The density of seawater at a depth where the pressure is 500 atm is 
Explanation:
The relationship between bulk modulus and pressure is the following:
where
B is the bulk modulus
is the density at surface
is the variation of pressure
is the variation of density
In this problem, we have:
is the bulk modulus
is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)
Therefore, we can find the density of the water where the pressure is 500 atm as follows:
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Answer: R = 0.131 m
Explanation: The formulae for the distance between a fringe and the first (center) is given by
y = R×mλ/d
Where y = distance between first and nth fringe = 4mm = 4×10^-3m
λ = wavelength of light = 441.1nm = 441.1×10^-9m
R = distance between slits and screen =?
d = distance between slits = 0.29mm = 2.9×10^-4m
4×10^-3 = R ×2 ×441.4×10^-9/ 2.9×10^-5
Hence R = (4×10^-3) ×(2.9×10^-5)/2×441.4×10^-9
R = 1.16 × 10^-7/8.828×10^-7
R = 1.16/8.828
R = 0.131 m
Answer:
(A)chimney
Explanation:
bc all the smoke is going into the chimney
Answer:
Normalization is a systematic approach of decomposing tables to eliminate data redundancy. It is a multi-step process that puts data into tabular form, removing duplicated data from the relation tables.
Explanation:
Answer:
Work done W 2.938*10^9 J
Explanation:
given data:
mass m = 944 Kg
Mass of moon M = 7*10^22 Kg
Radius of the moon R = 1.5*10^6 m
gravitational constant G = 6.67*10^{-11} Nm^2/Kg^2
we know that work done is given as
Work done 
= 2.938*10^9 J
