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gtnhenbr [62]
3 years ago
8

A third class lever has a mechanical advantage of <1. What is an example of a third class lever and why use it?

Physics
2 answers:
AVprozaik [17]3 years ago
7 0

B.) Baseball bat; increases velocity

All third class levers have a mechanical advantage less than 1. Since the output end has a longer distance from the fulcrum than the input point, the output end moves at a greater velocity than the input point. Because of this, third class levers are commonly used when trying to hit an object with as much velocity as possible.


34kurt3 years ago
5 0
<span>Baseball bat. The handle of the bat is the fulcrum. Exerting a force from the handle supplies the input force just near the middle, while the other end of the baseball bat pushes the ball with the output forces. The input force is greater than the output force but the output load is able to move farther, and this increases the ball's velocity.</span>
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Two loudspeakers are placed on a wall 2 m apart. A listener stands directly in front of one of the speakers, 81.7 m from the wal
OverLord2011 [107]

Answer:

The phase difference is       \Delta \phi = 1.9995 rad  

Explanation:

From the question we are told that

    The distance between the  loudspeakers is d = 2m

     The distance of the listener from the wall  D = 81.7 \ m

     The frequency of the  loudspeakers is  f = 4450Hz

      The velocity of sound is v_s = 343 m/s

     

The path difference of the sound wave that is getting to the listener is mathematically represented as

        \Delta z  =\sqrt{d^2 + D^2} -D

Substituting values

        \Delta z  =\sqrt{2^2 + 81.7^2 } -81.7

       \Delta z  =0.0245m

The phase difference is mathematically represented as

           \Delta \phi =  \frac{2 \pi}{\lambda } *  \Delta z

Where \lambda is the wavelength which is mathematically represented as

          \lambda  = \frac{v_s }{f}

substituting value  

          \lambda  = \frac{343 }{4450}

        \lambda  = 0.0770 m

Substituting value into the  equation for phase difference

      \Delta \phi =  \frac{2 * 3.142 * 0.0245}{0.0770}

      \Delta \phi = 1.9995 rad  

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3 years ago
Which is true of the buoyant force?
victus00 [196]

It acts in the upward direction.

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Use this free body diagram to help you find the magnitude of the force F2 needed to keep this block in static equilibrium. WILL
oksian1 [2.3K]
Static equilibrium means that all forces are equal, so make this easiest you want to break F1 into it's horizontal and vertical components. As there are no other forces acting in the horizontal, we know the horizontal component of F1 is 40N. This allows the vertical component to be found using pythagorus theorem. After finding the vertical and horizontal components, you just have to add the vertical components to find the difference between the up and down.

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skelet666 [1.2K]

Answer:

B is the best answer for the question

6 0
3 years ago
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Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea
melisa1 [442]

Answer:

m = 1.26*10²⁵ kg.

Explanation:

Assuming that the mass of the stone is much smaller than the mass of the planet, we can get the mass, applying the Universal Law of  Gravitation to both masses, as follows:

Fg = G* ms* mp / rp²

Now, if we apply Newton's 2nd Law to the mass of the stone, we can get the gravitational acceleration, as follows:

Fg = ms*a = ms*g ⇒ g = G*mp / rp²

First of all, we need to get the value of g.

Assuming that this acceleration is constant, we can appy the kinematic equations to this situation.

We know that the stone is thrown upward with an initial velocity vo = 15 m/s.

At the highest point in the trajectory, just before of changing direction, the stone comes momentarily to a stop.

At this point, applying the definition of acceleration, we can write:

vf = vo -g*t ⇒ 0 = vo -gt ⇒ g = vo/t (1)

We have the total time since the stone was thrown upwards, not the one used for the upward trajectory.

It can be showed, using the expression for the displacement (which is the same in both directions) that the time used for going up, it's the same used to go down, so the time that we need to put in (1). is just the half of the total time.

So, replacing in (1) we get the value of g, as follows:

g = 15 m/s / 4.5 s = 3.33 m/s²

Now, we can replace this value in the equation that gives us g based in the Universal Law of Gravitation, as follows:

g=G*mp / rp² (2)

Before solving for mp, however, we need to get the value of the radius of the planet.

Assuming that it's a perfect sphere, we can get this value from the value of the circumference at the planet's equator:

rp = 2*π*rp / 2*π ⇒ rp = 1.0*10⁵ km / 2*π = 15,915 km.

With this value for  rp, we can solve (2) for mp, as follows:

mp= g*rp² / G = 3.33 m/s² * (15,915 km)² / 6,67*10⁻¹¹ N.m²/kg²

mp = 1.26*10²⁵ kg.

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