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iren2701 [21]
3 years ago
9

Why does surface water disperse when touched with soap?

Physics
2 answers:
aivan3 [116]3 years ago
5 0

The end of the detergent molecule which attaches to fat (grease) repels water molecules. It is known as hydrophobic, meaning "water fearing." By attempting to move away from the water molecules, the hydrophobic ends of the detergent molecules push up to the surface. This weakens the hydrogen bonds holding the water molecules together at the surface. The result is a break in the surface tension of the water.

Doss [256]3 years ago
4 0

Answer:

In a soap-and-water solution the hydrophobic (greasy) ends of the soap molecule do not want to be in the liquid at all. Those that find their way to the surface squeeze their way between the surface water molecules, pushing their hydrophobic ends out of the water. This separates the water molecules from each other.

Explanation:

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Skeletal muscles must have help from the bone to be able to move.
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can cuclin concentration durng mitosis be explained by the fact that the cell divides in two and thus divides the material in th
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Can the change in cyclin concentration during mitosis be explained by the fact that the cell divides in two and thus divides the material in the cell into two smaller volumes?

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3 years ago
A challenge gaining prominence throughout the planet is the increased need for \"green\" or sustainable energy. in certain parts
Inga [223]

32.6% of 1.4 Mw

= (0.326) x (1,400,000 joules/second)

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1 year

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(456,000 joules/sec) x (31,536,000 sec)  =  1.438 x 10¹³ Joules 

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5 0
4 years ago
Read 2 more answers
Lawerence kohl berg developed the theory of
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7 0
2 years ago
When a 0.058 kg tennis ball is served, it accelerates from rest to a speed of 45.0 m/s. The impact with the racket gives the bal
AURORKA [14]

Answer:

Explanation:

Note : the questions do not come in the right order.

Data:

Mass = 0.0518kg

Velocity = 45 m / s

Distance (s) = 0.44m

C) what's the acceleration on the ball?

Using equation of motion,

V² = u² + 2as

V² - u² = 2as

a = (v² - u²) / 2s

a = (45² - 0) / 2 * 0.44 [the ball was at rest]

a = 2025 / 0.88

a = 2301.136m/s²

D) The net force on the ball?

Force = mass * acceleration

F = m*a

F = 0.0518 * 2301.136

F = 119.199N

The force acting on the ball was 133.465N

F = 133.47N

b) time period the ball was struck.

From the relationship between impulse and momentum,

Ft = m * v

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t = 2.61 / 133.47

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V = 22.50m/s.

e) if the ball was heavier and still experienced the same velocity, the applied would've been lesser than before.

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4 years ago
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