four students push carts filled with sports equipment across the gym. Each student pushes with the same amount of force. which c
2 answers:
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1) The buoyant force acting on an object immersed in a fluid is:
where
is the density of the fluid, 
is the volume of displaced fluid, and 
is the gravitational acceleration.
2) We must calculate the volume of displaced fluid. Since the titanium object is completely immersed in the fluid (air), this volume corresponds to the volume of 1 Kg of titanium, whose density is
. Using the relationship between density, volume and mass, we find
3) Now we can recall the formula written at step 1) and calculate the buoyant force. The air density is
, so we have
4) The weight of 1 Kg of titanium is instead:
So, the buoyant force is negligible compared to the weight.
where
2) We must calculate the volume of displaced fluid. Since the titanium object is completely immersed in the fluid (air), this volume corresponds to the volume of 1 Kg of titanium, whose density is
3) Now we can recall the formula written at step 1) and calculate the buoyant force. The air density is
4) The weight of 1 Kg of titanium is instead:
So, the buoyant force is negligible compared to the weight.
The height of the roof is <u>3.57m</u>
Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.
Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.
The length of the window s is given by,
The first drop is at the bottom and it takes 5t seconds to reach down.
The height of the roof h is the distance traveled by the first drop and is given by,
the height of the roof is 3.57 m
