Answer:
Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write
where
is the distance of the new object from the sun (orbital radius)
is the orbital period of the object
is the orbital radius of the Earth
is the orbital period the Earth
Solving the equation for , we find
![r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m](https://tex.z-dn.net/?f=r_o%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Br_e%5E3%7D%7BT_e%5E2%7DT_o%5E2%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.50%5Ccdot%2010%5E%7B11%7Dm%29%5E3%7D%7B%28365%20d%29%5E2%7D%28180%20d%29%5E2%7D%3D9.4%5Ccdot%2010%5E%7B10%7D%20m)
Explanation:
After some time t the current does not passing through the circuit
=>so the back emf is zero
=>here the inductor opposes decay of the circuit
- Ldi/dt = Ri
di/dt = - R/Li
di/i = - R/Ldt
now we applying the integration on both sides
log i=-R/Lt+C
here t=0=>i=io
Log io=C
=>Log i=-R/L*t + Log io
logi-Log io=-R/L*t
Log[i/io]=-R/L*t
i/io=e^-Rt/L
i=ioe^-Rt/L
the option D is correct
Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png
Solution:
We need to find the magnitude of the resultant on both x- and y-axis.
x-axis) The resultant on the x-axis is
in the positive direction.
y-axis) The resultant on the y-axis is
in the positive direction.
Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
from which we find
The valence electrons are the one furthest from the nucleus
