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Allisa [31]
3 years ago
11

As a rain storm passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury baromet

er drops by 21.6 mm from the normal height, what is the atmospheric pressure (in Pa)? Normal atmospheric pressure is 1.013 ✕ 105 Pa and the density of mercury is 13.6 g/cm3.
Physics
1 answer:
Pani-rosa [81]3 years ago
6 0

Answer:

new atmospheric pressure is 0.9838 × 10^{5}  Pa

Explanation:

given data

height = 21.6 mm = 0.0216 m

Normal atmospheric pressure = 1.013 ✕ 10^5 Pa

density of mercury = 13.6 g/cm³

to find out

atmospheric pressure

solution

we find first height of mercury when normal pressure that is

pressure p = ρ×g×h

put here value

1.013 × 10^{5}  = 13.6 × 10³ × 9.81 × h

h =  0.759 m

so change in height Δh = 0.759 - 0.0216

new height H = 0.7374 m

so new pressure = ρ×g×H

put here value

new pressure = 13.6 × 10³ × 9.81 × 0.7374

atmospheric pressure = 98380.9584

so new atmospheric pressure is 0.9838 × 10^{5}  Pa

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Determine the mass of the object below to the correct degree of precision.
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The correct degree of precision is 272.94 g.

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A spring with a spring constant of 25.1 N/m is attached to different masses, and the system is set in motion. What is its period
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1) 2.17 s

The period of a mass-spring oscillating system is given by

T=2 \pi \sqrt{\frac{m}{k}}

where k is the spring constant and m is the mass attached to the spring. In this problem, we have

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m = 3.0 kg

Substituting into the equation, we find

T=2 \pi \sqrt{\frac{3.0 kg}{25.1 N/m}}=2.17 s

2) 0.46 Hz

The frequency of the oscillating system is equal to the reciprocal of the period:

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Therefore, by substituting T=2.17 s, we find:

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3) 0.13 s

As before, the period of a mass-spring oscillating system is given by

T=2 \pi \sqrt{\frac{m}{k}}

where k is the spring constant and m is the mass attached to the spring. In this part of the problem, we have

k = 25.1 N/m

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Substituting into the equation, we find

T=2 \pi \sqrt{\frac{0.011 kg}{25.1 N/m}}=0.13 s

4) 7.69 Hz

The frequency of the oscillating system is equal to the reciprocal of the period:

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Therefore, by substituting T=0.13 s, we find:

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5) 1.59 s

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T=2 \pi \sqrt{\frac{m}{k}}

In this part of the problem, we have

k = 25.1 N/m

m = 1.6 kg

Substituting into the equation, we find

T=2 \pi \sqrt{\frac{1.6 kg}{25.1 N/m}}=1.59 s

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3 years ago
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