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Rus_ich [418]
4 years ago
7

A girl moves quickly to the center of a spinning merry-go-round, traveling along the radius of the merry-go-round. Which of the

following statements are true?Check all that apply.Check all that apply.The angular speed of the system increases.The moment of inertia of the system remains constant.The angular speed of the system decreases.The moment of inertia of the system increases.The moment of inertia of the system decreases.The angular speed of the system remains constant.
Physics
1 answer:
DochEvi [55]4 years ago
3 0

Answer:

The angular speed of the system increases.

The moment of inertia of the system decreases.

Explanation:

As we know that the girl is going towards the center of the circle so here the moment of inertia of the girl is given as

I = mr^2

here we know that

r = position of the girl from the center of the disc

now we know that the girl is moving towards the center so its distance will continuously decreasing

So the moment of inertia of the girl will decrease

Now we know that that with respect to the center of the disc there is no torque on the disc + girl system

So here we can use angular momentum conservation

So we have

I\omega = constant

since moment of inertia is decreasing for the system

so angular speed will increase

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What is the gradual process through which humans change from birth to<br> adulthood?
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Answer:

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3 years ago
How long must a flute be in order to have a fundamental frequency of 262 Hz (this frequency corresponds to middle C on the evenl
spin [16.1K]

Answer:

L=0.654 m

Explanation:

<u>Concepts and Principles  </u>

1- The speed of sound in air is expressed as a function of the temperature of air as follows:  

 v=(331 m/s)√(1+T_C/273°C)                        (1)

where 331 m/s is the speed of sound in air at temperature 0°C and Tc is the temperature of air in Celsius.  

<u>Standing Wave Patterns in Pipes:  </u>

A pipe open at both ends can have standing wave patterns with resonant frequencies:  

f=v/λ=nv/2L                     n=1,2,3.........

where v is the speed of sound in air.  

<u>Given Data </u>

f_1 (fundamental frequency of the flute) = 262 Hz

T (temperature of the air) = 20°C  

The flute is open at both ends.  

<u>Required Data </u>

We are asked to determine the length of the tube.  

<u>Solution</u><u>  </u>

The speed of sound in air at temperature T = 20°C is found from Equation (1):

 v=(331 m/s)√(1+T_C/273°C)  

 =342.91 m/s

The fundamental frequency of the flute is found by substituting n = 1 into Equation (2):  

f=v/2L

Solve for L:  

L=v/2f_1

L=0.654 m

7 0
3 years ago
A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?
rosijanka [135]

Answer:

a

     l_o  =52 \  m

b

      l = 37.13 \ LY

Explanation:

From the question we are told that

    The  speed of the spaceship is  v  =  0.800c

    Here  c is the speed of light with value  c =  3.0*10^{8} \ m/s

    The  length is  l = 31.2 \  m

     The  distance of the star for earth is d = 145 \  light \  years

     The  speed is v_s = 2.90 *10^{8}

     

Generally the from the length contraction equation we have that

       l  =  l_o  \sqrt{1 -[\frac{v}{c } ]}

Now the when at rest the length is  l_o

So  

      l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }

      l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }

      l_o=52 \  m

Considering b  

  Applying above equation

            l  =l_o \sqrt{1 -  [\frac{v}{c } ]}

Here l_o  =145 \  LY(light \ years )

So

           l=145 *  \sqrt{1 -  \frac{v_s^2}{c^2 } }

            l =145 *  \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }

            l = 37.13 \ LY

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Answer:

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Explanation:

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