As the roller coaster speeds up on the way down the hill, the potential energy of roller coaster will be converted to kinetic energy.
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What is Conservation of Energy ?</h3>
Conservation of energy state that energy is neither created nor destroy, they can only be transformed from one form to another. Energy of and object can transform from Potential energy to kinetic energy and vice versa
Given that at the top of a hill a roller coaster has gravitational potential energy due to its position. What will happen to this potential energy as the roller coaster speeds up on the way down the hill is that the potential energy to the roller coaster will start decreasing while the kinetic energy will start to increase.
The total energy of the roller coaster will be constant because of conservation of energy. As the roller coaster speeds up on the way down the hill, the potential energy will eventually reduce to zero where the total energy of the as the roller coaster will be equal to maximum kinetic energy.
Therefore, as the roller coaster speeds up on the way down the hill, the potential energy of roller coaster will be converted to kinetic energy.
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Answer:
Explanation:
You didn't last any of the variables. You have to list the variables to tell which are which.
Answer:
the shooting angle ia 18.4º
Explanation:
For resolution of this exercise we use projectile launch expressions, let's see the scope
R = Vo² sin (2θ) / g
sin 2θ = g R / Vo²
sin 2θ = 9.8 75/35²
2θ = sin⁻¹ (0.6)
θ = 18.4º
To know how for the arrow the tree branch we calculate the height of the arrow at this point
X2 = 75/2 = 37.5 m
We calculate the time to reach this point since the speed is constant on the X axis
X = Vox t
t2 = X2 / Vox = X2 / (Vo cosθ)
t2 = 37.5 / (35 cos 18.4)
t2 = 1.13 s
With this time we calculate the height at this point
Y = Voy t - ½ g t²
Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²
Y = 6.23 m
With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch
The thermal efficiency of an engine is
where
W is the work done by the engine
Q is the heat absorbed by the engine to do the work
In this problem, the work done by the engine is W=200 J, while the heat exhausted is Q=600 J, so the efficiency of the machine is
