2 questions that I am just curious about!!! WILL NAME YOU BRAINLIEST!!! Hope you can figure out!! Worth 11 points!!
1 answer:
You might be interested in
Answer:
- 4 times
Explanation:
Since the equation for the illumination of an object, i.e. the brightness of the light, is <em>inversely proportional to the square of the distance from the light source</em>, the form of the function is:
- f(x) = k.x⁻²
Where x is the distance between the object and the light force, k is the constant of proportionality, and f(x) is the brightness.
Then, if you move halfway to the lamp the new distance is x/2 and the new brightness (call if F) is :
Then, you have found that the light is 4 times as bright as it originally was.
We can use the law of conservation of energy to solve the problem.
The total mechanical energy of the system at any moment of the motion is:
where U is the potential energy and K the kinetic energy.
At the beginning of the motion, the ball starts from the ground so its altitude is h=0 and therefore its potential energy U is zero. So, the mechanical energy is just kinetic energy:
When the ball reaches the maximum altitude of its flight, it starts to go down again, so its speed at that moment is zero: v=0. So, its kinetic energy at the top is zero. So the total mechanical energy is just potential energy:
But the mechanical energy must be conserved, Ef=Ei, so we have
and so, the potential energy at the top of the flight is
The total mechanical energy of the system at any moment of the motion is:
where U is the potential energy and K the kinetic energy.
At the beginning of the motion, the ball starts from the ground so its altitude is h=0 and therefore its potential energy U is zero. So, the mechanical energy is just kinetic energy:
When the ball reaches the maximum altitude of its flight, it starts to go down again, so its speed at that moment is zero: v=0. So, its kinetic energy at the top is zero. So the total mechanical energy is just potential energy:
But the mechanical energy must be conserved, Ef=Ei, so we have
and so, the potential energy at the top of the flight is