Spherical lead shot are poured into a square tray, 10 centimeters on a side, until they completely cover the bottom. The shot are poured from the tray into a graduated cylinder, which they fill to the 20-cm3 mark. What is the diameter of a single shot? How many shot were in the tray? If the 20 cm3 of shot had a mass of 130 grams, what would be the mass of a single shot?
Let D be the diameter in cm. The number of shot in the tray is approximately 100/D^2. The equation is not accurate unless D<<10 cm. With a hexagonal arrangement of spheres, you can get about 119/D^2 speheres in. It depends upon how you arrange the spheres and how much loose space remains at the edges.
The same number fills the cylinder up to 20 cm^3. Calculate the number of spheres of diameter D that can occupy that volume with hexagonal close packing. According to a theorem of Gauss, 74% of the volume can be occupied by solid spheres, whose volume is pi*D^3/6, so the number of spheres also equals
0.74*20/(pi*D^3/6) = 28.3/D^3
28.3/D^3 = 100/D^2 ; therefore
D = 0.28 cm
You can complete the other questions with that information.
To find the diameter of a single shot, we can use the volume of a sphere formula. The formula for the volume of a sphere is:
V = (4/3) * π * r^3
where V is the volume and r is the radius of the sphere.
In this case, the volume of a single shot is equal to the volume of 20 cm^3, which we can convert to m^3 for consistency:
V_shot = 20 cm^3 * (1 m/100 cm)^3 = 0.00002 m^3
We need to solve for the radius, so we rearrange the equation:
V_shot = (4/3) * π * r^3
0.00002 = (4/3) * π * r^3
Divide both sides by (4/3) * π:
r^3 = 0.00002 / [(4/3) * π]
r^3 ≈ 0.000003189
Now, take the cube root of both sides to solve for the radius:
r ≈ ∛(0.000003189)
r ≈ 0.0148 m
Finally, to find the diameter (D) of the shot, multiply the radius (r) by 2:
D = 2 * r ≈ 2 * 0.0148 = 0.0296 m
So, the diameter of a single shot is approximately 0.0296 meters.
To find the number of shots in the tray, we can calculate the volume of the tray and divide it by the volume of a single shot.
Volume of the tray = length * width * height = 10 cm * 10 cm * height
Since the tray is filled completely with lead shots, its volume is equal to the volume of the shots poured into the graduated cylinder, which is 20 cm^3.
So, we have:
10 cm * 10 cm * height = 20 cm^3
Solving for height:
height = 20 cm^3 / (10 cm * 10 cm)
height = 0.2 cm
Now, we divide the volume of the tray by the volume of a single shot:
Number of shots in the tray = Volume of tray / Volume of a single shot
= (10 cm * 10 cm * 0.2 cm) / 20 cm^3
= 0.1
So, there is approximately 0.1 shot in the tray. Since we can't have a fraction of a shot, we can assume that there is either 0 shot or 1 shot in the tray.
To calculate the mass of a single shot, we can use the mass of the 20 cm^3 of shot given as 130 grams.
Mass of a single shot = Mass of 20 cm^3 of shot / Number of shots in the tray
= 130 grams / 0.1
= 1300 grams
So, the mass of a single shot is 1300 grams.