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notsponge [240]
3 years ago
13

the thermometer has its stem marked in millimeter instead of degree Celsius. the lower fixed point is 30mm and the upper fixed p

oint is 180mm. calculate the temperature in degrees Celsius when the thermometer reads 45mm​
Physics
1 answer:
Lorico [155]3 years ago
7 0

10°c

Explanation:

Given parameter;

Lower fixed point = 30mm

Upper fixed point = 180mm

Reading = 45mm

Unknown:

The degree celcuis temperature at 45mm = ?

Solution:

To solve this problem we simply compare the mm- scale to the celcius - scale that we know.

The upper fixed point is the boiling point of water

Lower fixed point is the freezing point of water

This shows that both the upper and lower fixed point of both thermometers are the same;

         mm-scale          °c scale

          180mm               100°c

            45mm                x

          30mm                  0°c

Solving;

                     \frac{x - 0}{100 -0} =  \frac{45 - 30 }{180- 30}

   x (150) = 100  x 15

   x = 10°c

learn more:

Temperature scales brainly.com/question/1603430

#learnwithBrainly

 

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Answer:

275.5 N

Explanation:

F_{1} = Force on one side of the door by first waiter = 257 N

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A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if
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Answer:

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b) If the mass is halved, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

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Explanation:

The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>

We have a block-spring system, whose angular frequency (\omega) is defined by the following formula:

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Where:

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T = \frac{2\pi}{\omega} (2)

By applying (1) in (2), we get the following formula:

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

a) If the mass is doubled, then the period is increased by \sqrt{2}. Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

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