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3 years ago

A girl with a mass of 40 kg is swinging from a rope with a length of 2.5 m. What is the frequency of her swinging?

Physics

2 answers:

Dmitrij [34]3 years ago

8 0

For a pendulum with a massless rope 'L' meters long, swinging through
a small arc, the period of the swing is

2 π √(L/g) seconds .

and it doesn't depend on the mass of the thing on the end of the rope ...
that could be a pebble or a bus.

For the girl on the 2.5-m rope,

Period = (2 π) √(2.5 / 9.8) = 3.17 seconds

Frequency = 1 / period = about <em> 0.315 Hz .</em>

blagie [28]3 years ago

8 0

If a girl with a mass of 40 kg is swinging from a rope with a length of 2.5 m , then the frequency of her swinging is 0.32 Hz

<h3>Further explanation</h3>

Simple Harmonic Motion is a motion where the magnitude of acceleration is directly proportional to the magnitude of the displacement but in the opposite direction.

The pulled and then released spring is one of the examples of Simple Harmonic Motion. We can use the following formula to find the period of this spring.

$\large { \boxed {T = 2 \pi\sqrt{\frac{m}{k}} } }$

T = Periode of Spring ( second )

m = Load Mass ( kg )

k = Spring Constant ( N / m )

The pendulum which moves back and forth is also an example of Simple Harmonic Motion. We can use the following formula to find the period of this pendulum.

$\large { \boxed {T = 2 \pi\sqrt{\frac{L}{g}} } }$

T = Periode of Pendulum ( second )

L = Length of Pendulum ( kg )

g = Gravitational Acceleration ( m/s² )

Let us now tackle the problem !

<u>Given:</u>

Mass of A Girl = m = 40 kg

Length of Rope = L = 2.5 m

Gravitational Acceleration = g = 10 m/s²

<u>Unknown:</u>

Frequency of Swinging = f = ?

<u>Solution:</u>

Recall the formula for calculating period as mentioned above.

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

$T = 2 \pi\sqrt{\frac{2.5}{10}}$

$T = 2 \pi\sqrt{\frac{1}{4}}$

$T = 2 \pi \frac{1}{2}$

$T = \pi ~ seconds$

$T \approx 3.1 ~ seconds$

Finally, we can calculate the magnitude of frequency with the following formula.

$f = \frac{1}{T}$

$f = \frac{1}{\pi} ~ Hz$

$f \approx 0.32 ~ Hz$

<h3>Learn more</h3>

Model for Simple Harmonic Motion : brainly.com/question/9221526
Force of Simple Harmonic Motion : brainly.com/question/3323600
Example of Simple Harmonic Motion : brainly.com/question/11892568

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Simple Harmonic Motion

Keywords: Simple , Harmonic , Motion , Pendulum , Spring , Period , Frequency

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a).

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Complete question:

Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.

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The magnitude of this field is 826 N/C

Explanation:

Given;

The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m

PEsinθ = T

where;

E is the magnitude of the electric field

P is the dipole moment

First, we determine the magnitude dipole moment;

Magnitude of dipole moment = q*r

P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m

Finally, we determine the magnitude of this field;

$E = \frac{T}{P*sin(\theta)}= \frac{7.3 X 10^{-9}}{1.476X10^{-11}*sin(36.8)}\\\\E = 825.6 N/C$

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From the question given above, the following data were obtained:

Total time (T) spent by the ball in air = 5 s

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Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:

Total time (T) spent by the ball in air = 5 s

Time (t) taken to reach the maximum height =.?

T = 2t

5 = 2t

Divide both side by 2

t = 5/2

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Thus, the time (t) taken to reach the maximum height is 2.5 s

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Time (t) taken to reach the maximum height = 2.5 s

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