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:
681.6/ms
Explanation:
A reconnaissance plane flies 545 km away from its base at 568 m/s. then flies back to its base at 852 m/s.
What is its average speed?
Answer in its of m/s
Avg speed of the round trip is
2*568*852/(568+852)= 681.6/ms
The question involves a ping-pong ball that is held submerged in a bucket by a string attached to the bottom of the bucket.
The answer is the tension of the string will increase. This is because making the water salty increases its density, and consequently, increases its buoyancy. This is why sea water is more buoyant than fresh water. Therefore the ping pong is pushed more upwards by the water when salt is added than initially. This gives the string more tension.
Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
