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.
The pressure exerted by a fluid solely relies on the depth or height of the fluid, its density, and the gravitational constant. These three are related in the equation:
Pressure = density x g x height
In the problem, point A is within the block inside the tank. The water above the block is assumed to be 0.6 meters. This gives a point A pressure of:
P = 1000 kg/m^3 * 9.81 m/s^2 * 0.6 m = 5,886 Pa or 5.88KPa
12.8 rad
Explanation:
The angular displacement 
through which the wheel turned can be determined from the equation below:
(1)
where
Using these values, we can solve for from Eqn(1) as follows:
or
Explanation:
The solution is be found in the attachment.
Answer:
E = 16.464 J
Explanation:
Given that,
Mass of tetherball, m = 0.8 kg
It is hit by a child and rises 2.1 m above the ground, h = 21. m
We need to find the maximum gravitational potential energy of the ball. The formula for the gravitational potential energy is given by :
E = mgh
g is acceleration due to gravity
E = 0.8 kg × 9.8 m/s² × 2.1 m
= 16.464 J
So, the maximum potential energy of the ball is 16.464 J.
