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24 June 2019 Physics Pre Lab 2

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Physics

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The University of Alabama

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Homework

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1 - Buoyancy

QUESTION 1:

Recall that the density of any object can be expressed as the ratio between its mass

and its volume:

 





(1)

From this equation, the mass m and density  are directly proportional to each other.

Therefore, if the mass of the object is increased, then its density will also increase.

QUESTION 2:

When the density of the object is less than that of the liquid, the upward buoyant force

acting on the object will be larger than the downward weight of the object.

Consequently, this will make the object float.

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QUESTION 3:

When the density of the object is greater than that of the liquid, we encounter the

opposite scenario. The downward weight of the object will now be larger than the

upward buoyant force. The larger downward weight will make the object sink.

QUESTION 4:

Using the recommended settings, the block-fluid system will look like:

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1 - Buoyancy QUESTION 1: Recall that the density of any object can be expressed as the ratio between its mass and its volume: 𝜌= 𝑚 𝑉 (1) From this equation, the mass m and density 𝜌 are directly proportional to each other. Therefore, if the mass of the object is increased, then its density will also increase. QUESTION 2: When the density of the object is less than that of the liquid, the upward buoyant force acting on the object will be larger than the downward weight of the object. Consequently, this will make the object float. QUESTION 3: When the density of the object is greater than that of the liquid, we encounter the opposite scenario. The downward weight of the object will now be larger than the upward buoyant force. The larger downward weight will make the object sink. QUESTION 4: Using the recommended settings, the block-fluid system will look like: The volume of the displaced liquid will be equal to the volume of the object itself. This volume comparison is always true. In fact, this particular principle is used to measure the volume of irregularly-shaped objects. QUESTION 5: In relation to question (4) that the displaced liquid volume is equal to the object volume, this means that the displaced liquid volume should also be: 𝑉 = 1,000 𝑐𝑚3 (2) The value of this volume was obtained directly from the simulation display. QUESTION 6: From the simulation display, we know that the mass of the object is: 𝑚 = 5 𝑘𝑔 (3) In order t ...
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