Given :
Mass of water, m = 2 grams.
The temperature of water drops from 31 °C to 29 °C .
The specific heat of water is 4.184 J/(g • °C).
To Find :
Amount of heat lost in this process.
Solution :
We know, heat lost is given by :
Therefore, amount of heat lost in this process is 16.736 J.
The diagram is showing a 3d model of an atom, with all of the electrons demonstrated in a rounded shape, which resembles a cloud, thus being called an electron cloud.
This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3
Answer:
270 μA
Explanation:
Use the magnetic field due to long, straight wire and solve for current I.
plug in the values
= 2.7×10^{-4)×10^6
=270 μA
The current that flows in the heart is 270 μA
