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3 years ago

Times will the kinetic one

Physics

1 answer:

mixer [17]3 years ago

3 0

Answer:

The new kinetic energy is 9 times of the initial kinetic energy.

Explanation:

The kinetic energy of an object is given by the formula as follows :

$E=\dfrac{1}{2}mv^2$

Where

m is mass and v is speed of the body

If the speed of the body is tripled. Let v' is the new speed, v' = 3v. The new kinetic energy is E' and it is given by :

$E'=\dfrac{1}{2}mv'^2\\\\\text{Put v' = 3v}\\\\E'=\dfrac{1}{2}m\times (3v)^2\\\\=\dfrac{1}{2}m\times 9v^2\\\\=9\times (\dfrac{1}{2}mv^2)\\\\E'=9\times E$

Hence, the new kinetic energy is 9 times of the initial kinetic energy.

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(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

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(4) The tension in the rope is 1.47 N.

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where;

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Torque equation for frictional force;

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solve (1) and (2)

$a = \frac{TR(R - r)}{I + MR^2}$

since the yoyo is pulled in vertical direction, T = mg $a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2$

<h3>Angular acceleration of the yoyo</h3>

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

<h3>Weight of the yoyo</h3>

W = mg

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<h3>Tension in the rope </h3>

T = mg = 1.47 N

<h3>Angular speed of the yoyo </h3>

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

Learn more about angular speed here: brainly.com/question/6860269

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