dimensions, the volume is ( 4.50 , \text{cm} \times 5.20 , \text{cm} \times 6.00 , \text{cm} = 140.4 , \text{cm}^3 ). The density of lead is approximately ( 11.34 , \text{g/cm}^3 ), so the mass of the block can be found by multiplying the volume by the density, resulting in a mass of about ( 1583.9 , \text{g} ). Thus, the density remains ( 11.34 , \text{g/cm}^3 ).
The volume ( V ) of a box can be calculated using the formula ( V = \text{length} \times \text{width} \times \text{height} ). For a box measuring 5 cm x 20 cm x 5 cm, the volume is ( V = 5 , \text{cm} \times 20 , \text{cm} \times 5 , \text{cm} = 500 , \text{cm}^3 ). Therefore, the volume of the box is 500 cubic centimeters.
The speed of the ball is ( \frac{120 \text{ cm}}{15 \text{ sec}} = 8 \text{ cm/sec} )
To find the density of the wood, we first calculate its volume using the formula for the volume of a rectangular prism: ( V = length \times width \times height = 3.0 , \text{cm} \times 6.0 , \text{cm} \times 4.0 , \text{cm} = 72.0 , \text{cm}^3 ). Then, we calculate the density using the formula ( \text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{80.0 , \text{g}}{72.0 , \text{cm}^3} \approx 1.11 , \text{g/cm}^3 ). Since the density of water is 1.0 g/cm³, the wood, having a density greater than that, would not float in water.
Density is calculated using the formula ( \text{Density} = \frac{\text{Mass}}{\text{Volume}} ). For an object with a mass of 60 grams and a volume of 8 cm³, the density would be ( \frac{60 , \text{g}}{8 , \text{cm}^3} = 7.5 , \text{g/cm}^3 ). Therefore, the density of the object is 7.5 g/cm³.
The area of a square is calculated by squaring the length of one of its sides. For a square with sides that are 10 cm long, the area is (10 , \text{cm} \times 10 , \text{cm} = 100 , \text{cm}^2). Therefore, the area of the square is 100 square centimeters.
To find the volume of a saucepan (which is shaped like a cylinder), you can use the formula for the volume of a cylinder: ( V = \pi r^2 h ). The radius ( r ) is half of the diameter, so ( r = 9.0 , \text{cm} ). Plugging in the values, ( V = \pi (9.0 , \text{cm})^2 (9.0 , \text{cm}) \approx 2261.95 , \text{cm}^3 ). Therefore, the volume of the saucepan is approximately 2262 cm³.
The volume of the cube is (3 \times 3 \times 3 = 27 , \text{cm}^3). Density is calculated by dividing mass by volume, so the density of the cube would be (27 , \text{g} \div 27 , \text{cm}^3 = 1 , \text{g/cm}^3).
if a radius is 5cm the diameter is 10cm rule- you double the radius
To find the volume of the brick, multiply its dimensions: (4.0 , \text{in} \times 2.7 , \text{in} \times 8.0 , \text{in} = 86.4 , \text{in}^3). To convert cubic inches to cubic centimeters, use the conversion factor (1 , \text{in}^3 = 16.387 , \text{cm}^3). Thus, the volume in cubic centimeters is (86.4 , \text{in}^3 \times 16.387 , \text{cm}^3/\text{in}^3 \approx 1415.4 , \text{cm}^3).
To find the perimeter of a square with an area of 100 cm², first, determine the length of one side. The area of a square is given by the formula ( \text{Area} = \text{side}^2 ), so ( \text{side} = \sqrt{100} = 10 ) cm. The perimeter is calculated using the formula ( \text{Perimeter} = 4 \times \text{side} ), which gives ( 4 \times 10 = 40 ) cm. Thus, the perimeter of the square is 40 cm.
A larger cube that is 1 meter on each side has a volume of (1 , \text{m}^3), which is equivalent to (1,000,000 , \text{cm}^3) (since (1 , \text{m} = 100 , \text{cm})). Each small cube is (1 , \text{cm}^3) in volume. Therefore, a total of (1,000,000) small cubes, each 1 cm on a side, will fit into the larger cube.