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Kruka [31]
3 years ago
15

An ice skater spins with her arms outstretched at 1 revolution per second. She pulls her arms in toward her body. Does her angul

ar momentum change? Ignore friction with the ice.
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
7 0

Answer:

Her angular momentum does not change.

Explanation:

Conservation of angular momentum: It states that there is no change in the angular momentum when no torque outside the system is applied to the body.

In the given problem, an ice skater spins her arm outstretched. She pulls her arms in towards her body.

The angular momentum of an ice skaters remains same as the there is no external torque applied on an ice skater. The initial angular momentum  and the final angular momentum of the given system are same. Therefore, the angular momentum is conserved.

Therefore, her angular momentum does not change.

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What is the acceleration of a 10.5 kg mass pushed by a 50.5 n force
Lina20 [59]

Answer:4.8 m/s^2

Explanation:

mass=10.5kg

Force=50.5N

Acceleration =force ➗ mass

Acceleration =50.5 ➗ 10.5

Acceleration =4.8 m/s^2

8 0
3 years ago
The waves shown below represent sound waves. Which of the waves would have the highest-pitched sound?
Ostrovityanka [42]
Humans hear frequencies from 20 Hz<span> (low) up to </span>20,000 Hz (high)

8 0
2 years ago
In a mail-sorting facility, a 2.50-kg package slides down an inclined plane that makes an angle of 20.0° with the horizontal. Th
lawyer [7]

Answer:

The coefficient of kinetic friction is 0.382.

Explanation:

Given:

Angle of inclination is, \theta=20.0°

Mass of package is, m=2.50\ kg

Initial speed of package is, u=2.00\ m/s

Final speed of the package at the bottom is, v=0\ m/s

Distance of travel along the incline is, d=12.0\ m

Acceleration due to gravity is, g=9.8\ m/s^2

Let the coefficient of kinetic friction be \mu.

Now, the frictional force will be acting along the incline but in the direction opposite to the direction of motion.

So, the net acceleration acting on the package will be up the incline and is equal to:

a=\mu g\cos\theta-g\sin\theta ----------------- 1

Now, using equation of motion, we have:

v^2-u^2=2ad\\\\0-(2.00)^2=2a(12.0)

Solving for 'a', we get:

-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2

Now, plug in the value of 'a' in equation (1). This gives,

\mu g\cos\theta-g\sin\theta=\frac{1}{6} ( Neglecting negative sign)

Plug in all the given values and solve for \mu. This gives,

9.8(-sin(20)+\mu cos(20))=\frac{1}{6}\\\\-0.342+\mu\times 0.94=0.017\\\\0.94\mu=0.342+0.017\\\\0.94\mu=0.359\\\\\mu=\frac{0.359}{0.94}=0.382

Therefore, the coefficient of kinetic friction is 0.382.

5 0
3 years ago
Can someone help me with this? (The answer marked in red text is a incorrect answer)
Tju [1.3M]

Answer:

the answer is A

Explanation:

because I just know

3 0
2 years ago
The coefficient of cubical expansion of a substance depends upon
zzz [600]
<span>I think that the coefficient of cubical expansion of a substance depends on THE CHANGE IN VOLUME.

Cubical expansion, also known as, volumetric expansion has the following formula:

</span>Δ V = β V₁ ΔT

V₁ = initial volume of the body
ΔT = change in temperature of the body
β = coefficient of volumetric expansion.

β is defined as the <span>increase in volume per unit original volume per Kelvin rise in temperature.
</span>
With the above definition, it is safe to assume that the <span>coefficient of cubical expansion of a substance depends on the change in volume, which also changes in response to the change in temperature. </span>
7 0
3 years ago
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