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3 years ago

In a cup of liquid water when would water molcules stop moving

2 answers:

6 0

Never. Molecules never stop moving; they only slow or speed up. They will only stop at absolute 0, which is impossible to get to,

Mandarinka [93]3 years ago

3 0

The the Water turns to ice. But even then they would never truly stop moving.<span />

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According to Kepler's Second Law the radius vector drawn from the Sun to a planet Multiple Choice is the same for all planets. s

Mademuasel [1]

Answer:

sweeps out equal areas in equal times.

Explanation:

As we know that there is no torque due to Sun on the planets revolving about the sun

so we will have

$\tau_{net} = 0$

now we have

$\frac{dL}{dt}= 0$

now we also know that

$Area = \frac{1}{2}r^2d\theta$

so rate of change in area is given as

$\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt}$

so we will have

$\frac{dA}{dt} = \frac{1}{2}r^2\omega$

$\frac{dA}{dt} = \frac{L}{2m}$

since angular momentum and mass is constant here so

all planets sweeps out equal areas in equal times.

4 0

3 years ago

A long. 1.0 kg rope hangs from a support that breaks, causing the rope to fall, if the pull exceeds 43 N. A student team has bui

raketka [301]

Answer:

6.8 m/s2

Explanation:

Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is

W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N

For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of

a = F/m = 13.6 / 2 = 6.8 m/s2

So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail

8 0

3 years ago

An object is initially moving at an unknown velocity. It accelerates at a rate of 2.0 m/s2 to a new velocity of 45 m/s in 10s. W

katrin2010 [14]

Answer:

Initial velocity=25m/s

Explanation:

acceleration(a)=2m/s^2

Final velocity(v)=45m/s

Time(t)=10seconds

Initial velocity(u)=?

V=u+axt

45=u+2x10

45=u+20

u=45-20

u=25m/s

6 0

3 years ago

1. When a particle moves in a circle with constant speed, its acceleration is

omeli [17]

Answer:

Its A.

Explanation:

8 0

3 years ago

During a goal-line stand, a 112-kg fullback moving eastward with a speed of 6 m/s

11111nata11111 [884]

Answer:

-1.24 m/s

Explanation:

Total momentum before collision = total momentum after collision

Total momentum before collision = (mass of full back * velocity of fullback) + (mass of lineman * velocity of line man).

Mass of full back = 112 kg, mass of line bag = 120 kg, velocity of full back 6 m/s (east), velocity of line back = -8 m/s (west). Hence:

Total momentum before collision = (112 * 6) + (120 * -8) = 672 - 960 = -288 kgm/s

The total momentum after collision = (mass of full back + mass of line back) * velocity after collision.

Let velocity after collision be v, hence:

The total momentum after collision = (112 + 120)v = 232v

Total momentum before collision = total momentum after collision

-288 = 232v

v = -288 / 232

v = -1.24 m/s

Therefore after collision, the two players would move at a velocity 1.24 m/s west (the same direction as the lineman).

7 0

3 years ago