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Tatiana [17]
3 years ago
11

Can y’all please help

Physics
1 answer:
TiliK225 [7]3 years ago
5 0

Answer:

<em>The cat's speed when it slid off the table was 2.53 m/s</em>

Explanation:

<u>Horizontal Motion</u>

When an object is thrown horizontally at a speed v, from a height h, it describes a curved path ruled by gravity, until it hits the ground.

the range or maximum horizontal distance traveled by the object can be calculated as follows:

\displaystyle d=v\cdot\sqrt{\frac  {2h}{g}}

The situation described in the problem fits into the concept of a horizontal launch when the cat slides off the table and gets to the ground some distance ahead.

It's given the range as d=1.3m and the height h=1.3 m, thus we can find the cat's speed by solving the equation for v:

\displaystyle v=d\cdot\sqrt{\frac  {g}{2h}}

Where g=9.81\ m/s^2

Substituting values:

\displaystyle v=1.3\cdot\sqrt{\frac  {9.81}{2\cdot 1.3}}

\displaystyle v=1.3\cdot\sqrt{\frac  {9.81}{2.6}}

Calculating:

v = 2.53 m/s

The cat's speed when it slid off the table was 2.53 m/s

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Answer:

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Explanation:

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Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

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