Answer:
L' = 555.95 lb
Explanation:
Analyzing the given conditions in the question, we get
The safe load, L is directly proportional to width (w) and square of depth (d²)
also,
L is inversely proportional length (l) i.e L = k/l
combining the above conditions, we get an equation as:
L = k(wd²/l)
now, for the first case we have been given
w = 3 in
d = 6 in
l = 11 ft
L = 1213 lbs
thus,
1213 lb = k ((3 × 6²)/11)
or
k = 123.54 lbs/(ft.in³)
Now,
Using the calculated value of k to calculate the value of L in the second case
in the second case, we have
w = 6 in
d =3 in
l = 12 ft
Final Safe load L' = 123.54 × (6 × 3²/12)
or
L' = 555.95 lb
Given that a car is in the road, there is only movement in the x-direction. There is no movement in the y-direction.
Looking at the y-direction for the normal force:
F = N - mg
0 = N - mg, (no movement in y-dir.)
N = mg
N = (990)(9.8)
N = 9702 newtons
The normal force exerted on the car by the road is 9702 newtons.
<span>The jump from 1966 to 16347 is the largest one or simply we can say it is hard to lose the 3rd electron.Whereas, it is relatively easy to lose the first two electrons.
So there will be only 2 electrons in the outer most shell.
According to the information mentioned above we can conclude the </span><span>unknown element likely belongs to the second group.
</span><span>I2 = 1752 kj/mol</span>
I think through convection I forget. Plus its enclosed (and, well closed)
