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Rainbow [258]
3 years ago
13

The type of function that describes the amplitude of damped oscillatory motion is _______. The type of function that describes t

he amplitude of damped oscillatory motion is _______. quadratic sinusoidal inverse exponential linear
Physics
1 answer:
Salsk061 [2.6K]3 years ago
8 0

Answer:

exponential

Explanation:

type of function that describes the amplitude of damped oscillatory motion is exponential because as we know that here function is

y = A × e^{\frac{-bt}{2m}}  × cos(ωt + ∅ )    ..................................... ( 1 )          

here function A × e^{\frac{-bt}{2m}}   is amplitude

as per equation ( 1 )it is exponential

so that we can say that amplitude of damped oscillatory motion is exponential

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You are riding a bicycle. If you apply a forward force of 125 N, and you and
erica [24]

Answer:

1.52g

Explanation:

Given parameters:

Force  = 125N

Mass combined  = 82kg

Unknown:

Acceleration of the bicycle  = ?

Solution:

From Newton second law of motion suggests that:

   Force = mass x acceleration

  Acceleration = \frac{force}{mass}  = \frac{125}{82}   = 1.52g

8 0
3 years ago
What element from the periodic table produces blue color in fireworks
sattari [20]
The answer is going to be element #29 Copper makes blue 
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4 0
3 years ago
Read 2 more answers
Normalize the equations
tatyana61 [14]

Answer:

Solution is in explanation

Explanation:

part a)

For normalization we have

\int_{0}^{\infty }f(x)dx=1\\\\\therefore \int_{0}^{\infty }ae^{-kx}dx=1\\\\\Rightarrow a\int_{0}^{\infty }e^{-kx}dx=1\\\\\frac{a}{-k}[\frac{1}{e^{kx}}]_{0}^{\infty }=1\\\\\frac{a}{-k}[0-1]=1\\\\\therefore a=k

Part b)

\int_{0}^{L }f(x)dx=1\\\\\therefore Re(\int_{0}^{L }ae^{-ikx}dx)=1\\\\\Rightarrow Re(a\int_{0}^{L }e^{-ikx}dx)=1\\\\\therefore Re(\frac{a}{-ik}[\frac{1}{e^{ikx}}]_{0}^{L})=1\\\\\Rightarrow Re(\frac{a}{-ik}(e^{-ikL}-1))=1\\\\\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1

\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1\\\\\frac{a}{k}Re(icos(-kL)+sin(kL)+\frac{1}{i})=1\\\\\frac{a}{k}sin(kL)=1\\\\a=\frac{k}{sin(kL)}

7 0
3 years ago
How does a car get energy?
Lorico [155]

Answer:

Explanation:

Just like your body converts food into energy, a car engine converts gas into motion. ... The process of converting gasoline into motion is called "internal combustion." Internal combustion engines use small, controlled explosions to generate the power needed to move your car all the places it needs to go.

6 0
3 years ago
Read 2 more answers
The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?
Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours.  So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes.  So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = <em>500 g</em> .

============================================

<u>Another way</u>:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = <em>500 g </em>.

5 0
2 years ago
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