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3 years ago

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3. Atmospheric pressure is often given in units of millimeters of mercury. This refers to the height of a mercury column above t

he mercury's surface at the base of a barometer . The force exerted by the atmosphere on the surface of the mercury in the reservoir equals the weight of the mercury in the column . If the mercury column extends 760 mm above the mercury's surface in the reservoir , what is the atmosphere's pressure over the mercury ? Use 13.6*10^ 3 kg/m^ 3 for the density of mercury

1 answer:

gregori [183]3 years ago

6 0

Answer:

101397.16 pa

Explanation:

The pressure recorded will be equal to pgh

Where p = density of mercury = 13.6x10^3 kg/m^ 3

g = acceleration due to gravity 9.81 m/s^2

h = height of mercury in the column = 760 mm = 760x10^-3 m

Pressure = 13.6x10^3 x 9.81 x 760x10^-3 = 101397.16 pa

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The height is $h_c = 42.857$

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From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

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