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sattari [20]
3 years ago
12

A uniform ladder of length 10.8 m is leaning against a vertical frictionless wall. The weight of the ladder is 323 N, and it mak

es an angle of 1.16 radians with the floor. A man weighing 734 N climbs slowly up the ladder. When he has climbed to a point that is 7.46 m from the base of the ladder (measured along the ladder), the ladder starts to slip. Find the coefficient of static friction between the floor and the ladder.
Physics
1 answer:
love history [14]3 years ago
8 0

Answer:

0.3625

Explanation:

From the given information:

Consider the equilibrium conditions;

On the ladder, net torque= 0

Thus,

\tau_{net} = 0; and

-fL \s in \theta +m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta = 0

However, by rearrangement;

fL \s in \theta =m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta  \\ \\  \mu(m_L + m) gL \ sin \theta = (323  \ N) ( 10.8 \ meters) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos  \  66.46^0

\mu= \dfrac{ (323  \ N) ( 10.8 \ m) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos  \  66.46^0}{\Big [(323 \ N)+(734 \ N) \Big] (10.8 \ m)}

\mathbf{\mu= 0.3625 }

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Answer:

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Given:

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To calculate the root mean squarer or rms speed of hydrogen molecule, we use the given formula:

v_{rms} = \sqrt{\frac{3TR}{m}}

where

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Putting the values in the above formula:

v_{rms} = \sqrt{\frac{3\times 3.13\times 8.314}{2.02\times 10^{-3}}}

v_{rms} =196.59 m/s

5 0
3 years ago
A car collides into a concrete wall going 25.0 m/s . It stops in 0.141 seconds and has a change in momentum of 39,400. What is t
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Answer:

Mass of the car is 1576 kg.

Explanation:

Let the mass of the car be m kg.

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As the car stops, final velocity of the car is, v=0\ m/s

Change in momentum is, \Delta p=39400

Now, we know that, momentum is given as the product of mass and velocity.

So, change in momentum is given as:

\Delta p=m(u-v)\\39400=m(25-0)\\39400=25m\\m=\frac{39400}{25}\\m=1576\ kg

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4 0
3 years ago
the time required for one cycle, a complete motion that returns to its starting point, it called the_____ medium frequency perio
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Answer:

The correct answer to the following question will be "Period".

Explanation:

The Period seems to be the time deemed necessary for such a perfect cycle of vibration to transfer a particular moment. Because as the amplitude of the wave raises, the wavelength falls.

It is denoted by "T" and its formula will be:

⇒  T  = \frac{1}{F}

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The other given choices are not related to the given circumstances. So that the above would be the right answer.

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3 0
3 years ago
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The free-fall acceleration on Mars is 3.7 m/s^2. What length of pendulum has a period of 1.0 s on Earth? What length of pendulum
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Answer:

Explanation:

Given

Free fall acceleration on mars g_{m}=3.7\ m/s^2

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Time period of Pendulum is given by

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L'=\frac{3.7}{39.48}

L'=0.0936\ m

L'=9.36\ cm                            

3 0
3 years ago
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