Physics, 28.11.2019 19:31, Fireburntbudder

Asimple pendulum has a 100.0 g mass tied to it and 23.00 seconds are required to complete 10 oscillations.
(a) if g = 9.80 m/s^2, what is the length of the pendulum?
(b) when the mass is released from a certain hight, its speed as it passes through its lowest point is measured to be 1.620 m/s. calculate the tension in the string at the pendulum's lowest point,
(c) when the 100.00 g mass is removed from the string forming the pendulum, it is attached to a spring. if 23.00 seconds is required to complete 25 oscillations on the spring, calculate the spring constant,
(d) if you are to move pendulum to another planet where g = 6.58 m/s^2, how will period change?
(e) if you are to move spring mass system to another planet where g = 6.58 m/s^2, how will period change. show detailed calculations.

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answered: xxtonixwilsonxx

a) The length of the pendulum is 1.31 m

b) The tension in the string is 1.18 N

c) The spring constant is 4.66 N/m

d) The period of the pendulum increases from 2.30 s to 2.80 s

e) The period of the spring does not change

Explanation:

The frequency of oscillation of the pendulum is

$f=\frac{N}{t}=\frac{10}{23.0}=0.435 Hz$

where N = 10 is the number of oscillations in a time of t=23.0 s .

The period of a simple pendulum is given by

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

where L is the length of the pendulum and g is the acceleration of gravity. Since the frequency is the reciprocal of the period,

$f=\frac{1}{T}=\frac{1}{2\pi} \sqrt{\frac{g}{L}}$

Since we know both f (the frequency) and g, we can solve the formula to find L, the length of the pendulum:

$L=\frac{g}{(2\pi f)^2}=\frac{9.80}{(2\pi (0.435))^2}=1.31 m$

The equation of motion for the pendulum when it passes through the lowest point is:

T-W = F_c

where

T is the tension in the string

W=mg is the weight of the pendulum, with m=100.0 g = 0.1 kg being the mass of the pendulum and g=9.80 m/s^2

$F_c = \frac{mv^2}{L}$ is the centripetal force, where

m = 0.1 kg is the mass of the pendulum

v = 1.620 m/s is the speed at the lowest point

L = 1.31 m is the radius of the circular trajectory, which is the length of the pendulum

Substituting and solving for T, we find

$T=mg+\frac{mv^2}{L}=(0.1)(9.80)+\frac{(0.1)(1.620)^2}{1.31}=1.18 N$

The frequency of oscillation of the spring is:

$f=\frac{N}{t}=\frac{25}{23.0}=1.087 Hz$

where N = 25 is the number of oscillations in a time of t=23.0 s .

The frequency of the spring can be written as

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

where

k is the spring constant

m = 0.1 kg is the mass attached to the spring

Since we know f and m, we can re-arrange the equation to find k, the spring constant:

$k=(2\pi f)^2 m = (2\pi (1.087))^2(0.1)=4.66 N/m$

The period of the a pendulum, as we said earlier, is given by

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

where

L is the length of the pendulum

g is the acceleration of gravity

The length of this pendulum is

L = 1.31 m

On Earth, g=9.80 m/s^2 , so the period on Earth is

$T=2\pi \sqrt{\frac{1.31}{9.80}}=2.30 s$

Here the pendulum is moved to another planet, where the acceleration of gravity is

g=6.58 m/s^2

Substituting, we find the new period of the pendulum:

$T=2\pi \sqrt{\frac{1.31}{6.58}}=2.80 s$

So, the period increases from 2.30 s to 2.80 s.

The period of the spring is equal to the reciprocal of its frequency of oscillation:

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

We know that the frequency of oscillation is

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

Therefore, we can write the period as

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

We see that the period does not depend on the value of g, the acceleration of gravity: therefore, if we move the spring on another planet, the value of its period does not change.

answered: Guest

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

hope this ! this is not copied! search it up, it's free of plagiarism!

answered: Guest

3. dark matter

empty space could also be a simple vacuum

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Asimple pendulum has a 100.0 g mass tied to it and 23.00 seconds are required to complete 10 oscilla...