The time required by the car to stop is 4.916 sec.
Since the car is moving with the constant deceleration we can apply the first equation of motion to calculate the time required by the car to stop.
The first equation of motion is given as
V=u+at
Here, V=final speed of the car=0 mi/h as the car stops
u =initial speed of the car=55 mi/hr=24.58 m/s
a= acceleartion =-5 m/s^2 (here negative sign indicates for deceleration)
Now applying the values in the first equation
V=u+at
0=24.58-5*t
t=4.916 sec
Therefore the car will stops in 4.916 sec.
Electrons carry a negative charge.
Answer: 70 N to the East
Explanation:
lets assume east is positive and west is negative, since they are in opposite directions the net external force = F1+F2
Net force = (-60) + 130
Net force = 70
or
Net force = 70 N in the east direction
Answer:
v_f = 30 m/s
Explanation:
v_f = v_I + a*t
v_f = 0m/s + (9.81m/s²)*(3s)
v_f = 29.43m/s
v_f = 30 m/s
Since the object is being launched horizontally off the cliff the only vertical force acting on the ball is gravity which is an acceleration of 9.81m/s². So we have all the flowing information to use the the kinematic equation: v_f = v_I + a*t.
