The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
Explanation:
So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.
So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then
v = 25.8 + (-1.66×8.3)
v =12.022 m/s.
So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
No because there must be an even # if their is an even amount one of the forces isn’t being cancelled
Answer:
Control of air–fuel ratio
Oxygen sensors tell the ECU whether the engine is running rich (too much fuel or too little oxygen) or running lean (too much oxygen or too little fuel) as compared to ideal conditions (known as stoichiometric).
Explanation:
Answer:
a = 9.94 m/s²
Explanation:
given,
density at center= 1.6 x 10⁴ kg/m³
density at the surface = 2100 Kg/m³
volume mass density as function of distance

r is the radius of the spherical shell
dr is the thickness
volume of shell

mass of shell


now,

integrating both side



we know,




a = 9.94 m/s²