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wlad13 [49]
4 years ago
7

Interactive Solution 9.63 illustrates one way of solving a problem similar to this one. A thin rod has a length of 0.620 m and r

otates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of 0.185 rad/s and a moment of inertia of 1.43 x 10-3 kg·m2. A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (whose mass is 5 x 10-3 kg) gets where it's going, what is the change in the angular velocity of the rod?
Physics
1 answer:
finlep [7]4 years ago
6 0

Answer:

w = 7.89 10⁻² rad/s

Explanation:

We will solve this exercise with the conservation of the annular moment, let's write it in two moments

Initial. With the insect in the center

      L₀ = I w₀

End with the bug on the edge

     L_{f}= I w +  I_{bug} w

The moments of inertia are

For a rod

       I = 1/3 M L²

For the insect, taken as a particle

       I = m L²

The system is formed by the rod and the insect, this is isolated, therefore the external torque is zero and the angular momentum is conserved

      L₀ =  L_{f}

      I w₀ = I w + I_{bug} w

      w = I / (I +  I_{bug}) w₀

 

      w = I / (I + m L²) w₀

Let's calculate

      w = 1.43 10⁻³ / (1.43 10⁻³ + 5 10⁻³ 0.620²)²   0.185

      w = 1.43 10⁻³ / 3.352 10³ 0.185

      w = 7.89 10⁻² rad/s

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Anvisha [2.4K]

The Moon s escape speed will be smaller than Earth's.

  • What is escape speed:

The minimum speed that is required for an object to free itself from the gravitational force exerted by a massive object.

The formula of escape speed is

  • v = \sqrt{\frac{2GM}{R} }

where

v is escape velocity

G is universal gravitational constant

M is mass of the body to be escaped from

r is distance from the center of the mass

we can say that,

Escape speed depends on the gravity of the object trying to hold the spacecraft from escaping.

we know that,

The Moon's surface gravity is about 1/6th as powerful or about 1.6 meters per second per second.

since, v ∝ g

The Moon s escape speed will be smaller than Earth's.

Learn more about escape speed here:

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2 years ago
A child rolls a ball on a level floor 4.0 m to another child. if the ball makes 10.0 revolutions, what is its diameter?
olga_2 [115]
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3 years ago
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What is the name of the quantity represented by the symbol ω?
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In physics the quantity represented by \omega is the <u>angular velocity, </u>the units of the angular velocity are radians/second.

This is a quantity that represents the speed when the movement of an object follows a circular path.

Mathematically defined as follows:

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Where T is the period, the time it takes to complete a lap.

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A sample of oxygen gas at 25.0°c has its pressure tripled while its volume is halved. What is the final temperature of the gas?
Marat540 [252]

Answer:

447 K

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25 C = 25 + 273 = 298 K

Assuming ideal gas, we can apply the ideal gas law

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

T_2 = T_1\frac{P_2}{P_1}\frac{V_2}{V_1}

Since pressure is tripled, then P_2 / P_1 = 3. Volume is halved, then V_2 / V_1 = 0.5

T_2 = 298*3*0.5 = 447 K

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3 years ago
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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 25
Ivanshal [37]

Answer:

a) v(2\,s) = 5.9\,\frac{m}{s}, v(4\,s) = -13.7\,\frac{m}{s}, b) t = 2.602\,s, c) h(2.602\,s) = 35.176\,m, d) t = 5.281\,s, e) v(2.602\,s) = -26.254\,\frac{m}{s}

Explanation:

a) The velocity function is determined by deriving the position function in time:

v(t) = 25.5-9.8\cdot t

Velocities after 2 seconds and 4 seconds are, respectively:

v(2\,s) = 5.9\,\frac{m}{s}

v(4\,s) = -13.7\,\frac{m}{s}

b) The maximum height is reached when velocity is equal to zero:

25.5-9.8\cdot t = 0

The time when the projectile reaches the maximum height:

t = 2.602\,s

c) The maximum height is:

h (2.602\,s) = 2 + 25.5\cdot (2.602\,s)-4.9\cdot (2.602\,s)^{2}

h(2.602\,s) = 35.176\,m

d) The projectile hits the ground when height is equal to zero:

-4.9\cdot t^{2}+25.5\cdot t + 2 =0

The roots of the second order polynomial are presented below:

t_{1} \approx 5.281\,s

t_{2} \approx -0.077\,s

The first one is the only reasonable solution in physical terms.

t = 5.281\,s

e) The velocity of the projectile when it hits the ground is:

v(2.602\,s) = 25.5-9.8\cdot (5.281\,s)

v(2.602\,s) = -26.254\,\frac{m}{s}

4 0
3 years ago
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