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Alexandra [31]
3 years ago
14

An experiment is conducted to determine how the arm length of a pendulum effects its period. The experimental design calls for v

arying the length of the arm and the mass of the bob. Which is true about this experiment, based on good experimental design?
a. Only the arm length should be changed. The mass should remain the same.
b. The force applied to the mass should be varied for trial.
c. The material the arm is made of should be varied for each trial.
d. Only the mass should be changed. The arm length should remain the same
Physics
2 answers:
Anestetic [448]3 years ago
8 0

Okay so don't quote me on this but I believe the answer is A) I'm saying this because B and C make no sense. and you can't change the mass of something without changing it totally.

Vanyuwa [196]3 years ago
6 0

Answer:

The answer is A

Explanation:

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A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax
Colt1911 [192]

Answer:

From the question we are told that

  The length of the rod is  L_o

    The  speed is  v  

     The angle made by the rod is  \theta

     

Generally the x-component of the rod's length is  

     L_x =  L_o cos (\theta )

Generally the length of the rod along the x-axis  as seen by the observer, is mathematically defined by the theory of  relativity as

       L_xo  =  L_x  \sqrt{1  - \frac{v^2}{c^2} }

=>     L_xo  =  [L_o cos (\theta )]  \sqrt{1  - \frac{v^2}{c^2} }

Generally the y-component of the rods length  is mathematically represented as

      L_y  =  L_o  sin (\theta)

Generally the length of the rod along the y-axis  as seen by the observer, is   also equivalent to the actual  length of the rod along the y-axis i.e L_y

    Generally the resultant length of the rod as seen by the observer is mathematically represented as

     L_r  =  \sqrt{ L_{xo} ^2 + L_y^2}

=>  L_r  = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}

=>  L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}

=>   L_r  = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}

=> L_r =  \sqrt{L_o^2 * cos^2(\theta)  [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}

=> L_r  =  \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }

=> L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

Hence the length of the rod as measured by a stationary observer is

       L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

   Generally the angle made is mathematically represented

tan(\theta) =  \frac{L_y}{L_x}

=>  tan {\theta } =  \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }

=> tan(\theta ) =  \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }

Explanation:

     

     

       

7 0
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When the balloon sticks to the wall (assuming it sticks to the wall). It is
inn [45]
The answer is false
8 0
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A point charge is used to determine the electric field around a charged particle. Why is it necessary that the point charge does
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The strength, and possibly the shape and direction, of the electric field
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If the process of measuring the field causes the particle to move, then
the measurement you get wouldn't mean anything. 

Your measurements wouldn't show the ACTUAL field around the particle.
They would show what the field is like AFTER something comes along
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It would be like carrying a flame thrower into a freezer when you go in
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Or if you had to measure how much light is leaking into a dark room,
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The velocity of a particle is described
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Answer:

hope you like it

Explanation:

To find velocity, we take the derivative of the original position equation. To find acceleration, we take the derivative of the velocity function. To determine the direction of the particle at t = 1 t=1 t=1, we plug 1 into the velocity function.

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