Answer:
I found the experience tasking
Explanation:
I wouldn't say it was hard, neither was it easy. I'd rather go for something like it being tasking. It's worthy of note that it was my first time, and I think it's very normal especially when one hasn't been doing something of that nature previously. Of course I did my draft, which unsurprisingly happened to be not good enough, and I had to look for templates to guide me through the acceptable way.
I still did it in my own way, but in the right way. Ever since then though, I have never stuttered when writing application letters, as it had since then seem inborn
Answer:3H2O + light-c3h603+302
Explanation:
Answer:
Magnitude is 25.8 m/s and direction is 
Explanation:
From the law of conservation of linear momentum
where 
and 
are masses of bullet and stones respectively, 
and 
are the initial velocities of bullet and stone respectively, 
and 
are the final velocities of bullet and stone respectively
Substituting 6g=0.006 Kg for mass of bullet, 0.1Kg for mass of stone, 350 m/s for initial velocity of bullet and 250 m/s as final velocity for stone but initial velocity of stone is zero
Substituting 21 for 
and 15 for 
To find direction
Therefore, magnitude is 25.8 m/s and direction is
Answer:
a) distance d = 293.36ft
b) acceleration a = 14.67ft/s^2
Explanation:
Acceleration is the change in velocity per unit time.
a = ∆v/t ....1
Given;
Initial velocity vi = 30mph × 5280ft/mile × 1/3600s/h
vi = 44ft/s
Final velocity vf = 70mph × 5280ft/mile × 1/3600s/h
vf = 102.67ft/s
time = 4.0s
From equation 1, acceleration is;
a = ∆v/t = (102.67-44)/4 = 14.67ft/s^2
Distance travelled can be given as;
d = ut + 0.5at^2 .....2
u = 44ft/s
t = 4
a = 14.67ft/s^2
Substituting into the equation 2
d = 44(4) + 0.5(14.67×4^2)
d = 293.36ft
Answer:
s (l = 0) only two electrons
p (l = 1) sublevel six (6) electrons fit
d (l = 2) sub-level has 10 electrons
f (l = 3) has 7 has 14 electrons
Explanation:
The electrons have an odd spin, so they must comply with the Pauli principle, therefore only two electrons fit in a given sublevel.
Sub level s (l = 0) only two electrons
Sub level p (l = 1) has three orbitals in which two electrons fit, therefore in the sublevel six (6) electrons fit
Sub-level d (l = 2) has 5 orbitals each with two electrons, the sub-level has 10 electors
Sub-level f (l = 3) has 7 orbitals and two electrons each, the sub-level has 14 electrons
