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geniusboy [140]
3 years ago
8

Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hourmiles/hour

in 2.22 ss starting from rest. Assuming that they have constant acceleration throughout that time, find their acceleration in meters per second squared. Ente
Physics
1 answer:
slavikrds [6]3 years ago
5 0

Answer:

The acceleration of the cheetahs is 10.1 m/s²

Explanation:

Hi there!

The equation of velocity of an object moving along a straight line with constant acceleration is the following:

v = v0 + a · t

Where:

v = velocity of the object at time t.

v0 = initial velocity.

a = acceleration.

t = time

We know that at t = 2.22 s, v = 50.0 mi/h. The initial velocity, v0, is zero.

Let's convert mi/h into m/s:

50.0 mi/h · (1609.3 m / 1 mi) · (1 h / 3600 s) = 22.4 m/s

Then, using the equation:

v = v0 + a · t

22.4 m/s = 0 m/s + a · 2.22 s

Solving for a:

22.4 m/s / 2.22 s = a

a = 10.1 m/s²

The acceleration of the cheetahs is 10.1 m/s²

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Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 2.88 kg and rotate with
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Answer:

(a) K_{small}=4839.3J

(b) K_{larger}=17406.4J

Explanation:

Given data

The angular velocity of two cylinders ω=257 rad/s

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The radius of larger cylinder r₂=0.605 m

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Where I is rotational of inertia of solid cylinder about its central axis.

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K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr^2)w^2

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K_{small}=\frac{1}{4}(2.88kg)(0.319)^2(257rad/s)^2 \\K_{small}=4839.3J

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K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr_{2}^2)w^2

Substitute the given values

K_{larger}=\frac{1}{4}mr_{2}^2w^2\\ K_{larger}=\frac{1}{4}(2.88kg)(0.605m)^2(257rad/s)^2\\ K_{larger}=17406.4J

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