Answer:
Elastic Collision
Inelastic Collision
The total kinetic energy is conserved. The total kinetic energy of the bodies at the beginning and the end of the collision is different.
Momentum does not change. Momentum changes.
No conversion of energy takes place. Kinetic energy is changed into other energy such as sound or heat energy.
Highly unlikely in the real world as there is almost always a change in energy. This is the normal form of collision in the real world.
An example of this can be swinging balls or a spacecraft flying near a planet but not getting affected by its gravity in the end.
The highest trophic level has the least available energy in kilojoules.
Even though the food web is not shown in the question, but we know that energy decreases steadily as it is passed on from one trophic level to the next according to the second law of thermodynamics.
Energy enters into the system from the sun. The primary producers utilize this energy to produce food. As plants are eaten by animals, this energy is transferred along the food web an diminishes at each higher trophic level.
At the highest trophic level, the the least available energy in kilojoules in this food web is found.
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Answer:
Looks like you have:
a = -.324 cos 2.5 t
In this case ω^2 A = .324
ω = 2.5
f = ω / (2 * pi) = 2.5 / 6.28 = .40 / sec
False. Inertia and mass is not described in Newton’s second law of motion but in Newton’s first law of motion. Newton’s first law of motion or sometimes referred to as the law of inertia. In Newton’s first law indicates that an object at rest will remain at rest unless acted by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
Answer:
20 N/m
Explanation:
From the question,
The ball-point pen obays hook's law.
From hook's law,
F = ke............................ Equation 1
Where F = Force, k = spring constant, e = compression.
Make k the subject of the equation
k = F/e........................ Equation 2
Given: F = 0.1 N, e = 0.005 m.
Substitute these values into equation 2
k = 0.1/0.005
k = 20 N/m.
Hence the spring constant of the tiny spring is 20 N/m
