Rough surface offer more resistance as compare to smooth surface.
Explanation:
Friction is defined as the force which offer resistance to a moving object i.e. hold back the movement of sliding object. It always act in opposite direction.
When the two object come into contact of one another then they rub against each other thereby, generating frictional force. For example, when a cart roll along the blacktop parking lot then the rough surface of parking lot offer resistance to cart which slow down the speed of cart, whereas, when the same cart roll along the smooth floor inside the grocery store then it will move more freely due to the less resistance offered by the smooth vinyl tile inside the store.
The most reasonable conclusion of the above phenomenon is that water is a poor conductor of heat. Basically water is an insulator. The heat from surface to the bottom of the beaker will take a lot of time. Moreover, no convection current is formed so, heat might not even reach the bottom surface. Hydrogen bonding also play a vital role in determining the thermal properties of water.
I'm pretty sure the answer would be deductive reasoning. Deductive reasoning relies on making logical premises and basing a conclusion around those premises. It does not rely on making inferences and then assuming those inferences to be true.
Assuming no friction present, the work-energy theorem says that the work done on the rock by the only force acting on it (gravity force, aiming downward), is equal to the change of kinetic energy of the rock, at any point.
When it's at 15.0 m high, the work done by gravity on the rock can be written as follows:
The change in kinetic energy of the rock can be expressed in this way:
where m= 2.04 kg, vf = 25.0 m/s.
As (1) and (2) are equal each other, we can solve for v₀, as follows:
b)
When the rock arrives to its highest point, it will momentarily come to a stop, before changing direction to start falling.
So, the final kinetic energy, will be zero.
Applying the work --energy theorem again, we can write the following equation:
Replacing by v₀ = 30.3 m/s, rearranging and simplifying common terms, we can solve for hmax, as follows:
