Hot water boils faster since its already hot and dosnt take much more time to get as hot
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:
3.76 m/s
Explanation:
Instantaneous velocity: This can be defined as the velocity of an object in a non uniform motion. The S.I unit is m/s.
v' = dx(t)/dt..................... Equation 1
Where v' = instantaneous velocity, x = distance, t = time.
Given the expression,
x(t) = 28.0 m + (12.4 m/s)t - (0.0450 m/s³)t³
x(t) = 28 + 12.4t - 0.0450t³
Differentiating x(t) with respect to t.
dx(t)/dt = 12.4 - 0.135t²
dx(t)/dt = 12.4 - 0.135t²
When t = 8.00 s.
dx(t)/dt = 12.4 - 0.135(8)²
dx(t)/dt = 12.4 - 8.64
dx(t)/dt = 3.76 m/s.
Therefore,
v' = 3.76 m/s.
Hence, the instantaneous velocity = 3.76 m/s
The correct answer of the given question above would be option B. The statement that is not correct is that, a steady magnetic field produces a steady current. The rest of the statements are all correct. <span>An unchanging/static magnetic field (relative to a wire/circuit) induces zero current.</span>
To answer the specific problem, the balloon contains 480kg
of helium. I am hoping that this answer has satisfied your query and it will be
able to help you, and if you would like, feel free to ask another question.
