A steel ball weighing 128 pounds is suspended from a spring. This stretches the spring 12865 feet. The ball is started in motion from the equilibrium position with a downward velocity of 8 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second) . Suppose that after t seconds the ball is y feet below its rest position. Find y in terms of t. (Note that this means that the positive direction for y is down.)

Answer and Step-by-step explanation::

Downward force mg = upward force kδg

Spring stiffness k = Load/extension = 128/(128/257) = 257 lb/ft

At instant t when spring extension from equilibrium position is y, downward force on the ball is weight mg and upward forces are (k(y+δ))g and air resistance.

But air resistance = (4v)g. Hence, by Newton's second law, my'' = mg - k(y+δ)g - 4vg

Further, v = dy/dt = y' and mg = kδg

So, my'' = -kyg - 4y'g

or, (m/g)y'' + 4y' + ky = 0,

Boundary conditions: y(0) = 0, y'(0) = 8 ft/s

Characteristic equation is (m/g)r^2 + 4r + k = 0

r = (g/2m)[-4 ± √(16-4mk/g)]

r = (32.2/2*128)[-4 ± √(16-4*128*257/32.2)] = (-0.503125 ± 8.024 i)

Roots are imaginary

So, y = e^(-0.503125*t) [C1*Cos (8.024*t) + C2*Sin (8.024*t)]

Putting y(0) = 0 we get, C1 = 0.

So, y' = C2*e^(-0.503125*t)* [8.024*Cos (8.024*t) + Sin(8.024*t)*(-0.503125)]

Putting y'(0) = 8 ft/s, we get C2 = 0.9969

Thus, y = 0.9969*e^(-0.503125*t) Sin (8.024*t)

i think it's ф or x=6,-6

step-by-step explanation:

-x^2=-36

x^2=-36

ф

or

if you mean by : -(x^2)=-36

it'll be: x^2=36

x=6,x=-6

hope it'll ! have a good day!

you for asking!

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐