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Complete Question
Three equal point charges are held in place as shown in the figure below
If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.
A) F1=2F2
B) F1=3F2
C) F1=4F2
D) F1=9F2
Answer:
D) F1=9F2
Explanation:
We are told in the question that there are three equal point charges.
q, Q1, Q2 ,
q = Q1 = Q2
From the diagram we see the distance between the points d
q to Q1 = d
Q1 to Q2 = nd
Assuming n = 2
= 2 × d = 2d
Sum of the two distances = d + 2d = 3d
F1 is the force on q due to Q1 and
F2 is the force on q due to Q2,
Since we have 3 equal point charges and a total sum of distance which is 3d
Hence,
F1 = 9F2
Answer:
Action and reaction are the same
Explanation:
Let us carefully analyze the situation in the context of Newton's third law, when the car hits the mosquito exerts a force that acts on the mosquito, ACTION and the mosquito responds to the force that is being applied with a force of equal magnitude and direction or opposite applied to the car (Reaction).
Consequently, the correct answer is: The force of the car on the mosquito is of the same magnitude as that of the mosquito on the car
A. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π
b. vy = 4π cos (4πt + π/2)
vy = 0
c. m = sin(4πt + π/2) / [<span>πt + cos(4πt + π/2)]
d. m = </span>sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]
e. t = -1.0
f. t = -0.35
g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax
h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for vymax
i. s(t) = [<span>x(t)^2 + y</span>(t)^2]^(1/2)
h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt
k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.
The big bang produced dark energy, which accounts for some of the energy needed to expand the universe.
The vastness of space may contain a type of matter known as “dark matter.”
The universe is currently expanding at an accelerating rate.
Hope this helps !