Physics
04.10.2022 20:21
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When an object is put into a liquid, it experiences a buoyant force that is equal
When an object is put into a liquid, it experiences a buoyant force that is equal to the weight of the liquid the object displaces. The force on the wire is given as the block is slowly lowered into the liquid (position is given in "centimeters", which you have to change to "meters") and force is given in "newtons"). Choose a mass of the block between 0.125 kg and 0.375 kg and a density of the liquid between 500 kg/m^3 and 1000 kg/m^3. The object is in static equilibrium when the clock stops.
Required:
a. What is the weight of the block and the tension, F, in the string when the block is in the liquid?
b. What is the volume of the block in the liquid—either the submerged part of the block if the block is partially submerged when you paused it or the entire block if it is completely submerged (the dimension of the block that is into the screen is 5.00 cm = 0.0500m)?
c. What is the volume of the water that is displaced by the block (the dimension of both water containers into the screen is 10.00 cm = 0.100m)?
Show Answers
a. What is the weight of the block and the tension, F, in the string when the block is in the liquid?
b. What is the volume of the block in the liquid—either the submerged part of the block if the block is partially submerged when you paused it or the entire block if it is completely submerged (the dimension of the block that is into the screen is 5.00 cm = 0.0500m)?
c. What is the volume of the water that is displaced by the block (the dimension of both water containers into the screen is 10.00 cm = 0.100m)?
sydney673
4,6(31 marks)
The correct answer to this question is 48.8N on a bearing of 134.2 degrees.
In order to solve this, first we need to find the resultant force of the men pulling in the north and south directions:
F = 40 - 6 = 34N [due south - 180]
Then, find the resultant force.
R^2 =34^2 + 35^2
R^2 = 2381
R = sqrt(2381)
R = 44.8N
The angle from the vertical is given by:
tan(x) = 35/34
tan(x) = 1.0294
angle = 45.8 degrees
Taking N as zero degrees, then it bears 134.2 degrees.
In order to solve this, first we need to find the resultant force of the men pulling in the north and south directions:
F = 40 - 6 = 34N [due south - 180]
Then, find the resultant force.
R^2 =34^2 + 35^2
R^2 = 2381
R = sqrt(2381)
R = 44.8N
The angle from the vertical is given by:
tan(x) = 35/34
tan(x) = 1.0294
angle = 45.8 degrees
Taking N as zero degrees, then it bears 134.2 degrees.
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