Answer:the
8/9 h
Explanation:
Height = 1/2 a T^2 now change to T/3
now height = 1/2 a (T/3)^2 =<u> 1/9</u> 1/2 a T^2 <===== it is 1/9 of the way down or 8/9 h
1)a 2)D 3)a. I think the answers are
Answer:
Dx = -0.5
Dy = -0.25
Explanation:
Two vectors are given in rectangular components form as follows:
A = i + 6j
B = 3i - 7j
It is also given that:
A - B - 4D = 0
so, we solve this to find D vector:
(i + 6j) - (3i - 7j) - 4D = 0
- 2i - j = 4D
D = - (2/4)i - (1/4)j
D = - (1/2)i - (1/4)j
<u>D = - 0.5i - 0.25j</u>
Therefore,
<u>Dx = -0.5</u>
<u>Dy = -0.25</u>
Work done = mgh where h varies from 0 (at the bottom of the pool) to 5 ft (at the sides of the pool).
m= density*volume = 62.5*pi*24^2*d/4 = 9000πd lbs
h = 5-d (the particles at the top of the pool require less work as required to those at the bottom).
Then,
WD (E) = 9000πd*9.81*(5-d) = 88290πd(5-d) = 441450πd-88290πd^2
Integrating from 0-4 with respect to depth of pool, d
Total E = {441450π*d^2/2 - 88290π*d^3/} for d between 0 and 4 ft
E=220725π*d^2 -29430π*d^3
Substituting for d=0 ft and d=4 ft;
= {220725*π*4^2-29430π*4^3} - {220725π*0^2-29430*pi*2*0^3)
=5177596.021 J= 5.17 MJ