Answer:
2.36 μ H
Explanation:
Given,
Number of turns= 90
diameter = 1.3 cm = 0.013 m
unscratched length = 57 cm = 0.57 m
Area, A = π r²
= π x 0.0065² = 1.32 x 10⁻⁴ m²
we know,


L = 2.36 μ H
Hence, the inductance of the unstretched cord is equal to 2.36 μ H
Answer:
Net Force Formula: F1+F2+F3...+FN
*(MAKE SURE YOU MAKE NOTE OF THE NEGATIVE FORCES AND SUBTRACT THEM!)*
Explanation:
Top left: 17N+25N+25N-42N= 25N
Top right: 65N+200N-65N-150N-200N= 150N
Bottom left: 189N+123N+284N-96N-188N-312N= 0N
Bottom right: 34N+77N-12N-34N= 65N
I hope this helps! :))
In a cathode ray tube, the number of electrons that reach the fluorescent screen is controlled by the grid. The correct option among all the options given in the question is option "B".When current is
supplied to the heater, it causes the cathode to emit electrons. The electrons
pass through opening of the grid before reaching the anode. By controlling the
number of electrons passing through the grid, the number of electrons reaching
the anode can also be controlled.<span>
</span>
<h2>
<em>Answer:</em></h2><h2>
<em>Regular </em><em>object</em></h2>
- <em>Those </em><em>substance </em><em>which </em><em>have </em><em>fixed </em><em>geometrical </em><em>shape </em><em>are </em><em>called </em><em>regular </em><em>object.</em>
- <em>For </em><em>example</em><em>:</em><em> </em><em>books,</em><em>pencils,</em><em> </em><em>basketball</em><em> </em><em>etc.</em>
<h2>
<em>Irregular </em><em>object</em></h2>
- <em>Those </em><em>substance </em><em>which </em><em>do </em><em>not </em><em>have </em><em>geometrical</em><em> </em><em>shape </em><em>are </em><em>called </em><em>irregular</em><em> </em><em>object</em><em>.</em>
- <em>For </em><em>example:</em><em> </em><em>a </em><em>piece </em><em>of </em><em>stone,</em><em>a </em><em>broken </em><em>piece </em><em>of </em><em>brick,</em><em>leaf </em><em>etc.</em>
<em>Hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em><em>.</em>
<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>