Answer:
Explanation:
a)
Ff = μmgcosθ
Ff = 0.28(1600)(9.8)cos(-84)
Ff = 458.9217...
Ff = 460 N
b) ignoring the curves required at top and bottom which change the friction force significantly, especially at the bottom where centripetal acceleration will greatly increase normal forces and thus friction force.
W = Ffd
W = 458.9217(-49.4/sin(-84)
W = 22,795.6119...
W = 23 kJ
c) same assumptions as part b
The change in potential energy minus the work of friction will be kinetic energy.
KE = PE - W
½mv² = mgh - (μmgcosθ)d
v² = 2(gh - (μgcosθ)(h/sinθ))
v = √(2gh(1 - μcotθ))
v = √(2(9.8)(49.4)(1 - 0.28cot84))
v = 30.6552...
v = 31 m/s
Answer:

Explanation:
Asúmase que la patinadora experimenta una aceleración constante. La fuerza neta experimentada por la patinadora:
![F_{net} = (50\,kg)\cdot \left[\frac{\left(15\,\frac{m}{s}\right)^{2}-\left(0\,\frac{m}{s}\right)^{2} }{2\cdot (3000\,m)} \right]](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20%2850%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%5Cfrac%7B%5Cleft%2815%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%5Cright%29%5E%7B2%7D-%5Cleft%280%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%5Cright%29%5E%7B2%7D%20%7D%7B2%5Ccdot%20%283000%5C%2Cm%29%7D%20%5Cright%5D)

Answer:
The “terminal speed” of the ball bearing is 5.609 m/s
Explanation:
Radius of the steel ball R = 2.40 mm
Viscosity of honey η = 6.0 Pa/s



While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

Substitute the given values to find "terminal speed"




The “terminal speed” of the ball bearing is 5.609 m/s
I would look this one up on Google