Answer:
I think the right answer is option B.
Explanation:
(i)
O is the object and I is the image.
The image formed is enlarged and it is erect. So the magnification will be positive (+) and greater than 1.
Refer above image. 1
(ii)
O is the object and I is the image.
The image formed is diminished and erect. So the magnification will be positive (+) and less than1.
Refer above image. 2
(iii)
The image will be formed as the 2F on the other side of the lens and it will be of same of the object.
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:
The answer is below
Explanation:
Let vₐ be the speed of airplane = 135 mph, vₙ be the speed of the wind = 70 mph and vₐₙ be the speed of the airplane relative to the wind.
The distance (d) = 135 miles, Δt = 1 hour, vₐₙ = 135 miles / 1 hour = 135 mph
vₐ = vₙ + vₐₙ
vₐ = vₐₙ
Therefore, vₐ, vₐₙ, vₙ can be represented by an isosceles triangle since vₐ = vₐₙ.
The direction of the wind θ is:
sin(θ / 2) = vₙ / 2vₐ
sin(θ / 2) = 70/ (2*135)
sin(θ / 2) = 0.2593
θ / 2 = sin⁻¹(0.2593) = 15
θ = 30⁰
2α = 180° - 30°
2α = 150°
α = 75°
a) The direction of the wind is 75° in the south east direction while the airplane is heading 30° in the north east direction.