The acceleration of gravity is
9.8 m/s^2 down.
When an object falls out of a hand, its speed after 1.8s is
(9.8)x(1.8) = 17.6 m/s down.
It doesn't matter what it is, how much it weighs, or how high it was dropped from.
If it's more than 17.6 m/s, then this happened on a different, bigger planet.
If it's less than 17.6 m/s, then it must have hit something on the way down, like some air or something.
Let the rod be on the x-axes with endpoints -L/2 and L/2 and uniform charge density lambda (2.6nC/0.4m = 7.25 nC/m).
The point then lies on the y-axes at d = 0.03 m.
from symmetry, the field at that point will be ascending along the y-axes.
A charge element at position x on the rod has distance sqrt(x^2 + d^2) to the point.
Also, from the geometry, the component in the y-direction is d/sqrt(x^2+d^2) times the field strength.
All in all, the infinitesimal field strength from the charge between x and x+dx is:
dE = k lambda dx * 1/(x^2+d^2) * d/sqrt(x^2+d^2)
Therefore, upon integration,
E = k lambda d INTEGRAL{dx / (x^2 + d^2)^(3/2) } where x goes from -L/2 to L/2.
This gives:
E = k lambda L / (d sqrt((L/2)^2 + d^2) )
But lambda L = Q, the total charge on the rod, so
E = k Q / ( d * sqrt((L/2)^2 + d^2) )
Answer: electrical energy
Explanation:
Electrical energy
Answer:
tcucugxojfjgfojcigxuogoudyifodtufukdutfuocyjxuogu
Answer:
Explanation:
given that
mass = 10kg
distance = 4m
force = 50N
to calculate the workdone when the force is applied in the same direction of displacement
mathematically,
workdone = force × distance
Workdone = 50 × 4
workdone = 200 joules
2) to calculate the workdone at an angle of 30° with the displacement we apply the formula
workdone = force × distance × cos Ф
workdone = 50 × 4 × cos 30°
workdone = 200 × 0.866
workdone = 173 . 2 joules
