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Bas_tet [7]
3 years ago
10

You can obtain a rough estimate of the size of a molecule by the following simple experiment. Let a droplet of oil spread out on

a smooth water surface. The resulting oil slick will be approximately one molecule thick. Given an oil droplet of mass of 7.61 × 10−7 kg and density of 912 kg/m3 that spreads out into a circle of radius 43.5 cm on the water surface, estimate the diameter of an oil molecule. Answer in units of m. Your answer must
Physics
1 answer:
Marat540 [252]3 years ago
6 0

Answer:

  • The diameter of the molecule of oil is 1.405 10 ^{-9} \ m

Explanation:

We define density as

\rho = \frac{mass}{volume}

So, the volume for our oil will be

volume = \frac{mass}{\rho}

volume = \frac{7.62 \ 10^{-7} \ kg}{ 912 \ \frac{kg}{m^3}  }

volume = \frac{7.62 \ 10^{-7} \ kg}{ 912 \ \frac{kg}{m^3}  }

volume = 8.355 \ 10 ^{-10} \ m^3

the volume for a cylinder with radius r and height h is

volume = \pi r^2 h

So, we can obtain the height of the droplet of oil as:

h = \frac{volume}{\pi r^2}

the radius is

r=43.5  \ cm = 0.435 \ m

h = 1.405 10 ^{-9} \ m^3

And this is the diameter of the oil molecule.

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Veseljchak [2.6K]
Given that a car is in the road, there is only movement in the x-direction. There is no movement in the y-direction.

Looking at the y-direction for the normal force:

F = N - mg
0 = N - mg, (no movement in y-dir.)
N = mg
N = (990)(9.8)
N = 9702 newtons

The normal force exerted on the car by the road is 9702 newtons.
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3 years ago
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What is the name of the process that is responsible for the formation of elements heavier than hydrogen? A) fusion B) fission C)
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5 0
3 years ago
Two objects (42.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h
Kazeer [188]

Answer:

a = 3.27 m/s²

T = 275 N

Explanation:

Given that:

Mass m₁ = 42.p0 kg

Mass m₂ = 21.0 kg

Consider both masses to be in a whole system, then:

The acceleration can be determined as:

(m_1+m_2)a = g(m_1-m_2)

Making acceleration the subject in the above formula;

a =\dfrac{g(m_1-m_2)}{(m_1+m_2)}

a =\dfrac{9.8(42.0-21.0)}{(42.0+21.0)}

a =\dfrac{9.8(21.0)}{(63.0)}

a =\dfrac{205.8}{(63.0)}

a = 3.27 m/s²

in the string, the tension is calculated using the formula:

T = \dfrac{2m_1m_2g}{(m_1+m_2)}

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T = 274.68 N

T ≅ 275 N

8 0
3 years ago
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