Newton's law of cooling states that d x d t = − k ( x − a ) where x is the temperature, t is time, a is the ambient temperature, and k > 0 is a constant. suppose that a = a 0 cos ( ω t ) for some constants a 0 and ω . that is, the ambient temperature oscillates (for example night and day temperatures). a) find the general solution. b) in the long term, will the initial conditions make much of a difference? why or why not?

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft