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vivado [14]
3 years ago
10

xFormula1" title="(\frac{z}{4} +30) + (\frac{z}{2} ) = 180" alt="(\frac{z}{4} +30) + (\frac{z}{2} ) = 180" align="absmiddle" class="latex-formula"> show work, please
Mathematics
1 answer:
Dovator [93]3 years ago
4 0

Answer:

  x = 200

Step-by-step explanation:

Multiply by 4:

  x + 120 + 2x = 720

  3x = 600 . . . . . . collect terms, subtract 120

  x = 200 . . . . . . . divide by 3

_____

<em>Check</em>

  (200/4 +30) +(200/2) = 180

  (50 +30) + 100 = 180

  80 + 100 = 180 . . . . . true

_____ _____

<em>Alternate solution</em>

If you like, you can simply work with the equation given.

  (3/4)x + 30 = 180 . . . . collect terms

  (3/4)x = 150 . . . . . . . . . subtract 30

  x = 200 . . . . . . . . . . . . multiply by 4/3

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A mass of 3.25 kg is attached to the end of a spring that is stretched 22 cm by a force of 15 N. It is set in motion with an ini
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Answer:

Step-by-step explanation:

Given that,

Mass of object=3.25kg

The extension e=22cm=0.22m

Force applied to cause extension F=15N

Initial position Xo=0

Initial velocity Vo=-12m/s

We can get the spring constant from Hooke's law

F=ke

Then, k=F/e

k=15/0.22

k=68.182N/m

Also our natural frequency w is given as

w=√(k/m)

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w=√20.98

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w=4.6rad/s

There is no damping in this situation, no outside force acting on the system and the equation that governs the system is

mx''+kx=0

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Divide through by 3.25

x''+20.98x=0

We can approximate 20.98 to 21

x"+21x=0

The solution to this differential equation using D operator

D²+21=0

D²=-21

D= ±√-21

D=±√21 •i

Then the solution is

x(t)=A•Sinwt +B•Coswt

x(t)=A•Sin√21 t +B•Cos√21 t

Note that x'(t)=v(t)

and at t=0 Vo=-12m/s

x(t)=A•Sin√21 t +B•Cos√21 t

x'(t)=v(t)=A√21•Cos√21 t - B√21•Sin√21 t

v(t)=A√21•Cos√21 t - B√21•Sin√21 t

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v(0)=-12

And X(0)=0

x(t)=A•Sin√21 t +B•Cos√21 t

X(0)=A•Sin√21•0 +B•Cos√21•0

X(0)=A•Sin0+B•Cos0

0=B

B=0

Also,, V(0)=-12m/s

v(t)=A√21•Cos√21 t - B√21•Sin√21 t

V(0)=A√21•Cos√21•0- B√21•Sin√21•0

V(0)=A√21•Cos0- B√21•Sin0

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Therefore,

A=-12/√21

A=-2.62

Therefore the general equation becomes

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a. The amplitude

Comparing x(t) to wave equations

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c. Period

Comparing the equation again

wt=√21t

Given that w=2πf

Therefore, 2πft=√21t

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f=√21/2π

f=0.73Hz

Then, period is the reciprocal of frequency

T=1/f

T=1/0.73

T=1.37seconds

The period is 1.37sec,

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