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Inga [223]
2 years ago
11

I need help in physics, pls hurry! I will give Brainliest for correct answers!!

Physics
1 answer:
Nana76 [90]2 years ago
6 0
Top left: slowing down
Top right: not moving
Bottom left: moving at a constant speed
Bottom right: speeding up
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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
lord [1]

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
A ball is thrown vertically upward with a speed of 1.86
Inessa05 [86]

Answer:

t = 1.09 s.

Explanation:

This is a one-dimensional kinematics question, so the equations of kinematics will be sufficient to solve the question.

y-y_0 = v_0t + \frac{1}{2}at^2\\0 - 3.82 = 1.86t +\frac{1}{2}(-9.8)t^2\\-3.82 = 1.86t - \frac{1}{2}9.8t^2\\4.9t^2 - 1.86t - 3.82 = 0

This quadratic equation can be solved using determinant.

\Delta = b^2 - 4ac\\t_{1,2} = \frac{-b \pm \sqrt{\Delta} }{2a}\\t_1 = 1.09~s\\t_2 = -0.71~s

Of course, we will choose the positive time.

4 0
3 years ago
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