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Andrew [12]
3 years ago
7

What is the energy transformation when a boy rides a bicycle​

Physics
1 answer:
castortr0y [4]3 years ago
8 0

Answer:

energy transform mechanical to motion

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This is the area where Earth's magnetic field is the strongest. *
Anna11 [10]

Answer:

1 is C : both poles

Explanation:

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4 years ago
In general, what is true about the frequency at which an object vibrates?.
zlopas [31]

In general s the wavelength of the vibration decreases, the frequency of the object increases and vice versa.

<h3>What is frequency?</h3>

This is the number of complete cycles per second made by a vibrating particle.

The frequency of a wave is calculated as follows;

v = f\lambda\\\\&#10;f = \frac{v}{\lambda}

where;

  • f is the frequency of the wave
  • v is the speed of the wave
  • λ is the wave length

Generally, as the wavelength of the vibration decreases, the frequency of the object increases and vice versa.

Learn more about frequency here: brainly.com/question/25699025

5 0
3 years ago
Resistance or opposite to a force refers to a
ryzh [129]

Answer:

Friction

Explanation:

7 0
4 years ago
Determine the acceleration of a boat<br> with a speed of 18 m/s for 20 seconds.
seropon [69]

Answer:

Explanation:

initial velocity, u = 0

final velocity, v = 18 m/s

time = 20 sec

acceleration, a = v - u / t

= 18 - 0 / 20

= 18 / 20

= 9 / 10

= 0.9 m/s^2

Hope this helps

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5 0
3 years ago
A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If
vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

  • The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
  • The traslational part can be written as follows:

       K_{trans} = \frac{1}{2}* M* v_{cm} ^{2}  (1)

  • The rotational part can be expressed as follows:

       K_{rot} = \frac{1}{2}* I* \omega ^{2}  (2)

  • where I = moment of Inertia regarding the axis of rotation.
  • ω = angular speed of the rotating object.
  • If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

       v = \omega * R (3)

  • For a solid cylinder, I = M*R²/2 (4)
  • Replacing (3) and (4)  in (2), we get:

       K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2}  (5)

  • Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

       K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2}  +\frac{1}{4}* M* v_{cmc}^{2}  =  \frac{3}{4}* M* v_{cmc} ^{2} (6)

  • Repeating the same steps for the spherical shell:

        I_{sph} = \frac{2}{3} * M* R^{2} (7)  

       K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2}  (8)

      K_{sph} = \frac{1}{2}* M* v_{cms} ^{2}  +\frac{1}{3}* M* v_{cms}^{2}  =  \frac{5}{6}* M* v_{cms} ^{2} (9)

  • Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
  • And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
  • Rearranging, and taking square roots on both sides, we get:

       \frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)

  • This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.
3 0
3 years ago
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