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ser-zykov [4K]
3 years ago
13

Suppose you are at the center of a large freely-rotating horizontal turntable in a carnival funhouse. As you crawl toward the ed

ge, the angular momentum of you and the turntable:
a. decreases.
b. increases.
c. remains the same, but the RPMs decrease.
d. decreases in direct proportion to your decrease in RPMs.
e. none of these
Physics
1 answer:
I am Lyosha [343]3 years ago
4 0

Answer:

c. remains the same, but the RPMs decrease.

Explanation:

Because there aren't external torques on the system composed by the person and the turntable it follows that total angular momentum (I) is conserved, that means the total angular momentum is a constant:

\overrightarrow{L}=constant

The total angular momentum is the sum of the individual angular momenta, in our case we should sum the angular momentum of the turntable and the angular momentum of a point mass respect the center of the turntable (the person)

\overrightarrow{L_{turnatble}}+\overrightarrow{L_{person}}=constant (1)

The angular momentum of the turntable is:

\overrightarrow{L_{turnatble}}=I\overrightarrow{\omega} (2)

with I the moment of inertia and ω the angular velocity.

The angular momentum of the person respects the center of the turntable is:

\overrightarrow{L_{person}}=\overrightarrow{r}\times m\overrightarrow{v} (3)

with r the position of the person respects the center of the turntable, m the mass of the person and v the linear velocity

Using the fact v=\omega r:

\overrightarrow{L_{person}}=\overrightarrow{r}\times rm\overrightarrow{\omega}(3)

By (3) and (2) on (1) and working only the magnitudes (it's all that we need for this problem):

I\omega+r^{2}m\omega=constant

\omega(I+r^{2}m)=constant

Because the equality should be maintained, if we increase the distance between the person and the center of the turntable (r), the angular velocity should decrease to maintain the same constant value because I and m are constants, so the RPM's (unit of angular velocity) are going to decrease.

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Answer:

Voltage-gated K+ channels

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3 years ago
un conductor de plata (p=1,6x10°-6ohm x m) tiene una seccion de 5x10°-6 m°2 y una longitud de 110 m . calcular la resistencia
Alexus [3.1K]

Responder:

35,2 ohm.

Explicación:

Dado:

La resistencia específica del conductor es,

La longitud del conductor es,

El área de la sección transversal del conductor es,

Sabemos que la resistencia de un conductor es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

Por lo tanto, la resistencia se puede expresar como:

R=\frac{\rho\times l}{A}

Ahora, conecte los valores dados y resuelva para 'R'. Esto da,

R=\frac{1.6\times 10^{-6}\ ohm\cdot m\times 110\ m}{5\times 10^{-6}\ m^2}\\\\R=35.2\ ohm

Por lo tanto, la resistencia del conductor es de 35,2 ohm.

4 0
3 years ago
A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the fri
omeli [17]

Answer:

4500 N

Explanation:

When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:

a = v^2/r

where v is the velocity of the body and r is the radius of the circumference:

Therefore, a body with mass m, will feel a force f:

f = m v^2/r

Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:

fs = μN = μmg

The car will not slide if     f = fs,   i.e.

fs = μmg =  m v^2/r

That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration

fs = (1000 kg)  * (30m/s)^2 / (200 m) = 4500 N

7 0
3 years ago
Read 2 more answers
Using the definition of moment of inertia, calculate icm, the moment of inertia about the center of mass, for this object.
Step2247 [10]
<span> The masses have no inertia about their own CM, and "the object" is the two masses. </span>
<span>1. Icm (at point A) = 2mr^2 
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3 years ago
He instruction booklet for your pressure cooker indicates that its highest setting is 11.1 psi . you know that standard atmosphe
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Given that,

Atmospheric Pressure = 14.7 psi

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Take, Atmospheric Temperature = 25 °C

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Since, we know that Gas equation is given by:

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14.7/ 25 = 25.8/ T2

T2 = 25*25.8/14.7

T2 = 43.87 °C

The cooking pressure will be 43.87 °C.

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3 years ago
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