Answer:
3.76 m/s
Explanation:
Instantaneous velocity: This can be defined as the velocity of an object in a non uniform motion. The S.I unit is m/s.
v' = dx(t)/dt..................... Equation 1
Where v' = instantaneous velocity, x = distance, t = time.
Given the expression,
x(t) = 28.0 m + (12.4 m/s)t - (0.0450 m/s³)t³
x(t) = 28 + 12.4t - 0.0450t³
Differentiating x(t) with respect to t.
dx(t)/dt = 12.4 - 0.135t²
dx(t)/dt = 12.4 - 0.135t²
When t = 8.00 s.
dx(t)/dt = 12.4 - 0.135(8)²
dx(t)/dt = 12.4 - 8.64
dx(t)/dt = 3.76 m/s.
Therefore,
v' = 3.76 m/s.
Hence, the instantaneous velocity = 3.76 m/s
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
,
and
, 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.
<span>The repelling of the support magnet decreases friction. is the answer you're looking for . :)
hope i helped - beanz</span>