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Svetradugi [14.3K]
3 years ago
15

A wheel with a weight of 388 N comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it i

s rotating at an angular velocity of 23.6 rad/s . The radius of the wheel is 0.591 m and its moment of inertia about its rotation axis is 0.800 MR2. Friction does work on the wheel as it rolls up the hill to a stop, at a height of h above the bottom of the hill; this work has a magnitude of 3482 J .
Calculate h. Use 9.81m/s2 for the acceleration due to gravity.
Physics
1 answer:
Alisiya [41]3 years ago
4 0

the correct answer is 14m

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Your bike starts with an initial velocity of 1 m/s and you accelerate to 4 m/s in 6 seconds. How far did you travel while accele
garri49 [273]

Answer:

78 m

Explanation:

Initial v = 1 m/s

a = 4 m/s²

t = 6s

x = v_{i} t + \frac{1}{2}at^{2} \\x = (1)(6) + \frac{1}{2} (4)(6)^{2} \\x = 6 + 72\\x = 78

5 0
3 years ago
g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the
MrRa [10]

Answer:

Approximately -1.58\; \rm rad \cdot s^{-2}.

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

v^2 - u^2 = 2\, a\, x,

where

  • v is the final velocity of the moving object,
  • u is the initial velocity of the moving object,
  • a is the (linear) acceleration of the moving object, and
  • x is the (linear) displacement of the object while its velocity changed from u to v.

The angular analogue of that equation will be:

(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta, where

  • \omega(\text{final}) and \omega(\text{initial}) are the initial and final angular velocity of the rotating object,
  • \alpha is the angular acceleration of the moving object, and
  • \theta is the angular displacement of the object while its angular velocity changed from \omega(\text{initial}) to \omega(\text{final}).

For this object:

  • \omega(\text{final}) = 0\; \rm rad\cdot s^{-1}, whereas
  • \omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}.

The question is asking for an angular acceleration with the unit \rm rad \cdot s^{-1}. However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}.

Rearrange the equation (\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta and solve for \alpha:

\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}.

7 0
3 years ago
Explain how that energy converts itself from POTENTIAL to KINETIC energy.​
Karolina [17]

Answer:

Potential energy can transfer into other forms of energy like kinetic energy. Kinetic energy is energy an object has because of its motion.

Explanation:

The ball was released, as the ball moves faster and faster toward the ground, the force of gravity will transfer the potential energy to kinetic energy.

8 0
3 years ago
HELP PLEASE STRUGGLING!
Assoli18 [71]

Answer:

To convert a milligram measurement to a kilogram measurement, divide the weight by the conversion ratio. The weight in kilograms is equal to the milligrams divided by 1,000,000.

Explanation:

5 0
2 years ago
Is water (H2O) a element? why or why not?
Dafna1 [17]
No it is not an element, it is a compound(two ore more elements bound together). It is one Oxygen and two Hydrogen bound together by covalent bonds.
8 0
3 years ago
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