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Naily [24]
3 years ago
7

Which one of the following indicates a drop in temperature? A. The alcohol in a thermometer expands. B. The brass on a bent bime

tallic strip is on the outside curve. C. The column of liquid in a thermometer moves up three degrees. D. A bimetallic strip bends so that the steel is on the outside curve.
Physics
2 answers:
solmaris [256]3 years ago
7 0

Answer:

A bimetallic strip bends so that the steel is on the outside curve.

Explanation:

A bimetallic strip is a combination of two different metals. The linear  coefficient of expansion of various metals are different, depending on their characteristics. For example, when a bimetallic strip of brass and steel are taken, thermal expansion coefficient of brass is more than that of steel.

That implies brass expands more than steel, for a given amount of heat supplied. So brass will curve outward since it becomes longer compared to steel. Steel will curve inwards.

Similarly, for cooling process where a drop in the temperature occurs, brass contracts more compared to steel and so Steel will be on outside curve.

Travka [436]3 years ago
3 0
D. A bimetallic strip bends so that the steel is on the outside curve.
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To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:

\rho_w =1 g/mL \rightarrow \text{Water Density}

\rho_o = 0.82g/mL \rightarrow \text{Object density}

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore

\sum F= 0

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The value of gravity is canceled because it is a constant

\frac{V_w}{V_o} = \frac{\rho_o}{\rho_w}

\frac{V_w}{V_o} = \frac{0.82}{1}

\frac{V_w}{V_o} = 0.82

The portion of the object that is submerged corresponds to 82%, while the portion that is visible, above the water level will be 18%

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1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position
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The value of the distance is \bf{14.52~cm}.

Explanation:

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v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)

where, \omega is the angular frequency, A is the amplitude of the oscillation and x is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., x = 0.

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\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)

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&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm

6 0
3 years ago
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Explanation:

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