A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, causing it to reverse its direction and giving it a velocity of +25 m/s the impulse the stick applies to the park is most nearly

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:

[tex]I = m\cdot (\vec{v}_{2} - \vec{v_{1}})[/tex] (1)

Where:

[tex]I[/tex] - Impulse, in kilogram-meters per second.

[tex]m[/tex] - Mass, in kilograms.

[tex]\vec{v_{1}}[/tex] - Initial velocity of the hockey park, in meters per second.

[tex]\vec{v_{2}}[/tex] - Final velocity of the hockey park, in meters per second.

If we know that [tex]m = 0.2\,kg[/tex], [tex]\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right][/tex] and [tex]\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right][/tex], then the impulse applied by the stick to the park is approximately:

[tex]I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right][/tex]

[tex]I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right][/tex]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

A nucleus with mass number 229 emits a 3.443 MeV alpha particle. Calculate the disintegration energy Q for this process, taking the recoil energy of the residual nucleus into account.

Which object will take the most force
to accelerate? *
4 kg
6 kg
8 kg
02 kg