A.kinetic energy for it an example of wood.
<span>A physical change is any change that alters the form or appearance of a substance but does not change it into another substance
-Some of the physical properties may be altered, but the chemical composition stays the same
-Examples: Bending, Crushing, Cutting, Melting, Freezing, Boiling</span>
Question
Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles travelled in a straight line and some were deflected at different angles.
Which statement best describes what Rutherford concluded from the motion of the particles?
A) Some particles travelled through empty spaces between atoms and some particles were deflected by electrons.
B) Some particles travelled through empty parts of the atom and some particles were deflected by electrons.
C) Some particles travelled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms.
D) Some particles travelled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.
Answer:
The right answer is C)
Explanation:
In the experiment described above, a piece of gold foil was hit with alpha particles, which have a positive charge. Alpha particles <em>α</em> were used because, if the nucleus was positive, then it would deflect the positive particles. The principles of physics posit that electric charges of the same orientation repel.
So most as expected some of the alpha particles went right through meaning that the gold atoms comprised mostly empty space except the areas that were with a dense population of positive charges. This area became known as the "nucleus".
Due to the presence of the positive charges in the nucleus, some particles had their paths bent at large angles others were deflected backwards.
Cheers!
Answer:
x = 17.88[m]
Explanation:
We can find the components of the initial velocity:
![(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]](https://tex.z-dn.net/?f=%28v_%7Bx%7D%29_%7Bo%7D%20%20%3D%2013.3%2Acos%2841.5%29%3D9.96%5Bm%2Fs%5D%5C%5C%28v_%7By%7D%29_%7Bo%7D%20%20%3D%2013.3%2Asin%2841.5%29%3D8.81%5Bm%2Fs%5D)
We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.
g = - 9.81[m/s^2]
Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.
![y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]](https://tex.z-dn.net/?f=y%3Dy_%7Bo%7D%20%2B%28v_%7By%7D%20%29_%7Bo%7D%20%2At-0.5%2Ag%2A%28t%29%5E%7B2%7D%20%5C%5C0%3D1.9%2B%288.81%2At%29-%284.905%2At%5E%7B2%7D%29%5C%5C-1.9%3D8.81%2At%2A%281-0.5567%2At%29%5C%5Ct%3D0%5C%5Ct%3D1.796%5Bs%5D)
With this time we can calculate the horizontal distance:
![x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]](https://tex.z-dn.net/?f=x%3D%28v_%7Bx%7D%29_%7Bo%7D%20%2At%5C%5Cx%3D9.96%2A1.796%5C%5Cx%3D17.88%5Bm%5D)