Answer:
Self inductance,
Explanation:
It is given that,
Length of the coil, l = 5 cm = 0.05 m
Area of cross section of the coil,
Number of turns in the coil, N = 130
The self inductance relates the magnetic flux linkage to the current through the coil and it is given by :
L = 0.000127 Henry
or
So, the self inductance of the coil is 127 microhenry. Hence, this is the required solution.
Answer:
Below
Explanation:
First draw the vectors that represent both electric fields.
E1 is the elictric field created by q1, E2 is the one created by q2.
● q1 is negative so E1 will point from P.
● q2 is positive so E2 will point out of P
(Picture below)
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The resulting electric field is equal to the sum of the two fields since both vectors are colinear.
Let E be the total field.
● E = E1 + E2
The formula of the electric field intensity is:
● E = K ×(q/d^2)
-K is Coulomb's constant
-d is the distance between the charge and the object ( here P)
-q is the charge
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● E1 = K × (q1/d1^2)
The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)
Coulombs constant is 9×10^9 m^2/C^2
● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]
● E1 = -359.43 N/C
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● E2 = K ×(q2/d^2)
The distance between q2 and P is 0.25 m.
● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]
● E2 = 463.68 N/C
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● E = E1 + E2
● E = -359.43+463.68
● E = 105.25 N/C
Answer:
The mass of the cube is 420.8 kg.
Explanation:
Given that,
Length of edge = 38.9 cm
Density
We need to calculate the volume of cube
Using formula of volume
We need to calculate the mass of the cube
Using formula of density
Hence, The mass of the cube is 420.8 kg.