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
Answer:
The rms voltage (in V) measured across the secondary coil is 459.62 V
Explanation:
Given;
number of turns in the primary coil, Np = 375 turns
number of turns in the secondary coil, Ns = 1875 turns
peak voltage across the primary coil, Ep = 130 V
peak voltage across the secondary coil, Es = ?
The rms voltage (in V) measured across the secondary coil is calculated as;
Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V
Answer:
So, we can assemble the options based or evaluation of there properties regarding getting an equilibrium or balanced state in a given time.We have the following rankings attributed to the elements:
- Silveringpot≥ aluminiumpot ≥ironskillet ≥glasscasseroldish ≥welldone steak ≥woodencuttingboard.
Explanation:
<u>Attaining equilibrium matters:</u>
When the materials are placed inside the oven, they attain a high temperature value causing it to be non touchable but some of the items has low value to attain the equilibrium state when comes in contact with other mediums. As these materials are also arranged based on that analyses.
I think it’s ultraviolet, if not i’m sorry!
The formula for Force is F = MA, or Force is equivalent to the product of Mass and Acceleration.
F = 128N.
M = 35.2kg.
128 = 35.2A
Divide both sides by 35.2 to solve for the acceleration.
A = ~3.636
The acceleration is 3.636 m/s^2.
I hope this helps!
