Answer:
25 m/s
Explanation:
from the question you van see that some detail is missing, however i found this same question using internet search engines on: 'https://www.chegg.com/homework-help/questions-and-answers/light-rail-passenger-trains-provide-transportation-within-cities-speed-slow-nearly-constan-q5808369'
here is the complete question:
'Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. A train travels through a congested part of town at 7.0m/s . Once free of this area, it speeds up to 12m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. What is the final Speed?'
SOLUTION
initial speed (u) = 7 m/s
final speed (v) = 13 m/s
initial acceleration time (t1) = 8 s
final acceleration time (t2) = 16 s
what is the higher cruising speed?
acceleration =
acceleration = = 0.75 m/s^{2}
since the train accelerates at the same rate, the increase in speed will be = acceleration x time (t2)
= 0.75 x 16 = 12 m/s
therefore the higher cruising speed = increase in speed + initial speed
= 12 + 13 = 25 m/s
Answer: False
Explanation: In order to explain this problem we have to use the Faraday law, which say
dФm/dt=-ε it means that the variation of the magnetic field flux with time is equal to the emf ( electromotive force). In our case the magnetic flux is constant then there is not a emf induced in a wire closed loop.
Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
A Parallel circuit has certain characteristics and basic rules: A parallel circuit has two or more paths for current to flow through. Voltage is the same across each component of the parallel circuit. The sum of the currents through each path is equal to the total current that flows from the source.