Explanation:
Given that,
Initial speed of the car, u = 88 km/h = 24.44 m/s
Reaction time, t = 2 s
Distance covered during this time, 
(a) Acceleration, 
We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :


s = 74.66 meters
s = 74.66 + 48.88 = 123.54 meters
(b) Acceleration, 


s = 37.33 meters
s = 37.33 + 48.88 = 86.21 meters
Hence, this is the required solution.
Answer:
The central blue square in between the lower pair of magnet has the least force of repulsion.
Explanation:
We can explain this using the dual nature of magnets.
Each magnet must have two poles namely:
-North pole
-South pole
We assume that the magnetic lines of forces enters from south pole and leaves from the north pole.
When brought together, like poles repel each other while opposite poles attract each other.
In the picture, the lower two magnets have opposite poles facing each other, hence the force of repulsion is minimum there and the force of attraction is maximum.
W=20 e(-kt)
A. Rearranging gives k= -(ln(w/20)/t
Substituting w= 10 and solving gives k=0.014
B. Using W=20e(-kt). After 0 hours, W=20. After 24 hours, W=14.29g. After 1 week (24x7=168h) W=1.9g
C. Rearranging gives t=-(ln(10/20)/k. Substituting w=1 and solving gives t=214 hours.
D. Differentiating gives dW/ dt = -20ke(-kt). Solving for t=100 gives dW/dt = 0.07g/h. Solving for t=1000 gives 0.0000002g/h
E. dW/dt = -20ke(-kt). But W=20e(-kt) so dW/dt = -kW
Answer:
(a)

(b) 

Explanation:
Let us take the north direction to be the positive y-axis and the east to be positive x-axis.
First day:
25.0 km southeast, which implies
south of east. The y-component will be negative and the x-component will be positive.


Second day:
She starts off at the stopping point of last day. This time, both the y- and x-components are positive.


Therefore, total displacements:


Magnitude of displacements,

Direction,

Answer:
32km per hour
Explanation:
Explanation:
In first case v = a t
==> a t = 40 km p h
Now distance covered S1 + S2 + S3
S1 = 1/2 a t^2 and S3 = 1/2 a t^2
But S2 = 3t * 40 = 120 t km
Hence total distance = at^2 + 120 t
Time taken (total) = t + 3t + t = 5 t
Hence average speed = at^2 + 120 t / 5 t
Cancelling t we have at + 120 / 5 = 40 + 120 / 5 = 160/5 = 32 km per hour