Answer:
The potential energy is transformed into kinetic energy
Explanation:
This particular case is defined as the principle of energy conservation since energy is not created or destroyed only transforms. When you have potential energy it can be transformed into kinetic energy or vice versa. In this problem, we have the case of a ball that sits on a desk and then falls to the ground. In this way the ground will be taken as a reference point, this is a point at which the potential energy will be equal to zero in such a way that when the ball is on the desktop that is above the reference line its potential energy will be maximum. As the ball drops its potential energy decreases, as the height relative to the ground (reference point) decreases. In contrast its kinetic energy increases and increases as it approaches the ground. So when it hits the ground it will have maximum kinetic energy and will be equal to the potential energy for when the ball was on the desk.
Therefore:
![E_{p} = potential energy [J] = E_{k} = kinetic energy [J]where:\\E_{p} =m*g*h\\m =mass [kg]\\g=gravity[m/s^2]\\h=elevation[m]\\E_{k} = \frac{1}{2} *m*v^{2} \\where:\\v=velocity [m/s]\\\frac{1}{2} *m*v^{2} = m*g*h](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3D%20potential%20energy%20%5BJ%5D%20%3D%20E_%7Bk%7D%20%3D%20kinetic%20energy%20%5BJ%5Dwhere%3A%5C%5CE_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5Cm%20%3Dmass%20%5Bkg%5D%5C%5Cg%3Dgravity%5Bm%2Fs%5E2%5D%5C%5Ch%3Delevation%5Bm%5D%5C%5CE_%7Bk%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cv%3Dvelocity%20%5Bm%2Fs%5D%5C%5C%5Cfrac%7B1%7D%7B2%7D%20%20%2Am%2Av%5E%7B2%7D%20%3D%20m%2Ag%2Ah)
In optics, a diffraction grating is an optical component with a periodic structure, which splits and diffracts light into several beams travelling in different directions.
Answer:
- <u>The water ballon that was thrown straight down at 2.00 m/s hits the ground first, 0.19 s before the other ballon.</u>
Explanation:
The motions of the two water ballons are ruled by the kinematic equations:
We are only interested in the vertical motion, so that equation is all what you need.
<u>1. Water ballon is thrown horizontally at sped 2.00 m/s.</u>
The time the ballon takes to hit the ground is independent of the horizontal speed.
Since 2.00 m/s is a horizontal speed, you take the initial vertical speed equal to 0.
Then:
<u>2. Water ballon thrown straight down at 2.00 m/s</u>
Now the initial vertical speed is 2.00 m/s down. So, the equation is:
To solve the equation you can use the quadratic formula.
You get two times. One of the times is negative, thus it does not have physical meaning.
<u>3. Conclusion:</u>
The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.
The momentum of a fast object compared to that of a slow object even if they both have the same mass, is their velocities.
Having same mass but different velocities results in different momentum.
Example: mass = 10kg
Velocity 1 = 50 Velocity 2 = 100
Momentum 1 = 10×50 = 500 Ns
Momentum 2 = 10×100 = 1000 Ns
Hope it helped!
Answer:
t = (ti)ln(Ai/At)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Explanation:
Let Ai represent the initial amount and At represent the final amount of beryllium-11 remaining after time t
At = Ai/2^n ..... 1
Where n is the number of half-life that have passed.
n = t/half-life
Half life = 14
n = t/14
At = Ai/2^(t/14)
From equation 1.
2^n = Ai/At
Taking the natural logarithm of both sides;
nln(2) = ln(Ai/At)
n = ln(Ai/At)/ln(2)
Since n = t/14
t/14 = ln(Ai/At)/ln(2)
t = 14ln(Ai/At)/ln(2)
Ai = 800
At = 50
t = 14ln(800/50)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Let half life = ti
t = (ti)ln(Ai/At)/ln(2)
