The formula for energy of motion is KE = .5 x m x v^2
Ke= Kinetic Energy in Joules
m = Mass in Kilograms
v = Velocity in Meters per Second
Solution :
Given weight of Kathy = 82 kg
Her speed before striking the water, = 5.50 m/s
Her speed after entering the water, = 1.1 m/s
Time = 1.65 s
Using equation of impulse,
Here, F = the force ,
dT = time interval over which the force is applied for
= 1.65 s
dP = change in momentum
dP = m x dV
= 82 x (1.1 - 5.5)
= -360 kg
∴ the net force acting will be
= 218 N
Answer:
a) 23.2 e V
b) energy of the original photon is 36.8 eV
Explanation:
given,
energy at ground level = -13.6 e V
energy at first exited state = - 3.4 e V
A photon of energy ionized from ground state and electron of energy K is released.
h ν₁ - 13.6 = K
K combine with photon in first exited state giving out photon of energy
= 26.6 e V
h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°
K + ( 3.4 ) = 26.6 e V
a) energy of free electron
K = 26.6 - 3.4 = 23.2 e V
b) energy of the original photon
h ν₁ - 13.6 = K
h ν₁ = 23.2 + 13.6
= 36.8 e V
energy of the original photon is 36.8 eV
They stay with the microscope as it moves around to different schools, and they are always located in the same classroom where the rest of the microscope is being used.