Answer:
3.53×10⁶ N/c due west
Explanation:
From the question
E = F'/q........................ Equation 1
Where E = Electric Field, F = Net Force, q = Charge.
But,
F' = F₂-F₁...................... Equation 2
Substitute equation 2 into equation 1
E = (F₂-F₁)/q................ Equation 3
Given: F₁ = 3 N due east, F₂ = 15 N due west, q = 3.4×10⁻⁶ C
Substitute these values into equation 1
E = (15-3)/(3.4×10⁻⁶)
E = 12/(3.4×10⁻⁶)
E = 3.53×10⁶ N/c due west
B. 60 cm
All parallel light rays are bent through the focal point of a convex lens, so the rays from the flashlight 150 cm above the floor must go through the same point on the principal axis as the rays from the flashlight 120 cm above the floor. The location of the focal point does not change when the position of the object is moved either vertically or horizontally.
Hope this helps !
<span>From the point of view of the astronaut, he travels between planets with a speed of 0.6c. His distance between the planets is less than the other bodies around him and so by applying Lorentz factor, we have 2*</span>√1-0.6² = 1.6 light hours. On the other hand, from the point of view of the other bodies, time for them is slower. For the bodies, they have to wait for about 1/0.6 = 1.67 light hours while for him it is 1/(0.8) = 1.25 light hours. The remaining distance for the astronaut would be 1.67 - 1.25 = 0.42 light hours. And then, light travels in all frames and so the astronaut will see that the flash from the second planet after 0.42 light hours and from the 1.25 light hours is, 1.25 - 0.42 = 0.83 light hours or 49.8 minutes.
Period and frequency are mutual reciprocals.
Period = 1 / frequency .
Frequency = 1 / period
(Frequency) x (Period) = 1
