The variety of species in an ecosystem is known as biodiversity.
Answer is
9.773m/s^2
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Given,
h=8848m
The value of sea level is 9.08m/s^2. So, Let g′ be the acceleration due to the gravity on Mount Everest.
g′=g(1 − 2h/h)
=9.8(1 - 6400000/17696)
=9.8(1 − 0.00276)
9.8×0.99724
=9.773m/s^2
Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s^2
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hope this helps :)
Answer:
![C_{pb}=0.501\ kJ/kg.K](https://tex.z-dn.net/?f=C_%7Bpb%7D%3D0.501%5C%20kJ%2Fkg.K)
Explanation:
Given that
![m_1=0.35 kg](https://tex.z-dn.net/?f=m_1%3D0.35%20kg)
![T_1=-27.5^oC](https://tex.z-dn.net/?f=T_1%3D-27.5%5EoC)
![m_2=0.214 kg](https://tex.z-dn.net/?f=m_2%3D0.214%20kg)
![T_2=25^oC](https://tex.z-dn.net/?f=T_2%3D25%5EoC)
![T=16.4^oC](https://tex.z-dn.net/?f=T%3D16.4%5EoC)
We know that
![C_{pw}=4.187 kJ/kg.K](https://tex.z-dn.net/?f=C_%7Bpw%7D%3D4.187%20kJ%2Fkg.K)
By using energy conservation
Heat lost by water = Heat gain by block
![m_2\times C_{pw}\times (T_2-T)=m_1\times C_{pb}\times (T-T_1)](https://tex.z-dn.net/?f=m_2%5Ctimes%20C_%7Bpw%7D%5Ctimes%20%28T_2-T%29%3Dm_1%5Ctimes%20C_%7Bpb%7D%5Ctimes%20%28T-T_1%29)
![0.214\times 4.187\times (25-16.4)=0.35\times C_{pb}\times (16.4+27.5)](https://tex.z-dn.net/?f=0.214%5Ctimes%204.187%5Ctimes%20%2825-16.4%29%3D0.35%5Ctimes%20C_%7Bpb%7D%5Ctimes%20%2816.4%2B27.5%29)
![C_{pb}=0.501\ kJ/kg.K](https://tex.z-dn.net/?f=C_%7Bpb%7D%3D0.501%5C%20kJ%2Fkg.K)
Therefore the specific heat of the block will be 0.501 kJ/kg.K
Answer:
8.9*10^6 V/m
Explanation:
The expression for electric field strength E is given as
![E=\frac{V}{d}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BV%7D%7Bd%7D)
where V= voltage
d= distance of separation
Given data
![voltage= 84 mV= 84*10^-^3\\distance= 9.4*10^-^9](https://tex.z-dn.net/?f=voltage%3D%2084%20mV%3D%2084%2A10%5E-%5E3%5C%5Cdistance%3D%209.4%2A10%5E-%5E9)
substituting our given data into the electric field strength formula we have
![E= \frac{84*10^-^3}{ 9.4*10^-^9} \\\\E= \frac{84}{9.4} *10^-^3^-^(^-^9^)\\\\E= \frac{84}{9.4} *10^-^3^+^9\\\\E= \frac{84}{9.4} *10^6\\\\E=8.9*10^6 V/m](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B84%2A10%5E-%5E3%7D%7B%209.4%2A10%5E-%5E9%7D%20%5C%5C%5C%5CE%3D%20%5Cfrac%7B84%7D%7B9.4%7D%20%2A10%5E-%5E3%5E-%5E%28%5E-%5E9%5E%29%5C%5C%5C%5CE%3D%20%5Cfrac%7B84%7D%7B9.4%7D%20%2A10%5E-%5E3%5E%2B%5E9%5C%5C%5C%5CE%3D%20%5Cfrac%7B84%7D%7B9.4%7D%20%2A10%5E6%5C%5C%5C%5CE%3D8.9%2A10%5E6%20%20V%2Fm)
Answer:
where m < M is delivered to the space station. Soon after, the space station's orbit is adjusted so that it is 50 km ...