Answer:
a.After
second Mr Comer's speed

b.Distance travelled by Mr.Comer in
seconds

Explanation:
a. Lets recall our first equation of motion 
Now we know that
,
and

Plugging the values we have.




Then Mr.Comer's speed after
sec

b.
Lets find the distance and recall our third equation of motion.

So
distance covered.
Dividing both sides with 2a we have.

Plugging the values.


So Mr.Comer will travel a distance of
.
It should be S since S stands for students. Giving the team with more members the upper hand. Hope this helps.
Answer:
a.
Explanation:
The air moves faster across the top of the surface and exerts less pressure on the air below and thus due to difference in air pressure the top roof is lifted.
hence the correct option is pressure underneath them is reduced.
Answer:
a). M = 20.392 kg
b). am = 0.56
(block), aM = 0.28
(bucket)
Explanation:
a). We got N = mg cos θ,
f = 
= 
If the block is ready to slide,
T = mg sin θ + f
T = mg sin θ +
.....(i)
2T = Mg ..........(ii)
Putting (ii) in (i), we get



M = 20.392 kg
b).
.............(iii)
Here, l = total string length
Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so


.....................(iv)
We got, N = mg cos θ

∴ 
................(v)
Mg - 2T = M

(from equation (iv))
.....................(vi)
Putting (vi) in equation (v),

![$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$](https://tex.z-dn.net/?f=%24%5Cfrac%7Bg%5Cleft%5B%5Cfrac%7BM%7D%7B2%7D-m%20%5Csin%20%5Ctheta-%5Cmu_K%20m%20%5Ccos%20%5Ctheta%5Cright%5D%7D%7B%28%5Cfrac%7BM%7D%7B4%7D%2Bm%29%7D%3Da_m%24)
![$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$](https://tex.z-dn.net/?f=%24%5Cfrac%7B9.8%5Cleft%5B%5Cfrac%7B20.392%7D%7B2%7D-10%28%5Csin%2030%2B0.5%20%5Ccos%2030%29%5Cright%5D%7D%7B%28%5Cfrac%7B20.392%7D%7B4%7D%2B10%29%7D%3Da_m%24)

Using equation (iv), we get,

The answer is “increasing wavelength near beach”