The volume of the object placed in the water is 23 ml. This can be calculated by subtracting the initial water level (30 ml) from the final water level (53 ml).
1.72 ml
weight the metal ball first. then fill a graduated cylinder with water- it doesnt really matter how much, and put the metal ball in the water. measure how much the water level has increased by in mL. take the mass, and divide by the mL of water and then you get the density. ++ If it's an accurate sphere you can also measure its diameter and so calculate the volume, from which and the mass you can calculate the density.
Knowing the diameter of the ball allows you to calculate it. Using displacement of water allows to directly calculate it. Using direct measurement eliminates the error caused by the "dimples" on the ball's surface. (The minimum legal size for a ball has a volume of about 2.482 cubic inches or 40.68 cm3.)How to computeV = 4/3 (pi) r3 (4 x 3.1416 x radius cubed)How to measureThe easiest way to check would be to put a golf ball in a beaker of water. Fill a 200 ml beaker with water, and place the ball in it. Catch the overflow in another measured beaker. Since water is almost exactly 1 ml = 1 cc, the liquid displaced by the completely submerged ball will provide a close estimate of its volume in cm3.
82 cubic cm 1 ml is equa to 82 cubic cm 82 ml=82 cm3
That is 10 ml of water.
It would be 60 ml
The weight of the object is equaled to the level of the displaced water minus the original water level before the object was placed in it.| I.e. Original water level was: 150 ml. When we placed a rock inside the water, it displaced the water and the water level now sits at 200ml. 200 ml - 150 ml = 50 ml. So the object weighs about 50 mL or 50g since 1 mL of water is 1g.
The smallest ball permitted by USGA rules is 1.68 inches in diameter, which would yield a minimum volume of about 2.48 cubic inches (40.68 cm3) -- the dimples vary in number and depth, so the displaced volume would be slightly smaller. Historically, there was a smaller "British" ball that yielded some advantage in distance.How to computeV = 4/3 (pi) r3 (4 x 3.1416 x radius cubed)How to measureThe easiest way to check would be to put a golf ball in a beaker of water. Fill a 200 ml beaker with water, and place the ball in. Catch the overflow in another measured beaker. Since water is almost exactly 1 ml = 1 cc, the liquid displaced by the completely submerged ball will provide a close estimate of its volume in cm3.
you can do a simple easy experiment to find the volume of a ping ball (water displacment). Place the ping pong ball in an amount of water with a known volume in a SI unit of volume (mL, liters). Notice the water has risen. Subtract the vslue of the original water from the new value of the water. This is the ping pong's volume.
The volume of aluminum is 1.130 ounces / 2.702 g/ml ≈ 0.418 ml. Since the aluminum is dropped into the water, the water level will rise by the same volume of aluminum added, so the final liquid level will be 15.90 ml + 0.418 ml = 16.318 ml.
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