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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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Answer:

When you ask a question, only two people can answer. When there are two answers, a little crown should appear at the bottom right hand corner. All you have to do is click the crown and it gives Brainliest. But you can only give it to one person per question

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3 years ago
At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is
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Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

\dfrac{1}{2}mv^2=mgl

v=\sqrt{2gl}

Put the value into the formula

u=\sqrt{2\times9.8\times50.0\times10^{-2}}

u=3.13\ m/s

The initial speed of the ball u_{1}=3.13\ m/s

The initial speed of the block u_{2}=0

(a). We need to calculate the speed of the ball after collision

Using formula of collision

v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13

v_{1}=-2.01\ m/s

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0

v_{2}=1.11\ m/s

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

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A zone reconnaissance involves a directed effort to obtain detailed information on all routes, obstacles, terrain, enemy forces,
son4ous [18]

Answer:

It is True

Explanation:.

A  commander assigns a zone reconnaissance mission when he seeks additional information on a zone before committing other forces in the zone. It is appropriate when the enemy situation is vague,  existing knowledge of the terrain is limited, or combat operations have altered the terrain. A zone  reconnaissance could include several route or area reconnaissance missions assigned to subordinate units.

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Ashley is flying a plane that has to reach a gradient of 360m/km in order to take off and not crash. Her goal is to travel from
KengaRu [80]

Answer:

She is likely to crash because her flight gradient is lesser than the flight gradient required gradient to avoid crashing

Explanation:

The given parameters are;

The required gradient of the plane Ashley is flying needs to reach in order to take off and not crash = 360 m/km

The initial elevation of the plane Ashley is flying = Sea level = 0 m

The goal Ashley intends to make = Elevation of 1000 m at 2.8 km. distance

∴ Ashley's goal = Traveling from sea level to 1000 m at 2.8 km horizontal distance

We have;

The gradient  = Rate of change of elevation/(Horizontal distance)

Therefore;

The gradient of Ashley's flight = (1000 - 0)/(2.8 - 0) = 357.143 m/km

The gradient of Ashley's flight ≈ 357.143 m/km which is lesser than the required 360 m/km in order to take off and not crash, therefore, she will crash.

6 0
3 years ago
Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
Alexus [3.1K]

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = \sqrt{2I/ce_{0} }

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9}  = <em>1.55 x 10^-8 v/m</em>

6 0
3 years ago
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