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Natali5045456 [20]
3 years ago
11

Fiora starts riding her bike at 18 mi/h. After a while, she slows down to 12 mi/h, and maintains that speed for the rest of the

trip. The whole trip of 69 miles takes her 4.5 hours. For how long did she travel at 18 mi/h?
Physics
1 answer:
lord [1]3 years ago
4 0

Answer:

t = 2.5 hours

Explanation:

given,

speed of the bike for t time= 18 mi/h

final speed of the bike after t time = 12 mi/h

total distance, D = 69 miles

total time, T= 4.5 hour

time for which speed of the bike is 18 mi/h = ?

we know distance = speed x time

now,

18 x t + 12 (4.5 - t) = 69

6 t + 54 = 69

6 t = 15

t = 2.5 hours

The bike was at the speed of 18 mi/h for 2.5 hours.

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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:

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  • 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.
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Answer:

True

Explanation:

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A bowling ball has a mass of 5 kg. A student rolls the bowling ball one time at 1 m/s and a second time and 2 m/s. Compare the m
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juin [17]

Answer:

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Explanation:

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From the question we have

PE = 30 × 9.8 × 1.5

We have the final answer as

<h3>441 J</h3>

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