Over time, the types of technology can vary and be improved upon so that more advanced techniques become more valued. This could be the situation with mining whereby back in the 1500's in underground mines the rock was broken by fire setting ie lighting a fire below the rock face to heat up the rock and then throwing cold water on it to crack it, so that it could be dug by hand. With the advent of explosives, this all changed so that the rock could be blasted. The increase in advance rates for an underground heading have thus gone from 5-20 feet per month to up to 300meters (984 ft) per month for a 24/7 mining operation, which is a huge improvement.
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.
Answer:
Explanation:
a ) V = 3 cos(0.5t)
differentiating with respect to t
dv /dt = -3 x .5 sin0.5t
= -1.5 sin0.5t.
acceleration = - 1.5 sin 0.5t
when t = 3 s
acceleration = - 1.5 sin 1.5
= - 1.496 ms⁻²
v = 3 cos.5t
b ) dx/dt = 3 cos 0.5 t
dx = 3 cos 0.5 t dt
integrating on both sides
x = 3 sin .5t / .5
x = 6 sin0.5t
At t = 2 s
x = 6 sin 1
x = 5.05 m
Answer: Temperature and humidity are the two characteristics used to classify air masses.
Explanation:
