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likoan [24]
3 years ago
15

The droplets on a glass of ice water are a result of water vapor releasing energy and changing to water droplets.

Physics
2 answers:
Anni [7]3 years ago
4 0
That is true.

In the solid state the molecules of water have the lowest energy content.

To pass from solid state to liquid stated water needs to gain some energy to increase vibration and motion of the molecules and reach a higher energy states.

To pass from liquid state to vapor (gaseous state) the molecules of water have to gain more energy to increase more the motion of the molecules and reach a more energetic stated.

Then you can think that when the water vapor becomes liquid (droplets) it needs to release some energy.

The droplets formed on a glass of ice water is an example of water condensation, one of the processes of the water cycle.

The droplets on the external surface of a cold glass are the result of condensation of water.

Condensation is the pass from vapor state to liquid state. As whe have stated, when water passes from vapor state to liquid state it has to release energy because liquid state is lower in energy than vapor (the molecules in water experiment slower motion than molecules of vapor).
Rina8888 [55]3 years ago
3 0

the answer is condensation

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Read 2 more answers
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati
Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

  • If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
  • Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
  • τ = F*r (2)
  • For a solid uniform disk, the rotational inertia regarding an axle passing through its center  is just I = m*r²/2 (3).
  • Replacing (2) and (3) in (1), we can solve for α, as follows:

       \alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)

  • Since the angular acceleration is constant, we can use the following kinematic equation:

        \omega_{f}^{2}  - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)

  • Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

       0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)

  • Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

       \omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)

  • Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

        v = \omega * r (8)

  • where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of  the disk, r becomes the radius of the disk, 0.200 m.
  • Replacing this value and (7) in (8), we get:

       v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)

b)    

  • There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

       a_{t} = \alpha * r (9)

  • where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
  • Replacing this value and (4), in (9), we get:

       a_{t}  = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)

  • Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
  • The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

       a_{c} = \omega^{2} * r  (11)

  • Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
  • Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
  • Replacing this value and (7) in (11) we get:

       a_{c} = \omega^{2} * r   = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)

  • The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
  • Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

       a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)

6 0
3 years ago
I need helpppppp asap
asambeis [7]
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If the north pole of a magnet in a compass is attracted to Earth's geographic north, what must be the polarity of Earth's magnet
nikitadnepr [17]
B south because north polarities line up with the opposite polarities
8 0
3 years ago
40 POINTS EASY
11111nata11111 [884]
We know, Mechanical Energy = K.E. + P.E.
As ball is at ground, P.E. would be zero. But as it is in motion, it must have some K.E. and that is:

K.E. = 1/2 mv²
K.E. = 1/2 * 1 * 2²
K.E. = 4/2
K.E. = 2 J

In short, Your Answer would be Option B

Hope this helps!
7 0
3 years ago
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