The position function x(t) of a particle moving along an x axis is 
a) The point at which particle stop, it's velocity = 0 m/s
So dx/dt = 0
0 = 0- 12t = -12t
So when time t= 0, velocity = 0 m/s
So the particle is starting from rest.
At t = 0 the particle is (momentarily) stop
b) When t = 0

SO at x = 4m the particle is (momentarily) stop
c) We have 
At origin x = 0
Substituting

t = 0.816 seconds or t = - 0.816 seconds
So when t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.
Answer:

Explanation:
Given:
Initial velocity of the vehicle, 
distance between the car and the tree, 
time taken to respond to the situation, 
acceleration of the car after braking, 
Using equation of motion:
..............(1)
where:
final velocity of the car when it hits the tree
initial velocity of the car when the tree falls
acceleration after the brakes are applied
distance between the tree and the car after the brakes are applied.

Now for this situation the eq. (1) becomes:
(negative sign is for the deceleration after the brake is applied to the car.)
Answer:
0.647 nC
Explanation:
The force experienced by a charge due to the presence of an electric field is given by

where
q is the charge
E is the magnitude of the electric field
In this problem, each antenna is modelled as it was a single point charge, experiencing a force of

Therefore, if the electric field magnitude is

Then the charge on each antenna would be

Answer:
Explanation:
total weight acting downwards
= 3g + 10g
13 g
volume of lead = 10 / 11.3 = .885 cm³
Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g
Buoyant force on lead = .885 x 1 x g
total buoyant force = vg + .885 g
For floating
vg + .885 g = 13 g
v = 12.115 cm³
total volume of bobber
= 4/3 x 3.14 x 2³
= 33.5 cm³
fraction of volume submerged
= 12.115 / 33.5
= .36
= 36 %
Answer:
<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>
Explanation:
The initial charge on 4 mF capacitor = 4 mf x 50 V = 200 mC
The initial Charge on 6 mF capacitor = 6 mf x 30 V =180 mC
Since the negative ends are joined together the total charge on both capacity would be;
q = 
q = 200 - 180
q = 20 mC
In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage
q = (4 x V) + (6 x V)
20 = 10 V
V = 2 V
For the final charge on 6.0 mF;
q = CV
q = 6.0 mF x 2 V
q = 12 mC
Therefore the final charge on the 6.0 mF capacitor would be 12 mC