Physics, 16.07.2020 17:01 fazegaze89

A spring is attached to the ceiling and pulled 9 cm down from equilibrium and released. The amplitude decreases by 7% each second. The spring oscillates 15 times each second. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

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Answer from: roseemariehunter12

$D(t) = 9(0.93)^t cos(30 \pi t)$

Explanation:

Amplitude begins at 9 cm, A_0 = 9 cm

The amplitude decreases by 7% (0.07) each second

The amplitude function can then be modeled as:

$A(t) = A_0(1 - 0.07)^t\\A(t) = 9(0.93)^t$

The spring oscillates 15 times each second, the period of oscillation (time to make 1 oscillation) will therefore be calculated as:

T = 1/15

$\frac{2\pi }{B} = \frac{1}{15} \\\\B = 30\pi$

The graphical equation of the system described is:

D(t) = A cos ( Bt - C) + D

Horizontal shift, C = 0

Vertical shift, D = 0

$D(t) = 9(0.93)^t cos(30 \pi t)$

Answer from: Quest

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm. the wire diameter is d = 0.9 mm. the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2) in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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A spring is attached to the ceiling and pulled 9 cm down from equilibrium and released. The amplitud...

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