Using the precise speed of light in a vacuum (
![299,792,458 \ \frac{m}{s}](https://tex.z-dn.net/?f=299%2C792%2C458%20%5C%20%5Cfrac%7Bm%7D%7Bs%7D%20)
), and your given distance of
![1.152 * 10^{8} km](https://tex.z-dn.net/?f=1.152%20%2A%2010%5E%7B8%7D%20km)
, we can convert and cancel units to find the answer. The distance in m, using
![\frac{1000 \ m}{1 \ km}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1000%20%5C%20m%7D%7B1%20%5C%20km%7D%20)
, is
![1.152 * 10^{11} m](https://tex.z-dn.net/?f=1.152%20%2A%2010%5E%7B11%7D%20m)
. Next, for the speed of light, we convert from s to min, using
![\frac{1 \ min}{60 \ s}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%20%5C%20min%7D%7B60%20%5C%20s%7D%20)
, so we divide the speed of light by 60. Finally, dividing the distance between the Sun and Venus by the speed of light in km per min, we find that it is
6.405 min.
Answer:
Explanation: the temperature range is at 1,400
Answer:
The distance between the two spheres is 914.41 X 10³ m
Explanation:
Given;
4 X 10¹³ electrons, and its equivalent in coulomb's is calculated as follows;
1 e = 1.602 X 10⁻¹⁹ C
4 X 10¹³ e = 4 X 10¹³ X 1.602 X 10⁻¹⁹ C = 6.408 X 10⁻⁶ C
V = Ed
where;
V is the electrical potential energy between two spheres, J
E is the electric field potential between the two spheres N/C
d is the distance between two charged bodies, m
![V = \frac{K*q}{d^2}*d = \frac{K*q}{d}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7BK%2Aq%7D%7Bd%5E2%7D%2Ad%20%3D%20%5Cfrac%7BK%2Aq%7D%7Bd%7D)
where;
K is coulomb's constant = 8.99 X 10⁹ Nm²/C²
d = (8.99 X 10⁹ X 6.408 X 10⁻⁶)/0.063
d = 914.41 X 10³ m
Therefore, the distance between the two spheres is 914.41 X 10³ m
Answer:
23932242.5 Pa
Explanation:
= Atmospheric pressure = ![1.013\times 10^5\ Pa](https://tex.z-dn.net/?f=1.013%5Ctimes%2010%5E5%5C%20Pa)
= Pressure of seawater
= Density of sea water = ![1.025\times 10^3\ kg/m^3](https://tex.z-dn.net/?f=1.025%5Ctimes%2010%5E3%5C%20kg%2Fm%5E3)
h = Depth of shipwreck = ![2.37\times 10^3\ m](https://tex.z-dn.net/?f=2.37%5Ctimes%2010%5E3%5C%20m)
g = Acceleration due to gravity = 9.81 m/s²
The absolute pressure is given by
![P_{ab}=P_a+P_w\\\Rightarrow P_{ab}=1.013\times 10^5+1.025\times 10^3\times 9.81\times 2.37\times 10^3\\\Rightarrow P_{ab}=23932242.5\ Pa](https://tex.z-dn.net/?f=P_%7Bab%7D%3DP_a%2BP_w%5C%5C%5CRightarrow%20P_%7Bab%7D%3D1.013%5Ctimes%2010%5E5%2B1.025%5Ctimes%2010%5E3%5Ctimes%209.81%5Ctimes%202.37%5Ctimes%2010%5E3%5C%5C%5CRightarrow%20P_%7Bab%7D%3D23932242.5%5C%20Pa)
The absolute pressure at the depth of the shipwreck is 23932242.5 Pa
oxygen's atomic number is 8
1s22s22p4 is the electron configuration
2+2=4 4+4=8