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
Main sequence stars are characterised by the source of their energy.They are all undergoing fusion of hydrogen into helium within their cores. The mass of the star is the main element for such process or phenomenon to take place for it is a determinant of both the rate at which they perform the said activity and the amount of fuel available.
To answer the question, the lower mass limit for a main sequence star is about 0.08. If the mass of a main sequence star is lower than the above-mentioned value, there would be a deficit or insufficiency of gravitational force to generate a standard temperature for hydrogen core fusion to take place and the underdeveloped star would form into a brown dwarf instead.
Answer:
A. 69.9m
Explanation:
Given parameters:
Initial velocity = 10.5m/s
Final velocity = 21.7m/s
Time = 4.34s
Unknown:
Distance traveled = ?
Solution:
Let us first find the acceleration of the car;
Acceleration = 
v is final velocity
u is initial velocity
t is the time
Acceleration = 
= 2.58m/s²
Distance traveled;
V² = U² + 2aS
21.7² = 10.5² + 2 x 2.58 x S
360.64 = 2 x 2.58 x S
S = 69.9m
Answer:
r = 20 m
Explanation:
The formula for the angular momentum of a rotating body is given as:
L = mvr
where,
L = Angular Momentum = 10000 kgm²/s
m = mass
v = speed = 2 m/s
r = radius of merry-go-round
Therefore,
10000 kg.m²/s = mr(2 m/s)
m r = (10000 kg.m²/s)/(2 m/s)
m r = 5000 kg.m ------------- equation 1
Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:
I = (1/2) m r²
where,
I = moment of inertia = 50000 kg.m²
Therefore,
50000 kg.m² = (1/2)(m r)(r)
using equation 1, we get:
50000 kg.m² = (1/2)(5000 kg.m)(r)
(50000 kg.m²)/(2500 kg.m) = r
<u>r = 20 m</u>
Answer:
they are both clocks i think please be right
