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:

Explanation:
given,
Radius of the solid rod, R = 5.3 cm
Electric field strength,E = 22 kN/C
Let the volume charge density be ρ
From Gauss law

ε₀ is the permitivity of free space
R is the radius of the rod
and also,

ρ is the volume charge density



Hence, the volume charge density is equal to 
Answer:
A. DT is given by Q= MCs DT
m = mass of the substances
Cs= is it's specific heat capacity
Ck= <u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>Q</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
Mk ×DTk
=<u>2</u><u>5</u><u>0</u><u> </u><u>×</u><u> </u><u>9</u><u> </u><u>×</u><u> </u><u>5</u><u> </u><u> </u>
129
=Dt = 180.1085271
answer is 180degree C.
Explanation:
B. = <u>2</u><u>5</u><u>×</u><u>1</u><u>0</u> ×100
1.082
=<u>2</u><u>5</u><u>0</u><u>0</u>
1.082
= 23105.360 g/kj.
Pluto is the last planet discovered in our solar system.
Answer:
proton and neutron has the same mass of 1amu