Force = Mass * Acceleration therefore the red ball with the higher mass will have more force and greater acceleration
Answer:
![F_a_v_g=7093333.33N*s](https://tex.z-dn.net/?f=F_a_v_g%3D7093333.33N%2As)
Explanation:
The impulse or average force in classical mechanics is the variation in the linear momentum that a physical object experiences in a closed system. It is defined by the following equation:
![F_a_v_g=m*\frac{\Delta v}{\Delta t}=m*\frac{v_2-v_1}{t_2-t_1}](https://tex.z-dn.net/?f=F_a_v_g%3Dm%2A%5Cfrac%7B%5CDelta%20v%7D%7B%5CDelta%20t%7D%3Dm%2A%5Cfrac%7Bv_2-v_1%7D%7Bt_2-t_1%7D)
Where:
![m=mass\hspace{3}of\hspace{3}the\hspace{3}object](https://tex.z-dn.net/?f=m%3Dmass%5Chspace%7B3%7Dof%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dobject)
![v_2=final\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}at\hspace{3}the\hspace{3}end\hspace{3}of\hspace{3}the\hspace{3}time\hspace{3}interval](https://tex.z-dn.net/?f=v_2%3Dfinal%5Chspace%7B3%7Dvelocity%5Chspace%7B3%7Dof%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dobject%5Chspace%7B3%7Dat%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dend%5Chspace%7B3%7Dof%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dtime%5Chspace%7B3%7Dinterval)
![v_1=initial\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}when\hspace{3}the\hspace{3}time\hspace{3}interval\hspace{3}begins.](https://tex.z-dn.net/?f=v_1%3Dinitial%5Chspace%7B3%7Dvelocity%5Chspace%7B3%7Dof%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dobject%5Chspace%7B3%7Dwhen%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dtime%5Chspace%7B3%7Dinterval%5Chspace%7B3%7Dbegins.)
![t_2=final\hspace{3}time](https://tex.z-dn.net/?f=t_2%3Dfinal%5Chspace%7B3%7Dtime)
![t_1=initial\hspace{3}time](https://tex.z-dn.net/?f=t_1%3Dinitial%5Chspace%7B3%7Dtime)
Asumming v1=0 and t1=0:
![F_a_v_g=m* \frac{v_2}{t_2} =(24.7)*\frac{784}{2.73*10^{*3} } =7093333.333N*s](https://tex.z-dn.net/?f=F_a_v_g%3Dm%2A%20%5Cfrac%7Bv_2%7D%7Bt_2%7D%20%3D%2824.7%29%2A%5Cfrac%7B784%7D%7B2.73%2A10%5E%7B%2A3%7D%20%7D%20%3D7093333.333N%2As)
Answer:
a)
& ![m_c.v_c=m_b.v_b\times \cos\theta](https://tex.z-dn.net/?f=m_c.v_c%3Dm_b.v_b%5Ctimes%20%5Ccos%5Ctheta)
b) ![v_c=0.0566\ m.s^{-1}](https://tex.z-dn.net/?f=v_c%3D0.0566%5C%20m.s%5E%7B-1%7D)
c) ![p_e=2.9218\ kg.m.s^{-1}](https://tex.z-dn.net/?f=p_e%3D2.9218%5C%20kg.m.s%5E%7B-1%7D)
Explanation:
Given:
mass of the book, ![m_b=1.35\ kg](https://tex.z-dn.net/?f=m_b%3D1.35%5C%20kg)
combined mass of the student and the skateboard, ![m_c=97\ kg](https://tex.z-dn.net/?f=m_c%3D97%5C%20kg)
initial velocity of the book, ![v_b=4.61\ m.s^{-1}](https://tex.z-dn.net/?f=v_b%3D4.61%5C%20m.s%5E%7B-1%7D)
angle of projection of the book from the horizontal, ![\theta=28^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D28%5E%7B%5Ccirc%7D)
a)
velocity of the student before throwing the book:
Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.
![u_c=0\ m.s^{-1}](https://tex.z-dn.net/?f=u_c%3D0%5C%20m.s%5E%7B-1%7D)
where:
initial velocity of the student
velocity of the student after throwing the book:
Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.
![m_c.v_c=m_b.v_b\times \cos\theta](https://tex.z-dn.net/?f=m_c.v_c%3Dm_b.v_b%5Ctimes%20%5Ccos%5Ctheta)
where:
final velcotiy of the student after throwing the book
b)
![m_c.v_c=m_b.v_b\times \cos\theta](https://tex.z-dn.net/?f=m_c.v_c%3Dm_b.v_b%5Ctimes%20%5Ccos%5Ctheta)
![97\times v_c=1.35\times 4.61\cos28](https://tex.z-dn.net/?f=97%5Ctimes%20v_c%3D1.35%5Ctimes%204.61%5Ccos28)
![v_c=0.0566\ m.s^{-1}](https://tex.z-dn.net/?f=v_c%3D0.0566%5C%20m.s%5E%7B-1%7D)
c)
Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.
![p_e=m_b.v_b\sin\theta](https://tex.z-dn.net/?f=p_e%3Dm_b.v_b%5Csin%5Ctheta)
![p_e=1.35\times 4.61\times \sin28^{\circ}](https://tex.z-dn.net/?f=p_e%3D1.35%5Ctimes%204.61%5Ctimes%20%5Csin28%5E%7B%5Ccirc%7D)
![p_e=2.9218\ kg.m.s^{-1}](https://tex.z-dn.net/?f=p_e%3D2.9218%5C%20kg.m.s%5E%7B-1%7D)
Answer:
Explanation:
There are two hypotheses she could test:
A cat's heart rate changes while it is napping.
A cat's heart rate does not change while it is napping.