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a_sh-v [17]
1 year ago
6

How do you find the oscillation period in seconds for different pendulum lengths?

Physics
1 answer:
GalinKa [24]1 year ago
3 0

To find:

The equation to find the period of oscillation.

Explanation:

The period of oscillation of a pendulum is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the acceleration due to gravity.

Thus the period of a pendulum is given by the equation,

T=2\pi\sqrt{\frac{L}{g}}

Where L is the length of the pendulum and g is the acceleration due to gravity.

On substituting the values of the length of the pendulum and the acceleration due to gravity at the point where the period of the pendulum is being measured, the above equation yields the value of the period of the pendulum.

Final answer:

The period of oscillation of a pendulum can be calculated using the equation,

T=2\pi\sqrt{\frac{L}{g}}

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Please help. I don’t understand this
skad [1K]

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>

and under constant acceleration,

<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2

According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so

∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2

∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2

∆<em>x</em> = 20.00 m

5 0
2 years ago
A 60-kg rollerblader rolls 10 m down a 30? incline. When she reaches the level floor at the bottom, she applies the brakes. The
VARVARA [1.3K]

Answer:

s = 20 m

Explanation:

given,

mass of the roller blader = 60 Kg

length = 10 m

inclines at = 30°

coefficient of friction = 0.25

using conservation of energy

\dfrac{1}{2}mu^2 = m g d sin \theta

u^2 = 2 g d sin30^0

u= \sqrt{2\times 9.8 \times 10 sin30^0}

u = 9.89 m/s

Using second law of motion  

ma =μ mg

a = μ g

a = 0.25 x 9.8

a = 2.45 m/s²

Using third equation of motion ,  

v² - u² = 2 a s

0² - 9.89² = 2 x 2.45 x s

s = 20 m

the distance moved before stopping is 20 m

3 0
3 years ago
A sailboat is traveling to the right when a gust of wind causes the boat to accelerate leftward at 2.5​​​​m​​/s2 for 4s. After t
Ber [7]

Answer:

+7.0 m/s

Explanation:

Let's take rightward as positive direction.

So in this problem we have:

a = -2.5 m/s^2 acceleration due to the wind (negative because it is leftward)

t = 4 s time interval

v = -3.0 m/s is the final velocity (negative because it is leftward)

We can use the following equation:

v = u + at

Where u is the initial velocity

We want to find u, so if we rearrange the equation we find:

u = v - at = (-3.0 m/s) - (-2.5 m/s^2)(4 s)=+7.0 m/s

and the positive sign means the initial direction was rightward.

6 0
3 years ago
How do I differentiate between final and initial velocity in Physics?​
pychu [463]

Answer:

Initial velocity describes how fast an object travels when gravity first applies force on the object. On the other hand, the final velocity is a vector quantity that measures the speed and direction of a moving body after it has reached its maximum acceleration.

Explanation:

3 0
2 years ago
A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju
denpristay [2]

Answer:

a) h=250\ m

b) \Delta h=0.0835\ m

Explanation:

Given:

  • upward acceleration of the helicopter, a=5\ m.s^{-2}
  • time after the takeoff after which the engine is shut off, t_a=10\ s

a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

h=u.t+\frac{1}{2} a.t^2

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

h=0+0.5\times 5\times 10^2

h=250\ m

b)

  • time after which Austin Powers deploys parachute(time of free fall), t_f=7\ s
  • acceleration after deploying the parachute, a_p=2\ m.s^{-2}

<u>height fallen freely by Austin:</u>

h_f=u.t_f+\frac{1}{2} g.t_f^2

where:

u= initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

t_f= time of free fall

h_f=0+0.5\times 9.8\times 7^2

h_f=240.1\ m

<u>Velocity just before opening the parachute:</u>

v_f=u+g.t_f

v_f=0+9.8\times 7

v_f=68.6\ m.s^{-1}

<u>Time taken by the helicopter to fall:</u>

h=u.t_h+\frac{1}{2} g.t_h^2

where:

u= initial velocity of the helicopter just before it begins falling freely = 0

t_h= time taken by the helicopter to fall on ground

h= height from where it falls = 250 m

now,

250=0+0.5\times 9.8\times t_h^2

t_h=7.1429\ s

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

<u>remaining time,</u>

t'=t_h-t_f

t'=7.1428-7

t'=0.1428\ s

<u>Now the height fallen in the remaining time using parachute:</u>

h'=v_f.t'+\frac{1}{2} a_p.t'^2

h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2

h'=9.8165\ m

<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>

\Delta h=h-(h_f+h')

\Delta h=250-(240.1+9.8165)

\Delta h=0.0835\ m

5 0
3 years ago
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