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lorasvet [3.4K]
3 years ago
12

1.) an object moves along the x axis, subject to the potential energy shown. The object has a mass of 1.1kg and starts at rest a

t point A. (a) what is the object’s speed at point B? (b) at point C? (c) at point D? (d) what are the turning points for this object?
2.) a 1250-kg car drives up a hill that is 16.2m high. During the drive, two nonconservative forces do work on the car: (i) the force of friction, and (ii) the force generated by the car’s engine. The work done by friction is -3.11x10^5J; the work done by the engine is +6.44x10^5J. find the change in the car’s kinetic energy from the bottom of the hill to the top of the hill.
3.) starting at rest at the edge of a swimming pool, a 72.0-kg athlete swims along the surface of the water and reaches a speed of 1.20m/s by doing the work Wncl=+161J. find the nonconservative work, wnc2, done by the water on the athlete.
4.) starting at rest at the edge of a swimming pool, a 72.0-kg athlete swims along the surface of the water and reaches a speed of 1.20m/s by doing the work Wncl=+161J. find the nonconservative work, wnc2, done by the water on the athlete.
Physics
1 answer:
MrRa [10]3 years ago
4 0
If I'm not wrong #1 should be C
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A vehicle travels a distance of 300km. It took the vehicle 6 hours to make the trip.
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Answer:

50 kmph

Explanation:

300 / 6 = 50

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What would be the magnitude of a
Pavlova-9 [17]

The magnitude of a velocity vector is 60 m/s.

To find the magnitude, the given values are,

x - components = 36 m/s

y - components = 48 m/s

<h3>What is magnitude?</h3>
  • The term magnitude can be defined as “ quantity ”.
  • For instance, the magnitude can be used for describing about the comparison of speeds.
  • It can also be used to explain the distance travelled by an object or to explain the amount of an object in terms of its magnitude.
  • Magnitude of the velocity vector is total value which is square root of the value.

The magnitude of the vector can be,

| a | = √ ( x² + y² )

Substituting the values of x component and y component given,

      = √ ( 36² + 48² )

      = √ ( 1296 + 2304)

      = √ 3600

      = 60 m/s

The magnitude is found to be 60 m/s.

Hence, Option B is the correct answer.

Learn more about magnitude,

brainly.com/question/14452091

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7 0
1 year ago
There is no difference between the question and hypothesis steps of scientific inquiry true or false
WARRIOR [948]
The answer would be false
7 0
3 years ago
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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}.

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A because I heard it is
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