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DIA [1.3K]
1 year ago
7

A 60.0-kg skater begins a spin with an angular speed of 6.0 rad/s. By changing the position of her arms, the skater decreases he

r moment of inertia to one-half its initial value. What is the skater's final angular speed
Physics
1 answer:
juin [17]1 year ago
7 0

The skater's final angular speed is equal to 12 rad/s.

When implemented to angular momentum, the regulation of conservation means that the momentum of a rotating item is no longer exchanged until some form of external torque is carried out. Torque, in this sense, can check with any outside pressure that acts upon the object for the purpose to twist or rotate.

The law of conservation of angular momentum states that once no external torque acts on an item, no trade of angular momentum will occur. The angular momentum of a machine is conserved as long as there may be no net external torque performing on the machine.

In angular kinematics, the conservation of angular momentum refers back to the tendency of a device to keep its rotational momentum inside the absence of outside torque. For a round orbit, the system for angular momentum is (mass) ×(pace) ×(radius of the circle): (angular momentum) = m × v × r.

Learn more about angular momentum here brainly.com/question/7538238

#SPJ4

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What is the electric potential at a point between the 2.5-ω and 5.5-ω resistors, if the electric potential at the positive termi
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2 years ago
A point charge Q is placed at the center of a cube of side l. what is the flux through one face of the cube?
dimaraw [331]

Answer:

\phi=\dfrac{q}{6\epsilon_o}

Explanation:

A point charge Q is placed at the center of a cube of side l.

We need to find the flux through one face of the cube. Let the flux is \phi.

There are 6 faces in a cube.

According to Gauss's law, net flux is given by :

\phi=\dfrac{q}{\epsilon_0}

As there are 6 faces of a cube, flux through one surface is :

6\phi=\dfrac{q}{\epsilon_o}\\\\\phi=\dfrac{q}{6\epsilon_o}

So, the flux through one face of the cube is \dfrac{q}{6\epsilon_o}.

8 0
3 years ago
Consider a satellite in a circular orbit around the Earth. If it were at an altitude equal to twice the radius of the Earth, 2RE
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Answer:

v=\sqrt{\frac{gR_E}{2}}

Explanation:

Satellites experiment a force given by Newton's Gravitation Law:

F=\frac{GMm}{r^2}

where M is Earth's mass, m the satellite's mass, r the distance between their gravitational centers and G the gravitational constant.

We also know from Newton's 2nd Law that <em>F=ma, </em>so putting both together we will have:

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If we are on the surface of the Earth, the acceleration would be g and r=R_E (Earth's radius):

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v^2=\frac{gR_E^2}{r}

v=\sqrt{\frac{gR_E^2}{r}}

And if this orbit has a radius r=2R_E we have:

v=\sqrt{\frac{gR_E^2}{2R_E}}=\sqrt{\frac{gR_E}{2}}

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2 years ago
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