It’s will be B because the circuit had a open or close so if that doesn’t work than it’s because it’s open
Answer:

Explanation:
As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from
to
is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.
The total distance covered by the train during the entire journey is the area of the speed-time graph.
Area


As velocity is in
and time is in
so the unit of area is 
Hence, the total distance is
.
Answer:
Net charge contained in the cubeq= 3.536×10^-6C
Explanation:
Formular for total flux in a cube is given as:
Total flux= E300Acos(180) + E200Acos(0)
Where A is crossectional area
Total flux= A(E200-E300)
Total flux= q/Eo
q= Eo×total flux
q=(8.84×10^-12)×(100)^2×(100-60)
q= 3.536×10^-6C
Answer:
-30=5(x+1) is -7
Explanation:
distribute flip subtract 5 from both sides divide both sides by 5