When you run around a track which type of energy are you not demonstrating ?

The index of refraction for flint glass is 1.5. This means that

Alexxx [7]

Answer:

Explanation:

This means that the ratio of the speed of the incident beam of light in a vacuum to the speed of light in the glass is 1.5

7 0

3 years ago

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What happens to the size of atoms as we go from left to right on the periodic table? T

Lady_Fox [76]

Answer:

Atomic size gradually decreases from left to right across a period of elements.

Explanation:

This is because, within a period or family of elements, all electrons are added to the same shell. However, at the same time, protons are being added to the nucleus, making it more positively charged.

8 0

3 years ago

A horizontal circular platform rotates counterclockwise about its axis at the rate of 0.841 rad/s.0.841 rad/s. You, with a mass

kvv77 [185]

Answer:

The total angular momentum of the system is $217.46\ kg-m^2/s$ .

Explanation:

Given that,

Angular speed = 0.841 rad/s

Mass of platform = 72.1 kg

Speed = 1.17 m/s

Mass of poodle = 20.3 kg

Mass of mutt = 18.5 kg

Distance =3/4 of the platform's radius from the center

Mass of disk = 92.5 kg

radius = 1.87 m

Angular momentum is the product of moment of inertia and angular speed.

We need to calculate the angular momentum of person

$L_{person} = I\omega$

$L_{person}=\dfrac{1}{2}mr^2\times\omega$

$L_{person}=\dfrac{1}{2}\times72.1\times(\dfrac{v}{\omega})^2\times\omega$

$L_{person}=\dfrac{1}{2}\times72.1\times(\dfrac{1.17}{0.841})^2\times0.841$

$L_{person}=58.679\ kg m^2/s$

We need to calculate the angular momentum of platform

$L_{platform}=\dfrac{1}{2}mr^2\times\omega$

Put the value into the formula

$L_{platform}=\dfrac{1}{2}\times92.5\times1.87^2\times0.841$

$L_{platform}=136.02\ kg m^2/s$

We need to calculate the angular momentum of poodle

$L_{poodle}=\dfrac{1}{2}mr^2\times\omega$

Put the value into the formula

$L_{poodle}=\dfrac{1}{2}\times20.3\times(\dfrac{1.87}{2})^2\times0.841$

$L_{poodle}=7.4625\ kg m^2/s$

We need to calculate the angular momentum of mutt

$L_{mutt}=\dfrac{1}{2}mr^2\times\omega$

$L_{mutt}=\dfrac{1}{2}\times18.5\times(\dfrac{3}{4}\times1.87)^2\times0.841$

$L_{mutt}=15.302\ kg m^2/s$

We need to calculate the total angular momentum

$L=L_{person}+L_{platform}+L_{poodle}+L_{mutt}$

$L=58.679+136.02+7.4625+15.302$

$L=217.46\ kg-m^2/s$

Hence, The total angular momentum of the system is $217.46\ kg-m^2/s$ .

3 0

3 years ago

Can one of yall help???

Art [367]

1,200 x 3.0 = 3600N

8 0

3 years ago

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he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?

Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

$\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}$

With the following condition :

$\sf{\omega}$ = angular frequency (rad/s)
$\sf{\theta}$ = change of angle value (rad)
t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

$\sf{\theta}$ = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
t = interval of the time = 54.9 s.

What was asked :

$\sf{\omega}$ = angular frequency = ... rad/s

Step by step :

$\sf{\omega = \frac{\theta}{t}}$

$\sf{\omega = \frac{2,000 \pi}{54.9}}$

$\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}$

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0

2 years ago