Hey! I'm working through the lab on Archimedes Principle and Buoyancy. I need someone to check my work, as my actual and theoretical values between question 2 and 3 are... pretty different from one another. should I be using a negative value for my buoancy? Thank you! Buoyant Force and Archimedes Principle3. Use the measured dimensions of the 250 g mass to calculate the volume of the mass basedon its shape (either hexagonal or cylindrical). State the shape of your mass, e.g., circularcylinder or hexagonal cylinder. Show your work.(Hint: If you do not know the formula for the volume of your shape, look it up on theinternet.)My mass shape is cylindrical, I will be using the formula V = (pi)(r^2)(h) for this reason.V = (pi)(1.25*10^-2)(4.4*10^-2) = 1.7*10^-3 → V = 1.7*10^-3 m^34. Determine the percent difference between the measured volume of the 250 g mass and thevalue calculated from the buoyant force measurement. Show your work.5. Using the fact that 1 mL = 1 x 10-6 m³, compare the volume of the displaced water to thecalculated volume of the mass from Post-Lab Question 3 with a percent error calculation.Show your work.6. Are the three experimental values of the volume of the 250 g mass (from Archimedes'Principle) larger or smaller than the calculated value using the caliper measurements? Why?Which of the three values for volume is more accurate (the two experimental values and thecaliper value)? Which of the two experimental volumes are more accurate? Explain. Data SheetBase Edge or Diameter (cm)Object250 g HangingMassTable 2. Dimensions of 250g Hanging MassHeight (cm)Rubber Stopper2.5Post-Lab QuestionsTable 3. 250 g Hanging Mass Buoyancy DataWeight in Air(N)2.45.25Weight inWater (N)2.154.4.05BuoyancyForce (N).30.20VolumeDisplaced(mL)165.51. What happens to the apparent weight when the objects are submerged in water?When submerged in water, the apparent weight is less than its actual weight. This is due tothe buoyant force acting on the object.2. Use the value of the buoyant force to calculate an experimental value of the volume of the250 g mass, using Fb =PLVDg. The density of water is approximately 1,000 kg/m³. Showyour work.Fb=PLVDg → Vo=Fbl (pl)(g) → VD= (.30)/(1000) (9.81)= VD= 3.06*10^-5m^3 MAY BENEGATIVE.30!! EMAILED!I

Name: Hey! I'm working through the lab on Archimedes Principle and Buoyancy.
Uploaded: 2024-03-20T06:56:55.000Z
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Description: Hey! I'm working through the lab on Archimedes Principle and Buoyancy. I need someone to check my work, as my actual and theoretical values between question 2 and 3 are... pretty different from one another. should I be using a negative value for my b