Red shift of distant galaxies
Answer: B)To the left of the charges.
Explanation: between the charges the electric field will not cancel but will be added since electric field lines from both charges point in the same direction. To the right of the charge the -4q will take over as it’s strength overcomes the strength of the +q charge. At this point the magnitude of +q will never reach a magnitude strong enough to cancel the -4q. To the left, it is further away from -4q and is closer to +q and electric field lines point in different direction
B, Studying how ones' character changes over time... That's the only one that has to deal with personality.
Answer:
Its diameter increases as it flows down from the pipe. Assuming laminar flow for the water, then Bernoulli's equation can be applied.
P1-P2 + (rho)g(h1 - h2) + 1/2(rho)(v1² - v2²) = 0
Explanation:
P1 = P2 = atmospheric pressure so, P1 - P2 = 0
h1 is greater than h2 so h1-h2 is positive. Rearranging the equation above 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho = v2²
From the continuity equation for fluids
A1v1 = A2v2
v2 = A1v1/A2
Substituting into the equation above
(A1v1/A2)² = 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho
Making A2² the subject of the formula,
A2² = (A1v1)²× rho/(2{ (rho)g(h1-h2) + 1/2(rho)v1²}
The denominator will be greater than the numerator and as a result the diameter of the flowing stream decreases.
Thank you for reading.
