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Answer:
To increase kinetic friction, the amount of fine water droplets sprayed before the game is limited.
To reduce kinetic friction. increase the amount of fine water droplets during pregame preparation and sweeping in front of the curling stones.
Explanation:
In curling sports, since the ice sheets are flat, the friction on the stone would be too high and the large smooth stone would not travel half as far. Thus controlling the amount of fine water droplets sprayed before the game is limited pregame is necessary to increase friction.
On the other hand, reducing ice kinetic friction involves two ways. The first way is adding bumps to the ice which is known as pebbling. Fine water droplets are sprayed onto the flat ice surface. These droplets freeze into small "pebbles", which the curling stones "ride" on as they slide down the ice. This increases contact pressure which lowers the friction of the stone with the ice. As a result, the stones travel farther, and curl less.
The second way to reduce the kinetic friction is sweeping in front of the large smooth stone. The sweeping action quickly heats and melts the pebbles on the ice leaving a film of water. This film reduces the friction between the stone and ice.
Answer:
The maximum speed of the car at the bottom of that drop is 26.34 m/s.
Explanation:
Given that,
The maximum vertical distance covered by the roller coaster, h = 35.4 m
We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :



v = 26.34 m/s
So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.
Answer:
Explanation:
Given that,
Spring constant = 16N/m
Extension of spring
x = 8cm = 0.08m
Mass
m = 5g =5/1000 = 0.005 kg
The ball will leave with a speed that makes its kinetic energy equal to the potential energy of the compressed spring.
So, Using conservation of energy
Energy in spring is converted to kinectic energy
So, Ux = K.E
Ux = ½ kx²
Then,
Ux = ½ × 16 × 0.08m²
Ux = 0.64 J
Since, K.E = Ux
K.E = 0.64 J
This question is in complete.The question is
A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.
Answer:
distance=0.124 m
Explanation:
