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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Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together. This is done because the friction force is going to have to be compensated for. We will need that much more force than we otherwise would to achieve the desired acceleration:
The friction force will be given by the normal force times the coefficient of friction. Here the normal force is just its weight, mg
Now the total force required is:
0.0702N+0.803N=0.873N
Answer:
They are both correct.
Explanation:
The density of an object is defined as the ratio of its mass to its volume. This implies that the density of the object is both proportional to the mass and also to the volume of the object. John only mentioned mass which is correct. Linda mentioned the second variable on which density depends which is the volume of the object.
Hence considering the both statements objectively, one can say that they are both correct.
Answer
given,
mass of ball, m = 57.5 g = 0.0575 kg
velocity of ball northward,v = 26.7 m/s
mass of racket, M = 331 g = 0.331 Kg
velocity of the ball after collision,v' = 29.5 m/s
a) momentum of ball before collision
P₁ = m v
P₁ = 0.0575 x 26.7
P₁ = 1.535 kg.m/s
b) momentum of ball after collision
P₂ = m v'
P₂ = 0.0575 x (-29.5)
P₂ = -1.696 kg.m/s
c) change in momentum
Δ P = P₂ - P₁
Δ P = -1.696 -1.535
Δ P = -3.231 kg.m/s
d) using conservation of momentum
initial speed of racket = 0 m/s
M u + m v = Mu' + m v
M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)
0.331 u' = 3.232
u' = 9.76 m/s
change in velocity of the racket is equal to 9.76 m/s
Answer:
Nucleus And electron cloud
Explanation:
Hope this helps
