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Basile [38]
3 years ago
8

Water is not used as thermometric liquid.why?​

Physics
1 answer:
Dima020 [189]3 years ago
8 0

Answer:

Water cannot be used in thermometer because of its higher freezing point and lower boiling point than other liquids .  If water is used in a thermometer , it will start phase change at 0degree\\C and 100degreeC and will not measure temperature , out of this range . This range is very small as compared  to other liquids as mercury , having freezing point about −39degreeC and boiling point 356degreeC.

Explanation:

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Your best choice would be to youtube it probably kahn acdemy!
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Is the formula for velocity the same as speed or different?
a_sh-v [17]

Answer:

always same

Explanation:

velocity and speed are same upto some extend but velocity is vector while speed is scalar quantity

5 0
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A boy weighs 50 kg and is running with 225 J of energy, what is his velocity?
Oksanka [162]

0.22

Explanation:

50 ÷ 225 = 0.22

<u>#CarryOnLearning</u>

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A bus starts to move from rest. if its velocity becomes 90 km per hour after 8s second. calculate acceleraton​
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7 0
3 years ago
1. An electron in an atom absorbs a photon with an energy of 2.38 eV and jumps from the n = 2 to n = 4 energy level in the atom.
oksian1 [2.3K]

1. 1.0\cdot 10^{-6}m

First of all, let's convert the energy of the absorbed photon into Joules:

E=2.38 eV \cdot (1.6\cdot 10^{-19}J/eV)=1.98\cdot 10^{-19} J

The energy of the photon can be rewritten as:

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength of the photon

Re-arranging the formula, we can solve to find the wavelength of the absorbed photon:

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.98\cdot 10^{-19} J}=1.0\cdot 10^{-6}m

2. 1.24 eV

In this case, when the electron jumps from the n=4 level to the n=3 level, emits a photon with wavelength

\lambda=1.66\cdot 10^{-6}m

So the energy of the emitted photon is given by the formula used previously:

E=\frac{hc}{\lambda}

and using

\lambda=1.66\cdot 10^{-6}m

we find

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.0\cdot 10^{-6}m}=1.99\cdot 10^{-19}J

converting into electronvolts,

E=\frac{1.99\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=1.24eV

EDIT: an issue in Brainly does not allow me to add the last 2 parts of the solution - I have added them as an attachment to this post, check the figure in attachment.

7 0
3 years ago
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