Based on Newton's first and second law of motion most people would find it less painful to catch a flying baseball than a bowling ball flying at the same speed as the baseball because the mass of the baseball is smaller and will require smaller force to be stopped.
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What is Newton's first law of motion?</h3>
Newton's first law of motion first law of motion states that a body at rest or uniform motion in a straight line will continue in that path unless acted upon by an external force.
Newton's first law of motion is also called law of inertia because it depends on mass of the object.
An object with a greater mass will require greater force to be stopped or get moving.
Based on Newton's first law of motion most people would find it less painful to catch a flying baseball than a bowling ball flying at the same speed as the baseball because the mass of the baseball is smaller and will require smaller force to be stopped.
Also according to Newton's second law of motion, the force applied to an object is proportional to the product of mass and acceleration of the object. Thus, a baseball with smaller mass will require smaller force to be stopped.
Learn more about Newton's first law of motion here: brainly.com/question/10454047
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<span>When you do work on a object some of your energy is "Transfers" to that object.
Hope this helps!</span>
Answer:
The new angular speed of the merry-go-round is 8.31 rev/min.
Explanation:
Because the merry-go-round is rotating about a frictionless axis there’re not external torques if we consider the system merry-go-round and child. Due that we can apply conservation fo angular momentum that states initial angular momentum (Li) should be equal final angular momentum (Lf):
(1)
The initial angular momentum is just the angular momentum of the merry-go-round (Lmi) that because it's a rigid body is defined as:
(2)
with I the moment of inertia and ωi the initial angular speed of the merry-go-round
The final angular momentum is the sum of the final angular momentum of the merry-go-round plus the final angular momentum of the child (Lcf):
(3)
The angular momentum of the child should be modeled as the angular momentum of a punctual particle moving around an axis of rotation, this is:
(4)
with m the mass of the child, R the distance from the axis of rotation and vf is final tangential speed, tangential speed is:
(5)
(note that the angular speed is the same as the merry-go-round)
using (5) on (4), and (4) on (3):
(6)
By (5) and (2) on (1):

Solving for ωf (12.0 rev/min = 1.26 rad/s):
![\omega_f= \frac{I\omega_i}{]I+mR^2}=\frac{(260)(1.26)}{260+(24.0)(2.20)^2}](https://tex.z-dn.net/?f=%5Comega_f%3D%20%5Cfrac%7BI%5Comega_i%7D%7B%5DI%2BmR%5E2%7D%3D%5Cfrac%7B%28260%29%281.26%29%7D%7B260%2B%2824.0%29%282.20%29%5E2%7D%20)
