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viktelen [127]
2 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]2 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:

Acceleration of gravity on Noveria = 4.4 m/s²

Explanation:

Commander Shepard, an N7 spectre for Earth, weighs 799 N on the Earth's surface.

We have weight, W = mg

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Mass of Shepard, m = 81.45 kg            

She lands on Noveria, a distant planet in our galaxy, she weighs 356 N.

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Assume a rectangular strip of a material with an electron density of n=5.8x1020 cm-3. The strip is 8 mm wide and 0.8 mm thick an
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Answer: I = 111.69 pA

Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.

The hall voltage of a semiconductor sensor is given below as

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Where V = hall voltage = 1.5mV =1.5/1000=0.0015V

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n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3

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By slotting in the parameters, we have that

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I = 111.69 pA

3 0
2 years ago
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