Average speed = distance traveled / time
average speed = (126.5 m * 3.5 laps) / (4.17 min)
= 106.2 m/min
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:
v₂=- 34 .85 m/s
v₁=0.14 m/s
Explanation:
Given that
m₁=70 kg ,u₁=0 m/s
m₂=0.15 kg ,u₂=35 m/s
Given that collision is elastic .We know that for elastic collision
Lets take their final speed is v₁ and v₂
From momentum conservation
m₁u₁+m₂u₂=m₁v₁+m₂v₂
70 x 0+ 0.15 x 35 = 70 x v₁ + 0.15 x v₂
70 x v₁ + 0.15 x v₂=5.25 --------1
v₂-v₁=u₁-u₂ ( e= 1)
v₂-v₁ = -35 --------2
By solving above equations
v₂=- 34 .85 m/s
v₁=0.14 m/s
Answer:
374.39 J/K
Explanation:
Entropy: This can be defined as the degree of disorder or randomness of a substance.
The S.I unit of entropy is J/K
ΔS = ΔH/T ..................................... Equation 1
Where ΔS = entropy change, ΔH = Heat change, T = temperature.
ΔH = cm................................... Equation 2
Where,
c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.
Substitute into equation 2
ΔH = 333000×0.3071
ΔH = 102264.3 J.
Also, T = 273.15 K
Substitute into equation 1
ΔS = 102264.3/273.15
ΔS = 374.39 J/K
Thus, The change in entropy = 374.39 J/K
The frictional force while the mass is sliding will be 46.2 N.
<h3>What is friction force?</h3>
Opposition forces on the surface cause heat loss during the motion of an object known as the friction force.
Given data:
m(mass)= 10.0-kg
Θ (Inclination angle)=25.0o
Coefficient of sliding friction,
=0.520
Coefficient of static friction,
The friction force, F=?
Resolve the force in the inclined plane;
Hence, the frictional force while the mass is sliding will be 46.2 N.
To know more about friction force refer to the link;
brainly.com/question/1714663
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