A 0.140 kg baseball is thrown horizontally with a velocity of 28.9 m/s. It is struck with a constant horizontal force that lasts for 1.85 ms, which results in a velocity of 37.0 m/s in the opposite direction. What was the magnitude of the horizontal force

4987N

Explanation:

Step 1:

Data obtained from the question include:

Mass (m) = 0.140 kg

Initial velocity (U) = 28.9 m/s

Time (t) = 1.85 ms = 1.85x10^-3s

Final velocity (V) = 37.0 m/s

Force (F) =?

Step 2:

Determination of the magnitude of the horizontal force applied. This can be obtained by applying the formula:

F = m(V + U) /t

F = 0.140(37+ 28.9) /1.85x10^-3

F = 9.226/1.85x10^-3

F = 4987N

Therefore, the magnitude of the horizontal force applied is 4987N

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.