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4 years ago

What happens to the molecules in water as it freezes?

Physics

2 answers:

Keith_Richards [23]4 years ago

6 0

The molecules give off heat.

Nostrana [21]4 years ago

3 0

I believe the answer would be A, because the water molecules slow down so much that the attractions can be be put into the hexagonal pattern

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A pipe of length 6.8 m is closed at one end and sustains a standing wave at its second overtone. determine the distance between

r-ruslan [8.4K]

For a pipe closed at one end, the second overtone is the 3rd natural frequency as shown in the diagram.

For the fundamental frequency,
λ = 4L
where
λ = wavelength
L = the length of the tube

For the second overtone (3rd frequency)
λ = (4/3)L

Because L = 6.8 m, therefore
λ = (4/3)*(6.8 m) = 9.0667 m

The distance between a node and an adjacent antinode is λ/2, which is
9.0667/2 = 4.533 m

Answer: 4.5 m

8 0

3 years ago

Which of the following is the water cycle process where the extra water that plants release is evaporated from their leaves?

ddd [48]

B evaporation since the water is going up

6 0

3 years ago

HELP PLEASEE ILL GIVE BRAINLIEST

ad-work [718]

If you start at 100% and divide it by 2 it will equal 50% then divide by 2 again and the answer will be 25%. Your answer should be 25%

7 0

3 years ago

Alan starts from his home and walks 1.3 km east to the library. He walks an additional 0.68 km east to a music store. From there

zepelin [54]

Answer: final Displacement = 0 km, total distance covered =7 km

Explanation:

Given the following :

From home to library = 1.3 km East

Library to music store = 0.68km East

Music store to friend's house = 1.1km North

Friend's house to grocery store = 0.42 km North

Displacement is the net change in distance traveled.

Eastward distance :

To = (1.3 + 0.68)km = 1.98km East

Fro = (0.68 + 1.3)km = 1.98 km East

Δ distance = (1.98 - 1.98) = 0

Northward direction:

To = (1.1 + 0.42)km = 1.52km north

Fro = (0.42 + 1.1)km = 1.52km North

Δ distance = (1.98 - 1.98) = 0

Hence final Displacement = 0

Total distance covered = 2 × (1.3 + 0.68 + 1.1 + 0.42) = 2 × 3.5

= 7km

3 0

4 years ago

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Yo

insens350 [35]

Full Question

Consider two point charges located on the x axis: one charge, Q1 = -12.0 nC , is located at x1 = -1.705m ; the second charge, Q2 = 36.5 nC, is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Your answer may be positive or negative, depending on the direction of the force.

Answer:

6.79E6 N

Explanation:

Given

Q1 is negative and to the left of Q3 the force will be to the left

Q2 is positive and to the right of Q3 the the force will also be to the left

Net Force is calculated as:

Using Coulomb's law

Coulomb's law: F = kqQ / r²

the constant k = 8.99 x 10^9 N m2 / C2

F = -kQ1*Q3/(r1)² -kQ2*Q3/(r2)²)

F = -kQ3(Q1/(r1)² + Q2/(r2)²)

Where

Q1 = -12nC = -12 * 10^-9C

Q2 = 36.5nC = 36.5 * 10^-9C

Q3 = 49.5nC = 49.5 * 10^-9C

x1 = -1.705m

x2 = x = 0

x3 = -1.170m

r1 = x3 - x1

r1 = -1.170 - -1.705

r1 = -1.170 + 1.705

r1 = 0.535

r1² = 0.286225

r2 = x3

r2 = -1.170

r2² = -1.170²

r2² = 1.3689

So,

F = (-8.99 * 10^9)(49.5 *10^-9) [-12 * 10^-9/0.286225 + 36.5 * 10^-9/1.3689]

F = -445.005 (−4.192505895711E−8 + 2.6663744612462E−8)

F = -445.005 * −1.5261314344648E−8

F = -(8.99 * 10^9) * (49.5 * 10^-9) * [ (-12 * 10^-9) /(-1.770 - -1.705)² + (36.5 * 10^-9)/(-1.170)²]

F = -445.005( −0.000002813572941777538)

F = 0.00000679136118994008324

F = 6.79E6 N

3 0

3 years ago