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Lyrx [107]
3 years ago
5

A. What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8cm from equilibrium and the

n released from rest, and whose period is 0.66s? Assume that the displacement at the start of the motion is positive. (express your answer in terms of t using two significant figures.)
B. What will be its displacement after 1.7s? (express your answer to two significant figures and include the appropriate units.)
Physics
2 answers:
Naddik [55]3 years ago
5 0

Answer:

a) x = 8.8 cm * cos (9.52 rad/s * t)

b) x = 8.45 cm

Explanation:

This is a Simple Harmonic Motion, and most Simple Harmonic Motion equations start from the equilibrium point. In this question however, we are starting from the max displacement the equations, and thus, it ought to be different.

From the question, we are given that

A = 8.8 cm = 0.088 m

t = 0.66 s

Now, we need to find the angular speed w, such that

w = 2π/T

w = (2 * 3.142) / 0.66

w = 6.284 / 0.66

w = 9.52 rad/s

The displacement equation of Simple Harmonic Motion is usually given as

x = A*sin(w*t)

But then, the equation starts from the equilibrium point at 0 sec, i.e x = 0 m

When you have to start from the max displacement, then the equation would be

x = A*cos(w*t).

So when t = 0 the cos(0) = 1, and then x = A which is max displacement.

Thus, the equation is

x = 8.8 cm * cos (9.52 rad/s * t)

At t = 1.7 s,

x = 8.8 cos (9.52 * 1.7)

x = 8.8 cos (16.184)

x = -8.45 cm

Jobisdone [24]3 years ago
3 0

Answer:  

(a)      $y(t) = 8.8cos(\frac{2\pi }{0.665} t)$

(b)  and after 1.7s the displacement would be

$y(1.7s) = 8.8cos(\frac{2\pi }{0.665s}1.7s)$ = 8.456461cm

Explanation:

This is periodic motion and the best equation to model this type of motion is sinusoidal equation.

We will chose cos function because it starts at the max or the amplitude.

A = 8.8cm, T = 0.66s in following general form gives the answer.

$y(t)=Acos(\frac{2\pi }{T} t)

and substituting t = 1.7s in it gives us 8.456461cm, just little bit lower than initial position.

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Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it
d1i1m1o1n [39]

Answer:

Q = 165.95 cm³ / s,  1)    v = \sqrt{0.55^2 + 19.6 y},  2)  v = 2.05 m / s,

3)  d₂ = 1.014 cm

Explanation:

This is a fluid mechanics exercise

1) the continuity equation is

         Q = v A

where Q is the flow rate, A is area and v is the velocity

         

the area of ​​a circle is

        A = π r²

radius and diameter are related

        r = d / 2

substituting

       A = π d²/4

       Q = π/4   v d²

let's reduce the magnitudes

       v = 0.55 m / s = 55 cm / s

let's calculate

       Q = π/4   55   1.96²

       Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

        v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

        v = \sqrt{0.55^2 + 2  \ 9.8\  y}

        v = \sqrt{0.55^2 + 19.6 y}

2) ask to calculate the velocity for y = 0.2 m

        v = \sqrt{0.55^2 + 19.6 \ 0.2}

        v = 2.05 m / s

3) We write the continuous equation for this point 2

        Q = v₂ A₂

        A₂ = Q / v₂

let us reduce to the same units of the SI system

        Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

        A₂ = 165.95 10⁻⁶ / 2.05

        A₂ = 80,759 10⁻⁶ m²

area is

        A₂ = π/4   d₂²

        d₂ = \sqrt{4  A_2 / \pi }

        d₂ = \sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }

        d₂ = 10.14 10⁻³ m

        d₂ = 1.014 cm

4 0
3 years ago
A car accelerates at 3 m/s*2. Assuming the car starts from rest, how much time does it need to
uranmaximum [27]
<h3><u>Given</u><u>:</u><u>-</u></h3>

Acceleration,a = 3 m/s²

Initial velocity,u = 0 m/s

Final velocity,v = 12 m/s

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the time take by a car.

<h3><u>Solution:-</u><u> </u></h3>

According to the first equation of motion:

v = u + at

★ Substituting the values in the above formula,we get:

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5 0
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Answer:

2. the volume of the square are the same

7 0
2 years ago
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A jogger runs at a constant rate of 10.0 m every 2.0 seconds. The jogger starts at the origin and runs in the positive direction
Elis [28]

Answer:

(a) 25 m

(b) 75 m

Explanation:

Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.

So, the speed of the jogger,

v=\frac{10}{2}=5m/s\;\cdots(i)

Let d be the distance covered by him in time, t s.

As distance=(speed) x (time)

So, d=vt

From equation (i)

\Rightarrow d=5t\;\cdots(ii)

As the jogger starts from origin, so, the distance, d, also represents the position of the jogger at the time t s.

The position-time graph has been shown.

(a) From equation (ii), for t=5.0 s

d=5\times 5=25 m

So, the jogger is at a distance of 25 m from the origin.

(b) Similarly, for t=15.0 s

d=5\times 15=75 m

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8 0
2 years ago
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AnnyKZ [126]
Hey there!

Here is your answer:

<u><em>The proper answer to this question is option C "</em></u><span><u><em>0.00349".</em></u>

Reason:

</span><span><u><em>1 L = 100 cL. Or 1 cL = 0.01 L</em></u>

</span><span><u><em>34.9 cL = 34.9 / 100 L = 0.349 L</em></u>

</span><span><u><em> 1 hL = 100 L. 0.349 L = 0.349 / 100 hL = 0.00349 hL</em></u>

<em>Therefore the answer is option C!</em>

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit</span>
4 0
3 years ago
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