Answer:
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Explanation:
The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:
(1)
Where:
- Impulse, in kilogram-meters per second.
- Mass, in kilograms.
- Initial velocity of the hockey park, in meters per second.
- Final velocity of the hockey park, in meters per second.
If we know that 
, ![\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right]](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D_%7B1%7D%20%3D%20-10%5C%2C%5Chat%7Bi%7D%5C%2C%5Cleft%5B%5Cfrac%7Bm%7D%7Bs%7D%5Cright%5D)
and ![\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right]](https://tex.z-dn.net/?f=%5Cvec%20%7Bv_%7B2%7D%7D%20%3D%2025%5C%2C%5Chat%7Bi%7D%5C%2C%5Cleft%5B%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%5D)
, then the impulse applied by the stick to the park is approximately:
![I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]](https://tex.z-dn.net/?f=I%20%3D%20%280.2%5C%2Ckg%29%5Ccdot%20%5Cleft%2835%5C%2C%5Chat%7Bi%7D%5Cright%29%5C%2C%5Cleft%5B%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%5D)
![I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]](https://tex.z-dn.net/?f=I%20%3D%207%5C%2C%5Chat%7Bi%7D%5C%2C%5Cleft%5B%5Cfrac%7Bkg%5Ccdot%20m%7D%7Bs%7D%20%5Cright%5D)
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Where they slide over each other.
Transform boundaries are formed or occur when two plates slide past each other in a sideways motion. They do not tear or crunch into each other (but the rock in between them may be ground up) and therefore none of the spectacular features are seen such as occur in divergent and convergent boundaries.
In a transform boundary, neither plate is added to at the boundary nor destroyed. They are marked in some places by features like stream beds that have been split in half and the two halves moved in opposite directions.
Answer:
Part A:
The proton has a smaller wavelength than the electron.
< 
Part B:
The proton has a smaller wavelength than the electron.
< 
Explanation:
The wavelength of each particle can be determined by means of the De Broglie equation.
(1)
Where h is the Planck's constant and p is the momentum.
(2)
Part A
Case for the electron:
But 
Case for the proton:
Hence, the proton has a smaller wavelength than the electron.
<em>Part B </em>
For part b, the wavelength of the electron and proton for that energy will be determined.
First, it is necessary to find the velocity associated to that kinetic energy:
(3)
Case for the electron:
but 
Then, equation 2 can be used:
Case for the proton :
But
Then, equation 2 can be used:
Hence, the proton has a smaller wavelength than the electron.
Work is closely related to energy. The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work.
So they are both closely related to each other.
HOPE THIS HELPS
