At point E
- the kinetic energy of the rollercoaster is small compared to the potential energy
- the potential energy is greater than the kinetic energy
- the total energy is a mixture of potential and kinetic energy
<h3>What is the energy of the roller coaster at point E?</h3>
The energy of a roller coaster could either be potential energy, kinetic energy or a combination of both potential and kinetic energy.
Using analogies, the energy of the roller coaster at point E can be compared to a falling fruit from a tree which falls onto a pavement and is the rolling towards the floor. Point E can be compared to the midpoint of the fall of the fruit.
At point E
- the kinetic energy of the rollercoaster is small compared to the potential energy
- the potential energy is greater than the kinetic energy
- the total energy is a mixture of potential and kinetic energy
In conclusion, the energy of the rollercoaster at E is both Kinetic and potential energy,
Learn more about potential and kinetic energy at: brainly.com/question/18963960
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Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
Answer:
The sport utility vehicle was traveling at V2= 11.5 m/s.
Explanation:
m1= 1090 kg
V1= 30.4 m/s
m2= 2880 kg
V2= ?
m1*V1 = m2*V2
V2= (m1*V1)/m2
V2= 11.5 m/s
Answer:
trees
Explanation:
referring to the tree to prove his/her point.
Answer:
KE = 4 mv2 m = 2xKE valami. V m.
Explanation:
