Answer:
![1060.41kg/m^3](https://tex.z-dn.net/?f=1060.41kg%2Fm%5E3)
Explanation:
Bulk modulus is defined as the relative change in the volume of a body produced by a unit of compressive acting uniformly over its surface:
![B=\rho _o \frac{\bigtriangleup P }{\bigtriangleup \rho}](https://tex.z-dn.net/?f=B%3D%5Crho%20_o%20%5Cfrac%7B%5Cbigtriangleup%20P%20%7D%7B%5Cbigtriangleup%20%5Crho%7D)
Hence the density of the seawater at a depth of 680atm is calculated as:-
![\rho=\rho_o +\bigtriangleup \rho=\rho_o(1+\frac{\bigtriangleup P}{B})\\\\=1030 \times (1+ \frac{(680-1)\times10^5}{2.3\times 10^9})\\=1060.41kg/m^3](https://tex.z-dn.net/?f=%5Crho%3D%5Crho_o%20%2B%5Cbigtriangleup%20%5Crho%3D%5Crho_o%281%2B%5Cfrac%7B%5Cbigtriangleup%20P%7D%7BB%7D%29%5C%5C%5C%5C%3D1030%20%5Ctimes%20%281%2B%20%5Cfrac%7B%28680-1%29%5Ctimes10%5E5%7D%7B2.3%5Ctimes%2010%5E9%7D%29%5C%5C%3D1060.41kg%2Fm%5E3)
<span>During 1970s, same observations were seen as what we have observed today pertaining to our climate. Journals were discussing that there would be warming because of greenhouse gases emissions. Also, it was observed between the years 1970 to 1990 that there was a steady surface temperature increase. Due to this, people are now fixated with global warming rather than on global cooling.</span>
This is your perfect answer
The base unit for time is the second (the other SI units are: metre for length, kilogram for mass, ampere for electric current, kelvin for temperature, candela for luminous intensity, and mole for the amount of substance). The second can be abbreviated as s or sec.
Answer:
![2.1\times 10^{-12} c](https://tex.z-dn.net/?f=2.1%5Ctimes%2010%5E%7B-12%7D%20c)
Explanation:
We are given that
Surface area of membrane=![5.3\times 10^{-9} m^2](https://tex.z-dn.net/?f=5.3%5Ctimes%2010%5E%7B-9%7D%20m%5E2)
Thickness of membrane=![1.1\times 10^{-8} m](https://tex.z-dn.net/?f=1.1%5Ctimes%2010%5E%7B-8%7D%20m)
Assume that membrane behave like a parallel plate capacitor.
Dielectric constant=5.9
Potential difference between surfaces=85.9 mV
We have to find the charge resides on the outer surface of membrane.
Capacitance between parallel plate capacitor is given by
![C=\frac{k\epsilon_0 A}{d}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7Bk%5Cepsilon_0%20A%7D%7Bd%7D)
Substitute the values then we get
Capacitance between parallel plate capacitor=![\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}](https://tex.z-dn.net/?f=%5Cfrac%7B5.9%5Ctimes%208.85%5Ctimes%2010%5E%7B-12%7D%5Ctimes%205.3%5Ctimes%2010%5E%7B-9%7D%7D%7B1.1%5Ctimes%2010%5E%7B-8%7D%7D)
![C=0.25\times 10^{-12}F](https://tex.z-dn.net/?f=C%3D0.25%5Ctimes%2010%5E%7B-12%7DF)
V=![85.9 mV=85.9\times 10^{-3}](https://tex.z-dn.net/?f=85.9%20mV%3D85.9%5Ctimes%2010%5E%7B-3%7D)
![Q=CV](https://tex.z-dn.net/?f=Q%3DCV)
![Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c](https://tex.z-dn.net/?f=Q%3D0.25%5Ctimes%2010%5E%7B-12%7D%5Ctimes%2085.9%5Ctimes%2010%5E%7B3%7D%3D2.1%5Ctimes%2010%5E%7B-12%7D%20c)
Hence, the charge resides on the outer surface=![2.1\times 10^{-12} c](https://tex.z-dn.net/?f=2.1%5Ctimes%2010%5E%7B-12%7D%20c)