Answer:
Due to the fact that the comb is in terms of mass bigger than pieces of paper
Explanation:
In terms of mass, the comb is way too bigger than the piece of paper so it is virtually impossible for it to move towards the pieces of paper
According to Newton third law action and reaction are equal and opposite
Mathematically
P=m*v
Where
P= impulse is Newton
m=mass of the body
V= velocity of the body
Now since the comb has a greater mass it will not move to the paper, else the paper must have to move to it.
OF2 -
<span>O has 6 electrons in outer shell and F has 7 in its outer shell </span>
<span>Therefore, you have to account for 20 electrons total in the </span>
<span>structure (7+7+6 = 20) </span>
<span>therefore draw it linear first. F ---- O-----F </span>
<span>The two bonds take care of 4 electrons now you have to add another 16. </span>
<span>Therefore 3 lone pairs on each F and 2 lone pair on O. </span>
<span>If you check for formal charges, all the atoms are neutral </span>
<span>F will have 3 lone pairs + 1 bond = 7 electrons (bond = 1/2 electron for formal charge distribution) therefore both the F's are neutral </span>
<span>Now look at the O: it should have 6.. it has two lone pair and 2 bonds = 4 electrons and 2 bonds = 1 electron each = 2 electrons from bonds = 6 total electrons for formal charge which is exactly the # it should have. There is no need for any double bond in this as there are no charges to be separated. </span>
<span>Now if u look at the # of domains around O you will see if you include the lone pairs it has a sp3 hybridization (4 domains) therefore a tetrahedron which has 2 lone pairs and 2 bonds.. since there are two lone pairs, the lone pair/bond pair repulsion is so high it is going to repel the two Fluorines and form a bent structure, looks a lot like H2O. </span>
The answer is: A compound can be separated by physical means.
Answer:
a) v = 54.7m/s
b) v = (58 - 1.66a) m/s
c) t = 69.9 s
d) v = -58.0 m/s
Explanation:
Given;
The height equation of the arrow;
H = 58t - 0.83t^2
(a) Find the velocity of the arrow after two seconds. m/s;
The velocity of the arrow v can be given as dH/dt, the change in height per unit time.
v = dH/dt = 58 - 2(0.83t) ......1
At t = 2 seconds
v = dH/dt = 58 - 2(0.83×2)
v = 54.7m/s
(b) Find the velocity of the arrow when t = a. m/s
Substituting t = a into equation 1
v = 58 - 2(0.83×a)
v = (58 - 1.66a) m/s
(c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s
the time when H = 0
Substituting H = 0, we have;
H = 58t - 0.83t^2 = 0
0.83t^2 = 58t
0.83t = 58
t = 58/0.83
t = 69.9 s
(d) With what velocity will the arrow hit the surface? m/s
from equation 1;
v = dH/dt = 58 - 2(0.83t)
Substituting t = 69.9s
v = 58 - 2(0.83×69.9)
v = -58.0 m/s
