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stiv31 [10]
3 years ago
9

A temperature scale has an upper fixed point of 260mm and the lower fixed point of 50mm.what will be the reading on this scale w

hen the thermometer read 125°F​
Physics
1 answer:
joja [24]3 years ago
7 0

Answer:

the reading on the scale is 158.5 mm.

Explanation:

Given;

upper fixed point of the temperature scale, x₁ = 260 mm

lower  fixed point of the temperature scale, x₂ = 50 mm

upper temperature scale, T₁ = 212 °F

lower temperature scale, T₂ = 32 °F

thermometer reading, t = 125 °F, let the reading on the scale = x

Interpolate as follows to determine the value of "x"

\frac{X_1 - X}{X_1-X_2} = \frac{T_1-t}{T_1-T_2} \\\\\frac{260-X}{260-50} = \frac{212-125}{212-32} \\\\\frac{260 -X}{210} =\frac{87}{180} \\\\180(260-X) = 87(210)\\\\46800 -180 X = 18270 \\\\180X = 46800-18270\\\\180X = 28530\\\\X = \frac{28530}{180} \\\\X = 158.5 \ mm

Therefore, the reading on the scale is 158.5 mm.

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Which source is considered a primary source of scientific information? an original research article in a scientific journal a re
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The primary source would be the original article published in a scientific journal. All other choices would be based on information from the original article.

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This questic concerns a tablet that
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Answer:

The remaining percentage of drug concentration is about 88.7% 2 years after manufacture.

Explanation:

Recall the formula for the decay of a substance at an initial N_0 concentration at manufacture:

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where k is the decay rate (in our case 0.06/year), and t is the elapsed time in years. Therefore, after 2 years since manufacture we have:

N=N_0\,e^{-0.06\,\,(2)}\\N=N_0\,e^{-0.12}\\N/N_0=e^{-0.12}\\N/N_0=0.8869

This in percent form is 88.7 %. That is, the remaining percentage of drug concentration is about 88.7% 2 years after manufacture.

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A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ
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The ocean is 6485.6m deep when measured from the vessel

Explanation:

v=1474m/s

t=8.88s

let d represent distance from the vessel to the ocean bottom.

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3 years ago
A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.
Luba_88 [7]

Answer:

The final equilibrium temperature of the system is T = 12.48^oC

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice M_{i} = 129 g = 0.129 kg

Mass of the steam M_s = 19 g = 0.019 kg

Initial temperature is  T_i = 0°C

Temperature of  steam  T_s = 100°C

Following the change of state of water in the question

 The energy required by ice to change to water is mathematically given as

          Q_A = M_iL_f

Where L_f is a constant known as heat of fusion  and the value is 334*10^3 J/kg

           Q_A = 0.129 *334 *10^3  = 43086 J

The energy been released when the steam changes to water is mathematically given as

            Q_B = M_s * L_v

           Where L_v is a constant known as heat of vaporization and the value is 2256*10^3J/kg

           Q_B = 0.019 * 2256*10^3 = 42864J

         The energy released when the temperature of water decrease from 100°C to 0°C is

                 Q_C = M_s *C_water (100°C)

Where C_{water} is the specific heat of water which has a value 4186J/kg \cdot K

                  Q_C = 0.019 *4186*100 = 7953.4

Looking at the values we obtained we noticed that ]

             Q_B + Q_C > Q_A

What this means is that the ice will melt

bearing in mind the conservation of energy

     looking the way at which water at different temperature were mixed according to the question

     Heat lossed by the vapor   = heat gained by ice

        Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T

                                               T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}

                                               T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}

                                              T = 12.48^oC

       

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