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.
Even though the wind "tries" to flow from high pressure to low pressure, the turning of the Earth causes the air flow to turn to the right (in the Northern Hemisphere), so the jet stream flows around the air masses, rather than directly from one to the other.
From the given information in the question, the correct option is Option 1: 14 cm.
A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:
PE = 
k 
Where PE is the energy, k is the spring constant and x is extension.
i. Given that: PE = 10 J and x = 10 cm, then;
PE = k
10 = k 
20 = 100k
k = 0.2 J/cm
ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.
PE = k
20 = (0.2)
40 = 0.2
= 200
x = 
= 14.1421
x = 14.14 cm
So that;
x is approximately 14.00 cm.
Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.
For more information, check at: brainly.com/question/1352053.
Answer:
e. all of these
Explanation:
The fatigue strength is improved by then high alloy steels , high yield steels , high hardened steel , high ultimate steel .
Due to the formation of the improved materials in alloy steels will increase the fatigue strength . Similarly for a high yield steels and hardened steels these cycles to failure will improve .
