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Alex
3 years ago
11

A glass vessel that can be repeatedly filled with precisely the same volume of liquid is called a pycnometer. A certain pycnomet

er, when empty and dry, weighed 25.296 g. When filled with water at 25 oC the pycnometer and water weighed 34.914 g. When filled with a liquid of unknown composition the pycnometer and its contents weighed 33.485 g. At 25 oC the density of water is 0.9970 g/ml. What is the density of the unknown liquid?
Physics
1 answer:
White raven [17]3 years ago
8 0

Answer:

density of liquid 0.848 g/ml

Explanation:

from the information given in the question

mass of water = 34.914 - 25.296 = 9.618 g

volume of pycnometer = volume of water

which will be equal to = \frac{ mass}{density}

= \frac{9.618}{0.9970} = 9.646 ml

mass of liquid =33.485-25.296 = 8.189 ml

density of liquid= \frac{mass}{volum\ of\ liquid}

                           = \frac{8.189}{9.646} =0.848 g/ml

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Answer:

liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

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Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be
Sedaia [141]

Answer:

The period will change a 0,036 % relative to its initial state

Explanation:

When the rod expands by heat its moment of inertia increases, but since there was no applied rotational force to the pendulum , the angular momentum remains constant. In other words:

ζ= Δ(Iω)/Δt, where ζ is the applied torque, I is moment of inertia, ω is angular velocity and t is time.

since there was no torque ( no rotational force applied)

ζ=0 → Δ(Iω)=0 → I₂ω₂ -I₁ω₁ = 0 → I₁ω₁ = I₂ω₂

thus

I₂/I₁ =ω₁/ω₂ , (2) represents final state and (1) initial state

we know also that ω=2π/T , where T is the period of the pendulum

I₂/I₁ =ω₁/ω₂ = (2π/T₁)/(2π/T₂)= T₂/T₁

Therefore to calculate the change in the period we have to calculate the moments of inertia. Looking at tables, can be found that the moment of inertia of a rod that rotates around an end is

I = 1/3 ML²

Therefore since the mass M is the same before and after the expansion

I₁ = 1/3 ML₁² , I₂ = 1/3 ML₂²  → I₂/I₁ = (1/3 ML₂²)/(1/3 ML₁²)= L₂²/L₁²= (L₂/L₁)²

since

L₂= L₁ (1+αΔT) , L₂/L₁=1+αΔT  , where ΔT is the change in temperature

now putting all together

T₂/T₁=I₂/I₁=(L₂/L₁)² = (1+αΔT) ²

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At what position or positions on the x-axis is the electric field zero?
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The electric field vector due to charge one

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Put the value into the formula

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