To be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving.
First, we manipulate the one of the kinematic equations
v^2 = v0^2 + 2 (a) (x) where v = 0 since the car stopped
Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,
<span>we get v0 = sqrt (2(a)(x))
Substituting the known values,
v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s
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Therefore, before stopping the car the original speed of the car would be 14.49 m/s
Answer:
The speed does it head toward the goal = 41.87 
Explanation:
Mass = 0.107 kg
Initial velocity ( u ) = 0
Force (F) = 28 N
Time = 0.16 sec
From newton's second law, Force = mass × acceleration
⇒ F = m × a
⇒ 28 = 0.107 × a
⇒ a = 261.7
--------- (1)
This is the value of acceleration.
Final speed of the mass is calculated by the equation V = U + at
⇒ U = 0 because mass in in rest position at start.
⇒ V = a t
Put the values of acceleration and time in above formula we get
⇒ V = 261.7 × 0.16
⇒ V = 41.87 
Therefore the speed does it head toward the goal = 41.87 
Answer:
25000 V
Explanation:
The formula for potential is
V = Kq/r
Potential at B due to the charge placed at origin O
V1 = K q / OB

V1 = 10000 V
Potential at B due to the charge placed at A
V2 = K q / AB

V2 = 15000 V
Total potential at B
V = V1 + V2 = 10000 + 15000 = 25000 V