Answer:
v=32.9m/s
Explanation:
The acceleration needed to mantain a circular motion of radius r and speed v is given by the equation ![a=v^2/r](https://tex.z-dn.net/?f=a%3Dv%5E2%2Fr)
This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).
We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:
![v=\sqrt{ar} = \sqrt{12.5gr}=\sqrt{(12.5)(9.8m/s)(8.84m)}=32.9m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bar%7D%20%3D%20%5Csqrt%7B12.5gr%7D%3D%5Csqrt%7B%2812.5%29%289.8m%2Fs%29%288.84m%29%7D%3D32.9m%2Fs)
The sum of the kinetic and potential energies of a system of objects is conserved only when no external force acts on the objects.
<h3>
Conservation of mechanical energy</h3>
The principle of conservation of mechanical energy states that the total mechanical energy of an isolated system (absence of external force) is always constant.
M.A = P.E + K.E
where;
P.E is potential energy
K.E is kinetic energy
Thus, the sum of the kinetic and potential energies of a system of objects is conserved only when no external force acts on the objects.
Learn more about conservation of mechanical energy here: brainly.com/question/24443465
Answer:
the object will travel 0.66 meters before to stop.
Explanation:
Using the energy conservation theorem:
![E_i+K_i+W_f=K_f+U_f](https://tex.z-dn.net/?f=E_i%2BK_i%2BW_f%3DK_f%2BU_f)
The work done by the friction force is given by:
![W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]](https://tex.z-dn.net/?f=W_f%3DF_f%2Ad%5C%5CW_f%3D%5C%C2%B5%2Am%2Ag%2Ad%5C%5CW_f%3D0.35%2A4%2A9.81%2Ad%5C%5CW_f%3D13.7d%5BJ%5D)
so:
![\frac{1}{2}1800*(10*10^{-2})+0-13.7d=0+0\\d=0.66m](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D1800%2A%2810%2A10%5E%7B-2%7D%29%2B0-13.7d%3D0%2B0%5C%5Cd%3D0.66m)
Answer:
The y-component of the normal force is 45.74 N.
Explanation:
Given that,
Mass of the crate, m = 5 kg
Angle with hill, ![\theta=21^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D21%5E%7B%5Ccirc%7D)
We need to find the y component of the normal force. We know that the y component of the normal force is given by :
![F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N](https://tex.z-dn.net/?f=F_y%3DF%5C%20%5Ccos%5Ctheta%5C%5C%5C%5CF_y%3Dmg%5C%20%5Ccos%5Ctheta%5C%5C%5C%5CF_y%3D5%5Ctimes%209.8%5C%20%5Ccos%2821%29%5C%5C%5C%5CF_y%3D45.74%5C%20N)
So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.