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V125BC [204]
2 years ago
14

A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance x from its center. It undergoes harm

onic oscillations by swinging back and forth under the influence of gravity. Randomized Variables M=2.2 kg x= 0.49 m 33% Part (a) In terms of M, L, and x, what is the rod' s moment of inertia / about the pivot point?
Physics
1 answer:
Alisiya [41]2 years ago
5 0

Answer:

The moment of inertia is 0.7500 kg-m².

Explanation:

Given that,

Mass = 2.2 kg

Distance = 0.49 m

If the length is 1.1 m

We need to calculate the moment of inertia

Using formula of moment of inertia

I=\dfrac{1}{12}ml^2+mx^2

Where, m = mass of rod

l = length of rod

x = distance from its center

Put the value into the formula

I=\dfrac{1}{12}\times2.2\times(1.1)^2+2.2\times(0.49)^2

I=0.7500\ kg-m^2

Hence, The moment of inertia is 0.7500 kg-m².

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A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Volgvan

Answer:

A=0.199

Explanation:

We are given that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

Frequency of oscillation=\nu=1.2Hz

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

A=\sqrt{0.0398}=0.199

Hence, the amplitude of oscillation=A=0.199

4 0
2 years ago
Person lifting a chair convert from what energy to another
vagabundo [1.1K]
A person lifting a chair is converting chemical energy to mechanical energy.
4 0
2 years ago
Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i
SCORPION-xisa [38]

Answer:

The value is  r =  5.077 \  m

Explanation:

From the question we are told that

   The  Coulomb constant is  k =  9.0 *10^{9} \  N\cdot  m^2  /C^2

   The  charge on the electron/proton  is  e =  1.6*10^{-19} \  C

    The  mass of proton m_{proton} =  1.67*10^{-27} \  kg

    The  mass of  electron is  m_{electron } =  9.11 *10^{-31} \ kg

Generally for the electron to be held up by the force gravity

   Then    

       Electric force on the electron  =  The  gravitational Force

i.e  

            m_{electron} *  g  = \frac{ k *  e^2  }{r^2 }

         \frac{9*10^9 *  (1.60 *10^{-19})^2  }{r^2 }  =     9.11 *10^{-31 }  *  9.81

         r =  \sqrt{25.78}

         r =  5.077  \  m

7 0
2 years ago
An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in ser
s2008m [1.1K]

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

V_L=V_oe^{-\frac{Rt}{L}}

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s

hence, after 1.6s the inductor will have a potential difference of 24V

3 0
2 years ago
A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri
vodka [1.7K]
We know that tangential acceleration is related with radius and angular acceleration according the following equation:  
at = r * aa  
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)  
So the radius is r = d/2 = 1.2/2 = 0.6 m  
Then at = 0.6 * 5 = 3 m/s2  
Tangential acceleration of a point on the flywheel rim is 3 m/s2
5 0
2 years ago
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