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3 years ago

Which branch of physics deals with the study of force, energy, and motion?

Physics

2 answers:

Strike441 [17]3 years ago

8 0

The answer is mechanics

dexar [7]3 years ago

4 0

The correct answer is Mechanics

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20 pts.

RUDIKE [14]

The cup is acted upon by an unbalanced force which is the cars acceleration, but before it was an object at rest that stayed at rest. This jet propels their body forward.

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3 years ago

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why can scientists ignore the forces of attraction between particles in a gas under ordinary conditions?

Anna35 [415]

<span>The particles in a gas are apart and moving fast, so the forces of attraction are too weak to have a noticeable effect.</span>

5 0

3 years ago

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An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

8 0

3 years ago

Where is a divergent boundary most likely to be found?

Alex787 [66]

Answer:

mid-ocean oceanic ridges

Explanation:

3 0

2 years ago

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A golf ball is hit with a golf club. While the ball flies through the air, which forces act on the ball? Neglect air resistance.

34kurt

To solve this problem it is necessary to apply the equation related to the Gravitational Force, the equation describes that

$F = \frac{GMm}{r^2}$

Where,

G = Gravitational Universal Constant

M = Mass of Earth (or Bigger star)

m = Mass of Object (or smallest star)

r = Radius

From the statement we know that once the impact is made, the golf ball is subjected to the forces that are exerted in nature. Since the air resistance, which would represent the drag force, is ignored. Only the forces related to gravity remain.

The gravitational force carries 'pushes' or 'attracts' the body towards the earth, while the speed decreases as it reaches its maximum height.

When the ball has reached its maximum height only the force of gravity begins to act on it, generating the attraction to the earth in parabolic motion.

Therefore the correct answer is B.

6 0

3 years ago

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