.A particle moving with a constant acceleration has a velocity of 20 cm/s when its position is x = 10 cm. Its position 7.0 s later is x = –30 cm. What is the acceleration of the particle?

Explanation:

The second equation of a uniformly accelerated motion could be used to solve this problem. This is given by equation (1);

where u is the particle's initial velocity, t is the time taken, a is the acceleration and s is the distance travelled.

Given;

u = 20cm/s

t = 7s

a = ?

s = ?

The particle moved from one point to another point along the x-axis, where and . This information could be used to find the distance s covered by the object as follows;

We the make appropriate substitutions into equation (1) and then solve for the acceleration.

The negative sign is an indication that the particle is decelerating.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j