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Bess [88]
2 years ago
13

What forces are equal and opposite sides

Physics
1 answer:
maria [59]2 years ago
5 0

Action and reaction forces from Newton's Third Law.

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Why is water considered a good solvent?
dem82 [27]

Answer:

D: Water is polar, so it can form attractions to and pull apart molecules of many types of substances.

Explanation:

Water is an excellent solvent simply because its polar water molecules usually form hydrogen bonds with ions and polar molecules, thereby allowing the ionic and polar covalent compounds to easily disperse easily in water.

Thus, the correct option is; D

7 0
3 years ago
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Which model shows how a comet's tail changes during its orbit?
pentagon [3]

Third model shows how a comet's tail changes during its orbit...

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When you turn brownie mix into brownies is that a physical change?
kotegsom [21]
No because for example you get a brownie mix and you want to make them well you put all of the mixture in the blender and you finish making them. Well now the brownies are done. well are they still the brownie mix?
5 0
3 years ago
5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc
ludmilkaskok [199]

Answer:

Δθ₁ =  172.5 rev

Δθ₁h =  43.1 rev

Δθ₂ =   920 rev

Δθ₂h = 690 rev

Explanation:

  • Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

       \Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2}  (1)  

  • Now, we need first to find the value of  the angular acceleration, that we can get from the following expression:

       \omega_{f1}  = \omega_{o} + \alpha * \Delta t  (2)

  • Since the machine starts from rest, ω₀ = 0.
  • We know the value of ωf₁ (the operating speed) in rev/min.
  • Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

       3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)

  • Replacing by the givens in (2):

       57.5 rev/sec = 0 + \alpha * 6 s  (4)

  • Solving for α:

       \alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)

  • Replacing (5) and Δt in (1), we get:

       \Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev  (6)

  • in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

       \Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev  (7)

  • In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

       \Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2}  (8)

  • First of all, we need to find the value of the angular acceleration during the second period.
  • We can use again (2) replacing by the givens:
  • ωf =0 (the machine finally comes to an stop)
  • ω₀ = ωf₁ = 57.5 rev/sec
  • Δt = 32 s

       0 = 57.5 rev/sec + \alpha * 32 s  (9)

  • Solving for α in (9), we get:

       \alpha_{2}  =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)

  • Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

        \Delta \theta_{2}  = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)

  • In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
  • \Delta \theta_{2h}  = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)
7 0
2 years ago
What system will break the chewed food down to a form your cell can use
horsena [70]
If I'm not mistaken it should be the digestive system due to the fact that our mouths and stomachs break down food and our intestines absorb any water and nutrients
4 0
3 years ago
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