The Law of Conservation of Energy
The answer is to this question D
Answer:v=3.28 m/s
Explanation:
Given
mass of rock 
diameter of circle 
radius 
At highest Point

At highest Point N=0 because mass is just balanced by centripetal Force
thus 




Answer:
B. A collision scene
Explanation:
It could have been a parade ceremony, but, if you notice the vehicle's hazard lights or an emergency vehicle ahead, it is common sense to figure that they is a collision scene nearby.
Answer:
Therefore the ratio of diameter of the copper to that of the tungsten is

Explanation:
Resistance: Resistance is defined to the ratio of voltage to the electricity.
The resistance of a wire is
- directly proportional to its length i.e

- inversely proportional to its cross section area i.e

Therefore

ρ is the resistivity.
The unit of resistance is ohm (Ω).
The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m
The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m
For copper:


......(1)
Again for tungsten:

........(2)
Given that
and 
Dividing the equation (1) and (2)

[since
and
]



Therefore the ratio of diameter of the copper to that of the tungsten is
