Well, since the question GIVES you the initial velocity, the acceleration, and the time, and ASKS for the final velocity, you'd be smart to find an equation that USES the initial velocity, the acceleration, and the time, and FINDS the final velocity.
Have a look at equation B ..... vf = vi + a*t . That's pretty durn close !
vf = (initial velocity) + (acceleration)*(time)
vf = (3 m/s) + (5 m/s²)*(4 sec)
vf = (3 m/s) + (20 m/s)
vf = 23 m/s
Answer:
9.5 m/s
Explanation:
Distance, S = 150m
Acceleration, a = 0.3 m/s^2
Initial velocity, u = 0 m/s
Final velocity, v
Use kinematics equation
v^2 - u^2 = 2aS
v^2 - 0 = 2*0.3*150 = 90
v = sqrt(90) = 9.49 m/s
Answer/ Explanation
the loop has no force and the wires cancel each other but the wire on the loop has a force
The segment a get the same force for both but in the loop segment b get an opposite force so the net force on the loop is smaller
The loop wire has forces that all cancel out while the other straight wire doesn't.
Answer:
In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before.
Explanation:
