Answer:
3) Ep = 13243.5[J]
4) v = 17.15 [m/s]
Explanation:
3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.
m = mass = 90 [kg]
h = elevation = 15 [m]
Potential energy is defined as the product of mass by gravity by height.
![E_{p}=m*g*h\\E_{p}=90*9.81*15\\E_{p}=13243.5[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3Dm%2Ag%2Ah%5C%5CE_%7Bp%7D%3D90%2A9.81%2A15%5C%5CE_%7Bp%7D%3D13243.5%5BJ%5D)
This energy will be transformed into kinetic energy.
Ek = 13243.5 [J]
4) The velocity can be determined by defining the kinetic energy, as shown below.
![E_{k}=\frac{1}{2} *m*v^{2} \\v = \sqrt{\frac{2*E_{k} }{m} }\\ v= \sqrt{\frac{2*13243.5 }{90} }\\v=17.15[m/s]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%20%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B2%2AE_%7Bk%7D%20%7D%7Bm%7D%20%7D%5C%5C%20v%3D%20%5Csqrt%7B%5Cfrac%7B2%2A13243.5%20%7D%7B90%7D%20%7D%5C%5Cv%3D17.15%5Bm%2Fs%5D)
Answer: a) the greater speed for the ball is getting with the large radius of the circle. b) 1.68* 10 ^3 m/s^2 c) 1.25*10^3 m/s^2
Explanation: In order to solve this problem firstly we have to consider that speed in a of the circular movement is directly the angular rotation multiply the radius of the circle so by this we found that the second radius get large speed.
Secondly to calculate the centripetal acceleration for the ball we have to considerer the relationship given by:
acceleration in a circular movement= ω^2*r
so
a1= (8.44 *2*π)^2*r1=1.68 *10^3 m/s^2
a2= (5.95*2*π)^2*r2=1.25*10^3 m/s^2
Answer:
The mass of a dust particle is 7.53 x 10^-10 kg. (Hope it helped.)
Explanation:
