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3 years ago

The Zero Gravity Research Facility at the NASA Glenn Research Center includes a

Physics

1 answer:

denis23 [38]3 years ago

4 0

Answer: (a) t = 5.44 sec

(b) vf = 53.31 m/s

Explanation: from the question, given data

the Height of the tower, h = 145m

from question

(a)

the initial velocity, v₁ = 0 m/s

s = v₁t + 1/2 gt²

-145 m = 0(t) + 1/2 (-9.8t²)

t² = 145/4.9

t² = 29.59

t = 5.44 sec

(b)

the speed of the sphere at the bottom of the tower is

vf² = vi² +2as

vf² = 0 + 2(-9.8 × -145)

vf² = 2842

vf = 53.31 m/s

(c)

when caught, the sphere experiences a deceleration of;

a = -29.0g

the time it would take to decelerate becomes;

vf = vi + at

0 = (53.31) + (-29 ×9.8)t

where t = 53.31 / 284.2

t = 0.1876 sec

∴ the distance travelled during the deceleration becomes;

vf² = vi² + 2as

s = (vf² - vi²) / 2a

s = (0 - 53.31²) / 2×-29×9.8

s = -2841.9561 / -568.4

s = 4.99 ≈ 5.0m

i hope this helps, cheers

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Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

$\Delta P \propto \rho \cdot \Delta h$

$\Delta P = k \cdot \rho \cdot \Delta h$

Where:

$\Delta P$ - Manometric pressure difference, measured in kilopascals.

$\rho$ - Fluid density, measured in kilograms per cubic meter.

$\Delta h$ - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

$\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}$

$\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}$

Where:

$\Delta h_{air}$ - Height difference of the air column, measured in meters.

$\Delta h_{Hg}$ - Height difference of the mercury column, measured in meters.

$\rho_{air}$ - Density of air, measured in kilograms per cubic meter.

$\rho_{Hg}$ - Density of mercury, measured in kilograms per cubic meter.

If $\Delta h_{Hg} = 0.015\,m$ , $\rho_{air} = 1.29\,\frac{kg}{m^{3}}$ and $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$ , the height difference of the air column is: