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:
Explanation:
a) Fundamental frequency
A harmonic is an integral multiple of the fundamental frequency.
b) Wave speed
(i) Calculate the wavelength
In a fundamental vibration, the length of the string is half the wavelength.
(b) Calculate the speed
s
(a) -2451 N
We can start by calculating the acceleration of the car. We have:
is the initial velocity
v = 0 is the final velocity of the car
d = 125 m is the stopping distance
So we can use the following equation
To find the acceleration of the car, a:
Now we can use Newton's second Law:
F = ma
where m = 1100 kg to find the force exerted on the car in order to stop it; we find:
and the negative sign means the force is in the opposite direction to the motion of the car.
(b) 
We can use again the equation
To find the acceleration of the car. This time we have
is the initial velocity
v = 0 is the final velocity of the car
d = 2.0 m is the stopping distance
Substituting and solving for a,
So now we can find the force exerted on the car by using again Newton's second law:
As we can see, the force is much stronger than the force exerted in part a).
Answer:
* The first thing we observe is that the frequency response does not change
* The current that circulates in the circuit decreases due to the new resistance at the resonance point,
Z = R + R₂
Explanation:
The impedance of a series circuit is
Z₀² = R² + (X_L-X_C) ²
when we place another resistor in series the initial resistance impedance changes to
Z² = (R + R₂) ² + (X_L - X_C) ²
let's analyze this expression
* The first thing we observe is that the frequency response does not change
* The current that circulates in the circuit decreases due to the new resistance at the resonance point,
Z = R + R₂
