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konstantin123 [22]
3 years ago
6

Describe how work is done by a skater pulling in her arms during a spin. In particular, identify the force she exerts on each ar

m to pull it in and the distance each moves, noting that a component of the force is in the direction moved. Why is angular momentum not increased by this action?
Physics
1 answer:
liberstina [14]3 years ago
4 0

Answer:

 ∑ τ =0,  L₀ = L_{f}

Explanation:

In a circular turning movement, when the arms are extended and then contracted in two possibilities:

- They are lowered the force of gravity is what pulls them, the tension of the muscle becomes zero to allow this movement.

    In this movement the force is vertical(gravity) and the movement of the center of mass of each arm is vertical, so that the work is the weight value of the arm by the distance traveled by the center of mass.

- Another possibility is that the arms have stuck to the body, in this case the person's muscles perform the force, this force is horizontal and the displacement is the horizontal of the center of mass of the arms from the extended position to the contracted

 

In these movements the torque of the external force is equal for each arm, but in the opposite direction, so they are canceled where a net torque of zero, this causes the angular momentum to be preserved, which changes is the moment of inertia of the system and therefore you must also change the angular velocity to keep your product constant

          ∑ τ =0

                L₀ = L_{f}

              I₀ w₀ = I w

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svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s

velocity at the exit=20m/s

for entry

V=\frac{5}{1} =5m/s

2.

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha  =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81  =p2\\P2=-37.5kPa

the pressure at exit is -37.5kPa

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3 years ago
I NEED HELP ON THIS QUESTION!
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The second option is the correct one. m/s^2

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if the efficiency of an electric furnace is 96%, then 96% of the input energy is transformed into thermal energy. what is the ot
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A force of 8.0 N is along x direction, another force of 6.0 N is along +y direction. If both forces are acting on a point object
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Answer:

Resultant force, R = 10 N

Explanation:

It is given that,

Force acting along +x direction, F_x=8\ N

Force acting along +y direction, F_y=6\ N

Both the forces are acting on a point object located at the origin. Let the resultant force of the object is given by R. So,

R=\sqrt{F_x^2+F_y^2+F_xF_y\ cos\theta}

Here \theta=90^{\circ}

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