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Harlamova29_29 [7]
3 years ago
12

"it was so cold yesterday that the temperature only reached 275." which temperature scale is being used? what would be the corre

sponding temperature on the other two scales?
Physics
2 answers:
kow [346]3 years ago
7 0
Remembering back to science I believe that temperature is on the Kelvin scale. The corresponding temperature on the Celsius scale would be very close to zero degrees which is the freezing point. The corresponding temperature on the Fahrenheit scale would be 32 degree which again is the freezing point.
Luba_88 [7]3 years ago
7 0
Answer is: Kelvin.
1) <span>The temperature </span>in degrees Celsius (°C) is equal to the temperature<span> in Kelvin (K) minus 273,15.
</span>T(°C)<span> = </span>T(K) - 273.15 = 275 - 273,15 = 1,85°C.
2) The temperature in degrees Fahrenheit (°F) is equal to the temperature in Kelvin (K) times 9/5, minus 459.67:
T(°F) = T(K) · 9/5 - 459,67 = 275 · 1,8 - 459,67 = 35,33°F.

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Read 2 more answers
Please solve the Problem.
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From the graph, the average acceleration between 5.0 s and 8.0 s is 2.33 ms^2.

<h3>What is acceleration?</h3>

The term acceleration has to do  with the change in velocity with time. We can see that the graph shown is a graph of velocity against time, the slope of the graph is the acceleration.

Thus, the average acceleration between 5.0 s and 8.0 s can be read off from the graph as 20 m/s - 13m//8.0 s - 5.0 s = 2.33 ms^2.

Learn more about acceleration:brainly.com/question/12550364

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8 0
2 years ago
A high-speed K0 meson is traveling at β = 0.90 when it decays into a π + and a π − meson. What are the greatest and least speeds
san4es73 [151]

Answer:

greatest speed=0.99c

least speed=0.283c

Explanation:

To solve this problem, we have to go to frame of center of mass.

Total available energy fo π + and π - mesons will be difference in their rest energy:

E_{0,K_{0} }-2E_{0,\pi }  =497Mev-2*139.5Mev\\

                       =218 Mev

now we have to assume that both meson have same kinetic energy so each will have K=109 Mev from following equation for kinetic energy we have,

K=(γ-1)E_{0,\pi }

K=E_{0,\pi}(\frac{1}{\sqrt{1-\beta ^{2} } } -1)\\\frac{1}\sqrt{1-\beta ^{2} }}=\frac{K}{E_{0,\pi}}+1\\   {1-\beta ^{2}=\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta ^{2}=1-\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta = +-\sqrt{\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\\\\beta =+-\sqrt{1-\frac{1}{(\frac{109Mev}{139.5Mev+1)^2}}

u'=+-0.283c

note +-=±

To find speed least and greatest speed of meson we would use relativistic velocity addition equations:

u=\frac{u'+v}{1+\frac{v}{c^{2} } } u'\\u_{max} =\frac{u'_{+} +v_{} }{1+\frac{v}{c^{2} } } u'_{+} \\u_{max} =\frac{0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } 0.828\\ u_{max} =0.99c\\u_{min} =\frac{u'_{-} +v_{} }{1+\frac{v}{c^{2} } } u'_{-}\\u_{min} =\frac{-0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } -0.828c\\u_{min} =0.283c

6 0
3 years ago
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