Answer:
h >5/2r
Explanation:
This problem involves the application of the concepts of force and the work-energy theorem.
The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.
Vertically two forces are acting on the rider, the weight and normal force of the cart on the rider. The normal force and weight are acting downwards at the top. For the rider not to fall out of the cart at the top, the normal force on the rider must be zero. This brings in a design requirement for the roller coaster to move at a minimum speed such that the cart exerts no force on the rider. This speed occurs when the normal force acting on the rider is zero (only the weight of the rider is acting on the rider)
So from newton's second law of motion,
W – N = mv²/r
N = normal force = 0
W = mg
mg = ma = mv²/r
mg = mv²/r
v²= rg
v = √(rg)
The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop
So
ΔPE = ΔKE = 1/2mv²
The height at the roller coaster starts is usually higher than the top of the loop by design. So
ΔPE =mgh - mg×2r = mg(h – 2r)
2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.
In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.
So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.
ΔPE > ΔKE
mg(h–2r) > 1/2mv²
g(h–2r) > 1/2(√(rg))²
g(h–2r) > 1/2×rg
h–2r > 1/2×r
h > 2r + 1/2r
h > 5/2r
C. Clear, dry weather. A good way to remember is H for high pressure = H for happy weather; L for low pressure = L for lousy weather (Glad I had someone to tell me this)
Answer:
Explanation:
We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.
1 m =100 cm
Surface area =S=
We have to find the charge Q on the positive plates of the capacitor.
V=Initial voltage between plates
d=Initial distance between plates
Initial Capacitance of capacitor
Capacitance of capacitor after moving plates
Potential difference between plates after moving
Hence, the charge on positive plate of capacitor=
When we set something down on the ground, the kind of work that our arms doing is : negative apex
It's happen whenever we do works that are align with the force of Gravity (to the bottom)
hope this helps
Answer:
Explanation:
The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:
The work done on the ball due to drag is:
![W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]](https://tex.z-dn.net/?f=W_%7Bdrag%7D%20%3D%20%280.599%5C%2Ckg%29%5Ccdot%20%289.807%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%29%5Ccdot%20%282.18%5C%2Cm-3.10%5C%2Cm%29%2B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%280.599%5C%2Ckg%29%5Ccdot%20%5B%287.05%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D-%284.19%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D%5D)
