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: 40 Newton
Explanation:
Mass of baseball = 20 kg
Velocity of baseball = 20 m/s.
Time taken for throwing = 10 seconds
Since acceleration is the rate of change of velocity per unit time
i.e Acceleration = velocity / time taken
Acceleration = (20 m/s / 10seconds)
Acceleration = 2m/s^2
Now, force required to throw the baseball is a product of mass and acceleration
i.e Force = Mass x acceleration
Force = 20 kg x 2m/s^2
Force = 40Newton
Thus, 40 newton of force is required to throw baseball to reach the plate
Answer:
Explanation:
a ) When the box starts to slip
static friction = mg sinα
mg cosα x μ = mg sinα ( μ is coefficient of static friction )
Tanα = μ = .35
α = 19.2°
b ) Once box starts moving , kinetic friction will start applying on it
kinetic friction = mg cos19.2 x .25
= 2.31m
net force downward = mgsin19.2 - mgcos19.2 x .25
= m ( 3.22 - 2.31 )
= .91 m
acceleration downward ( a ) = .91 m / s²
c )
v² = u² + 2 a s
= 0 + 2 x .91 x 5
= 9.1
v = 3 m / s
Answer:
Secondary structure
The secondary structure arises from the hydrogen bonds formed between atoms of the polypeptide backbone. The hydrogen bonds form between the partially negative oxygen atom and the partially positive nitrogen atom
