Answer:
<u>10 J</u>
Explanation:
work = force x distance
work = 5x2
work = 10 J
Joule is the unit for work
Volume=0.09m³
Density=4000kg/m³
Force=?
Density=mass/volume ⇒mass=volume×density
m=0.09×4000=360kg
Force=mass×accelaration
F=360×9.8
F=3528N
<span>They deliver oxygen to the blood.
For the regular body activities to be carried out.
Acid-base balance or body temperature control.</span>
You have to inform your coach and get his signature in the fitness log.
Explanation:
Every time you finish your target and complete the task given by your coach, you have to make an entry in the fitness log given.
For the coach to know that you have done the task for the day, you have to get a signature from the coach to show him the evidence that the task given has been completed.
This is the first thing that an individual should do after making an entry in the fitness log.
After that he can plan out the next level work out for the next day and inform his coach of the same.
Answer:
Part 1)
![L_1 = 185.2 kg m^2/s^2](https://tex.z-dn.net/?f=L_1%20%3D%20185.2%20kg%20m%5E2%2Fs%5E2)
Part 2)
![L_2 = 663.07 kg m^2/s^2](https://tex.z-dn.net/?f=L_2%20%3D%20663.07%20kg%20m%5E2%2Fs%5E2)
Part 3)
![L = 663.07 kg m^2/s^2](https://tex.z-dn.net/?f=L%20%3D%20663.07%20kg%20m%5E2%2Fs%5E2)
Part 4)
![\omega = 1.83 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%201.83%20rad%2Fs)
Part 5)
![F_c = 453.6 N](https://tex.z-dn.net/?f=F_c%20%3D%20453.6%20N)
Explanation:
Part a)
Initial angular momentum of the merry go round is given as
![L_1 = I \omega](https://tex.z-dn.net/?f=L_1%20%3D%20I%20%5Comega)
here we know that
![I = 194 kg m^2](https://tex.z-dn.net/?f=I%20%3D%20194%20kg%20m%5E2)
![\omega = 1.47 rad/s^2](https://tex.z-dn.net/?f=%5Comega%20%3D%201.47%20rad%2Fs%5E2)
now we have
![L_1 = 194 \times 1.47](https://tex.z-dn.net/?f=L_1%20%3D%20194%20%5Ctimes%201.47)
![L_1 = 185.2 kg m^2/s^2](https://tex.z-dn.net/?f=L_1%20%3D%20185.2%20kg%20m%5E2%2Fs%5E2)
Part b)
Angular momentum of the person is given as
![L = mvR](https://tex.z-dn.net/?f=L%20%3D%20mvR)
so we have
![m = 68 kg](https://tex.z-dn.net/?f=m%20%3D%2068%20kg)
![v = 4.9 m/s](https://tex.z-dn.net/?f=v%20%3D%204.9%20m%2Fs)
R = 1.99 m
so we have
![L_2 = (68)(4.9)(1.99)](https://tex.z-dn.net/?f=L_2%20%3D%20%2868%29%284.9%29%281.99%29)
![L_2 = 663.07 kg m^2/s^2](https://tex.z-dn.net/?f=L_2%20%3D%20663.07%20kg%20m%5E2%2Fs%5E2)
Part 3)
Angular momentum of the person is always constant with respect to the axis of disc
so it is given as
![L = 663.07 kg m^2/s^2](https://tex.z-dn.net/?f=L%20%3D%20663.07%20kg%20m%5E2%2Fs%5E2)
Part 4)
By angular momentum conservation of the system we will have
![L_1 + L_2 = (I_1 + I_2)\omega](https://tex.z-dn.net/?f=L_1%20%2B%20L_2%20%3D%20%28I_1%20%2B%20I_2%29%5Comega)
![185.2 + 663.07 = (194 + 68(1.99^2))\omega](https://tex.z-dn.net/?f=185.2%20%2B%20663.07%20%3D%20%28194%20%2B%2068%281.99%5E2%29%29%5Comega)
![848.27 = 463.28 \omega](https://tex.z-dn.net/?f=848.27%20%3D%20463.28%20%5Comega)
![\omega = 1.83 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%201.83%20rad%2Fs)
Part 5)
Force required to hold the person is centripetal force which act towards the center
so we will have
![F_c = m\omega^2 R](https://tex.z-dn.net/?f=F_c%20%3D%20m%5Comega%5E2%20R)
![F_c = 68(1.83^2)(1.99)](https://tex.z-dn.net/?f=F_c%20%3D%2068%281.83%5E2%29%281.99%29)
![F_c = 453.6 N](https://tex.z-dn.net/?f=F_c%20%3D%20453.6%20N)