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viktelen [127]
3 years ago
14

A 30 cm scale has one end broken. The mark at the broken end is 2.6 cm. How would you use it to measure the length of your penci

l
pls tell correct ans
Physics
1 answer:
lidiya [134]3 years ago
6 0

Answer:

the mark of the broken end is 2.6 cm so, we use the scale from the next full mark i.e. 3cm

Explanation:

<em>we </em><em>now </em><em>measure</em><em> </em><em>the </em><em>length</em><em> </em><em>of </em><em>the </em><em>pencil</em><em> </em><em>by </em><em>keeping </em><em>the </em><em>3</em><em> </em><em>c</em><em>m</em><em> </em><em>mark </em><em>of </em><em>the </em><em>scale</em><em> </em><em>at </em><em>it's</em><em> </em><em>left </em><em>end.</em>

<em>The </em><em>3</em><em> </em><em>cm </em><em>value </em><em>is </em><em>then </em><em>subtracted</em><em> </em><em>from </em><em>the </em><em>scale</em><em> </em><em>reading</em><em> </em><em>at </em><em>the </em><em>right</em><em> </em><em>side </em><em>end </em><em>of </em><em>the </em><em>pencil</em><em> </em><em>to </em><em>obtain </em><em>the </em><em>correct</em><em> </em><em>length</em><em> </em><em>of </em><em>the </em><em>pencil.</em><em> </em><em>✏️</em>

<em>(</em><em>i </em><em>i </em><em>)</em><em> </em>place the scale in the contact with object along it's length

(2) Your eyes must be exactly in front of the point where the measurements to be taken.

Hope_it_helps_mga_ka_joiners_mwehehe

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Answer: It will remain the same

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According to <u>Coulomb's Law:</u>    

<em>"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them".  </em>

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Mathematically this law is written as:  

F_{E}=K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:  

F_{E}  is the electrostatic force  

K is the Coulomb's constant  

q_{1}=q_{2}=q are the electric charges , which in this case have the same positive charge

d is the separation distance between the charges

Rewritting we have:

F_{E}=K\frac{q^{2}}{d^{2}}    (2)

Now, if the first charge is doubled:

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And the second is reduced to a half:

q_{2}=\frac{1}{2}q

We will have the following:

F_{E}=K\frac{(2q)(\frac{1}{2}q)}{d^{2}} (3)

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Answer:

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The angular distance it covers when starting from rest:

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The angular distance it covers when coming to complete stop:

0 - \omega^2 = 2\alpha_o\theta_o

\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad

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