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4 years ago

5

Ben franklin showed that 1 teaspoon of oil would cover about 0.50 acre of still water. if you know that 1.0 x 104m2 = 2.47 acres

, and that there are 5.0 cm3 in a teaspoon, what is the thickness (in cm) of a later of oil?

1 answer:

Dmitry_Shevchenko [17]4 years ago

3 0

To solve for the thickness of the latter of oil, we can simply divide the volume by the total surface area:

Thickness = Volume / Surface Area

We are both given the volume and the surface area, all we have to do now is to convert the units in like terms so that we can cancel those out.

Surface Area = 0.50 acres (1.0 x 10^4 m^2 / 2.47 acres)

Surface Area = 2,024.29 m^2

Further converting this into cm:

Surface Area = 2,024.29 m^2 (100 cm / m)^2

Surface Area = 20,242,914.98 cm^2

Therefore the thickness is:

Thickness = 5.0 cm^3 / 20,242,914.98 cm^2

Thickness = 2.47 x 10^-7 cm

or

<span>Thickness = 2.47 x 10^-9 m = 2.47 nm = 24.7 A</span>

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3 years ago

A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0

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3 years ago

A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu

Montano1993 [528]

Answer:

The angle of separation is $\Delta \theta = 0.93 ^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta _ i = 56^o$

The refractive index of violet light in diamond is $n_v = 2.46$

The refractive index of red light in diamond is $n_r = 2.41$

The wavelength of violet light is $\lambda _v = 400nm = 400*10^{-9}m$

The wavelength of red light is $\lambda _r = 700nm = 700*10^{-9}m$

Snell's Law can be represented mathematically as

$\frac{sin \theta_i}{sin \theta_r} = n$

Where $\theta_r$ is the angle of refraction

=> $sin \theta_r = \frac{sin \theta_i}{n}$

Now considering violet light

$sin \theta_r__{v}} = \frac{sin \theta_i}{n_v}$

substituting values

$sin \theta_r__{v}} = \frac{sin (56)}{2.46}$

$sin \theta_r__{v}} = 0.337$

$\theta_r__{v}} = sin ^{-1} (0.337)$

$\theta_r__{v}} = 19.69^o$

Now considering red light

$sin \theta_r__{R}} = \frac{sin \theta_i}{n_r}$

substituting values

$sin \theta_r__{R}} = \frac{sin (56)}{2.41}$

$sin \theta_r__{R}} = 0.344$

$\theta_r__{R}} = sin ^{-1} (0.344)$

$\theta_r__{R}} = 20.12^o$

The angle of separation between the red light and the violet light is mathematically evaluated as

$\Delta \theta = \theta_r__{R}} - \theta_r__{V}}$

substituting values

$\Delta \theta =20.12 - 19.69$

$\Delta \theta = 0.93 ^o$

6 0

4 years ago