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MrMuchimi
2 years ago
4

Max (15 kg) pushes maya (12 kg) on a swing so that maya moves in simple harmonic motion. the swing is 1.8 meters long and has a

mass of 4 kg, concentrated in the seat. if maya and the swing form a simple pendulum, what is the period of maya's motion?
Physics
2 answers:
AnnyKZ [126]2 years ago
8 0
The period of a simple pendulum depends only on the length of the pendulum and the gravitational acceleration:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the pendulum length and g the gravitational acceleration.

The problem says that Maya and the swing form a simple pendulum, so we can use this formula to calculate the period of Maya's motion, using the length of the swing (L=1.8 m):
T=2 \pi  \sqrt{ \frac{1.8 m}{9.81 m/s^2} }=2.69 s
const2013 [10]2 years ago
8 0
<h2>Answer:</h2>

The period is 2.69 seconds.

<h3>Explanation:</h3>

<u>The period of the pendulum can be defined as the time taken by the pendulum to motion back from the equilibrium position and then forth the equilibrium position and back to equilibrium position.</u>

Briefly it is time for one complete oscillation.

It is dependent on the length of pendulum. The mass of pendulum has no effect on the period.

T = 2π√L/g

Here g is gravitational acceleration, its value is constant on the earth and is 9.8 m/s2.

By putting values in formula:

T = 2(3.14)√1.8/9.8

T = 2.69 s

Hence the period is 2.69 seconds.

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Answer:

The speed of the cart after 8 seconds of Low fan speed is  72.0 cm/s

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Explanation:

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A certain part of a flat screen TV has a thickness of 150 nanometers. How<br> many meters is this?
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Answer:

1.5e-7 meters

.00000015 meters

Explanation:

.000000001 meters = 1 nanometer. Multiply that by 150 and an answer is there.

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2 years ago
A compact, dense object with a mass of 2.90 kg is attached to a spring and is able to oscillate horizontally with negligible fri
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(a) 80 N/m

The spring constant can be found by using Hooke's law:

F=kx

where

F is the force on the spring

k is the spring constant

x is the displacement of the spring relative to the equilibrium position

At the beginning, we have

F = 16.0 N is the force applied

x = 0.200 m is the displacement from the equilibrium position

Solving the formula for k, we find

k=\frac{F}{m}=\frac{16.0 N}{0.200 m}=80 N/m

(b) 0.84 Hz

The frequency of oscillation of the system is given by

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k = 80 N/m is the spring constant

m = 2.90 kg is the mass attached to the spring

Substituting the numbers into the formula, we find

f=\frac{1}{2\pi}\sqrt{\frac{80 N/m}{2.90 kg}}=0.84 Hz

(c) 1.05 m/s

The maximum speed of a spring-mass system is given by

v=\omega A

where

\omega is the angular frequency

A is the amplitude of the motion

For this system, we have

\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s

A=0.200 m (the amplitude corresponds to the maximum displacement, so it is equal to the initial displacement)

Substituting into the formula, we find the maximum speed:

v=(5.25 rad/s)(0.200 m)=1.05 m/s

(d) x = 0

The maximum speed in a simple harmonic motion occurs at the equilibrium position. In fact, the total mechanical energy of the system is equal to the sum of the elastic potential energy (U) and the kinetic energy (K):

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2

where

k is the spring constant

x is the displacement

m is the mass

v is the speed

The mechanical energy E is constant: this means that when U increases, K decreases, and viceversa. Therefore, the maximum kinetic energy (and so the maximum speed) will occur when the elastic potential energy is minimum (zero), and this occurs when x=0.

(e) 5.51 m/s^2

In a simple harmonic motion, the maximum acceleration is given by

a=\omega^2 A

Using the numbers we calculated in part c):

\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s

A=0.200 m

we find immediately the maximum acceleration:

a=(5.25 rad/s)^2(0.200 m)=5.51 m/s^2

(f) At the position of maximum displacement: x=\pm 0.200 m

According to Newton's second law, the acceleration is directly proportional to the force on the mass:

a=\frac{F}{m}

this means that the acceleration will be maximum when the force is maximum.

However, the force is given by Hooke's law:

F=kx

so, the force is maximum when the displacement x is maximum: so, the maximum acceleration occurs at the position of maximum displacement.

(g) 1.60 J

The total mechanical energy of the system can be found by calculating the kinetic energy of the system at the equilibrium position, where x=0 and so the elastic potential energy U is zero. So we have

E=K=\frac{1}{2}mv_{max}^2

where

m = 2.90 kg is the mass

v_{max}=1.05 m/s is the maximum speed

Solving for E, we find

E=\frac{1}{2}(2.90 kg)(1.05 m/s)^2=1.60 J

(h) 0.99 m/s

When the position is equal to 1/3 of the maximum displacement, we have

x=\frac{1}{3}(0.200 m)=0.0667 m

so the elastic potential energy is

U=\frac{1}{2}kx^2=\frac{1}{2}(80 N/m)(0.0667 m)^2=0.18 J

and since the total energy E = 1.60 J is conserved, the kinetic energy is

K=E-U=1.60 J-0.18 J=1.42 J

And from the relationship between kinetic energy and speed, we can find the speed of the system:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.42 J)}{2.90 kg}}=0.99 m/s

(i) 1.84 m/s^2

When the position is equal to 1/3 of the maximum displacement, we have

x=\frac{1}{3}(0.200 m)=0.0667 m

So the restoring force exerted by the spring on the mass is

F=kx=(80 N/m)(0.0667 m)=5.34 N

And so, we can calculate the acceleration by using Newton's second law:

a=\frac{F}{m}=\frac{5.34 N}{2.90 kg}=1.84 m/s^2

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