fyt as per the question the magnitude of two like charges are given as q1 and q2.the separation distance is given as r unit.hence the potential energy is given as-
here the potential energy is positive which means the force between two charges is repulsive.the potential energy is maximum which indirectly denotes that the system is unstable.due to this repulsion the smaller charge may accelerate.
coming to the same charges of opposite nature i.e unlike charges-here the magnitude of charges are same and separation distance is also are,so the potential energy will be given as-
here the potential energy is negative .so the system of two charges are attracted by each other.
Answer:
A pre-existing weather disturbance: A hurricane often starts out as a tropical wave.
Warm water: Water at least 26.5 degrees Celsius over a depth of 50 meters powers the storm.
Thunderstorm activity: Thunderstorms turn ocean heat into hurricane fuel.
For part a we can use the kinematic formula distance= velocity(initial)*time+1/2*acceleration*time^2
we know there is only velocity in the horizontal and none in the vertical so v initial in this case is 0, acceleration is gravity =9.81m/s^2
plug in .95=(1/2)*9.81*t^2 solve for t
t=.44seconds
Now for part b we know there is no horizontal acceleration so distance= velcoity* time
we will use horrizontial velocity and the time we just found
distance= .8m/s*.44seconds
=.35 meters
As per given conditions there are two directions along which forces are acting
1. Net force along left direction is given as
2. Net force towards right direction is given as
now since the two forces here in opposite direction so here we will have net force given as
so here net forces must be 440 N towards right
Answer:
W = 1.44 10⁻⁷ J
Explanation:
The expression for the job is
W = ∫ F. dx
Where the point is the scalar product in this case the direction of the meteor and the depth is parallel, whereby the scalar product is reduced to the ordinary product
W = 630 ∫ x³ dx
W = 630 x⁴ / 4
Let's evaluate between the lower limit x = 0, w = 0 to the upper limite the point at x = 5.5 10⁻³ m
W = 157.5 ((5.5 10⁻³)⁴ -0)
W = 1.44 10⁻⁷ J
