Before we find impulse, we need to find the initial and final momentum of the ball.
To find the momentum of the ball before it hit the floor, we need to figure out its final velocity using kinematics.
Values we know:
acceleration(a) - 9.81m/s^2 [down]
initial velocity(vi) - 0m/s
distance(d) - 1.25m [down]
This equation can be used to find final velocity:
Vf^2 = Vi^2 + 2ad
Vf^2 = (0)^2 + (2)(-9.81)(-1.25)
Vf^2 = 24.525
Vf = 4.95m/s [down]
Now we need to find the velocity the ball leaves the floor at using the same kinematics concept.
What we know:
a = 9.81m/s^2 [down]
d = 0.600m [up]
vf = 0m/s
Vf^2 = Vi^2 + 2ad
0^2 = Vi^2 + 2(-9.81)(0.6)
0 = Vi^2 + -11.772
Vi^2 = 11.772
Vi = 3.43m/s [up]
Now to find impulse given to the ball by the floor we find the change in momentum.
Impulse = Momentum final - momentum initial
Impulse = (0.120)(3.43) - (0.120)(-4.95)
Impulse = 1.01kgm/s [up]
It's commonly referred to as an electric current.
Answer:
E = 0.18 J
Explanation:
given,
Potential of the battery,V = 9 V
Charge on the circuit, Q = 20 m C
= 20 x 10⁻³ C
energy delivered in the circuit
E = Q V
E = 20 x 10⁻³ x 9
E = 180 x 10⁻³
E = 0.18 J
Energy delivered in the circuit is equal to E = 0.18 J
