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Cloud [144]
3 years ago
6

What is the elevation at point A?

Physics
1 answer:
horsena [70]3 years ago
6 0

Answer:

i think that says 4800

its the same as the number at the bottom

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A 10 kg ball strikes a wall with a velocity of 3 m/s to the left. The ball bounces off with a velocity of 3 m/s to the right. If
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Answer:

The force is 272.73 newtons

Explanation:

We're going to use impulse-momentum theorem that states impulse is the change on the linear momentum this is:

\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i} (1)

Impulse is also defined as average force times the time the force is applied:

\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t) (2)

By (2) on (1):

\overrightarrow{F}_{avg}(\varDelta t)= \overrightarrow{p}_{f}-\overrightarrow{p}_{i}

solving for \overrightarrow{F}_{avg}:

\overrightarrow{F}_{avg}=\frac{\overrightarrow{p}_{f}-\overrightarrow{p}_{i}}{\varDelta t} (3)

We already know Δt is equal to 0.22 s, all we should do now is to find \overrightarrow{p}_{f}-\overrightarrow{p}_{i} and put on (3) (\overrightarrow{p_{i}} the initial momentum and \overrightarrow{p_{f}} the final momentum). Linear momentum is defined as \overrightarrow{p}=m\overrightarrow{v} , using that on (3):

\varDelta\overrightarrow{p}=m \overrightarrow{v_{f}}-m \overrightarrow{v_{i}} (4)

Velocity (v) are vectors so direction matters, if positive direction is the right direction and negative direction left \overrightarrow{v_{i}}=+3\, \frac{m}{s} and \overrightarrow{v_{f}}=-3\, \frac{m}{s} so (4) becomes:

\varDelta\overrightarrow{p}=m(-3\frac{m}{s}- (+3\frac{m}{s}))=-(10kg)(6\frac{m}{s})

\varDelta\overrightarrow{p}=-60\, \frac{mkg}{s} (5)

Using (5) on (3):

\overrightarrow{F}_{avg}=\frac{-60\, \frac{mkg}{s}}{0.22s}

F_{avg}=272.73N

8 0
4 years ago
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