Answer:
Explanation:
Electric field due to a charge Q at a point d distance away is given by the expression
E = k Q / d , k is a constant equal to 9 x 10⁹
Field due to charge = 3 X 10⁻⁹ C
E = E = 
Field due to charge = 4 X 10⁻⁹ C
![E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}](https://tex.z-dn.net/?f=E%20%3D%20%5Btex%5D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes4%5Ctimes10%5E%7B-9%7D%7D%7B%282-d%29%5E2%7D)
These two fields will be equal and opposite to make net field zero
= ![[tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}](https://tex.z-dn.net/?f=%5Btex%5D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes4%5Ctimes10%5E%7B-9%7D%7D%7B%282-d%29%5E2%7D)
[/tex]
d = 0.928
Answer:
36.408cm3
Explanation:
Since we acknowledge that density is d= m/v, once we switch it up to maintain v as the number to be found it will change to v=m/d. Therefore, 275.32/7.562 is 36.408 and the unit is cm cube!
Hope that helped!!
<span>The answers are as follows:
(a) how many meters are there in 11.0 light-years?
11.0 light years ( 365 days / 1 year ) ( 24 h / 1 day ) ( 60 min / 1 h ) ( 60 s / 1 min ) ( 2.998x10^8 m/s ) = 1.04x10^17 m
(b) an astronomical unit (au) is the average distance from the sun to earth, 1.50 × 108 km. how many au are there in 11.0 light-years?
1.04x10^17 m ( 1 au / </span>1.50 × 10^8 km <span>) ( 1 km / 1000 m) = 693329.472 au
(c) what is the speed of light in au/h? au/h
</span>2.998 × 10^8 m/s ( 1 au / 1.50 × 10^8 km ) ( 1 km / 1000 m) ( 3600 s / 1 h ) = 7.1952 au/h
