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MariettaO [177]
2 years ago
12

Bicycle A with mass 40 Kg is traveling with a velocity of 4 m/s and Bicycle B with mass 20 Kg is traveling with a velocity of 2m

/s. The two bicycles have kinetic energy
Calculate it and state which bicycle has higher energy and why?
Physics
1 answer:
andrew-mc [135]2 years ago
7 0

Answer:

bicycle A has a greater K.E.

Explanation:

K.E = 1/2mv²

bicycle A = 1/2 × 40 × 4² = 320J

bicycle B = 1/2 × 20 × 2² = 100J

bicycle A has a greater K.E. because it has bigger mass and moves with faster velocity

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Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. Ass
Naddika [18.5K]

Answer:

L = 4.711 *10^{-6} kg m2/s

Explanation:

i =\frac{ml^3}{3}

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\omega = \frac{2\pi}{60}

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3 0
2 years ago
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he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?
Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}

With the following condition :

  • \sf{\omega} = angular frequency (rad/s)
  • \sf{\theta} = change of angle value (rad)
  • t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

  • \sf{\theta} = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
  • t = interval of the time = 54.9 s.

What was asked :

  • \sf{\omega} = angular frequency = ... rad/s

Step by step :

\sf{\omega = \frac{\theta}{t}}

\sf{\omega = \frac{2,000 \pi}{54.9}}

\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

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