Answer:
No. 67
Peter Street
12th Road
Chennai
24th June 201_
Dear Amrish
I have come to know that since your school has closed for the Autumn Break you have plenty of free time at your disposal at the moment. I would like to tell you that even I am having holidays now.
It has been a long time since we have spent some time together. If you are free, I would welcome to have your company this weekend. Why don’t you come over to my house and spend a day or so with me?
I am anxiously waiting for your reply.
Yours affectionately
your name
I did not see the post. But if she actually needs help She can dm me. keep y'all's head up❤
Explanation:
Given that,
Number of turns in the coil, N = 100
Area of the coil, A = 100 cm² = 0.01 m²
It is placed at an angle of 70°.
Magnetic field, B = 0.1 Wb/m²
We need to find the magnetic flux through the coil and the emf is induced in the coil after 10⁻³ s.
Magnetic flux is given by :

So, the magnetic flux through the coil is 0.1 Wb.
Emf induced in the coil is :

So, 34V of emf is induced in the coil.
Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N

Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
Answer:
Number of electrons, 
Explanation:
It is given that,
Resistance, R = 4 ohms
Current, I = 3 A
Time, t = 5 min = 300 s
We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.
i.e. q = n e ...............(1)
And current, 


e is the charge of an electron


So, the number of electrons pass through the resistor is
. Hence, this is the required solution.