A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a
1 answer:
You might be interested in
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
Double
Explanation:
Since the period T of a pendulum is given by
By increasing the length of the pendulum by 4, the period becomes
You can see that the period doubles when we increase the length by a factor of 4.
Answer:
v=20m/S
p=-37.5kPa
Explanation:
Hello! This exercise should be resolved in the next two steps
1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed
Q=VA
for he exitt
Q=flow=5m^3/s
A=area=0.25m^2
V=Speed
solving for V
velocity at the exit=20m/s
for entry
2.
To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.
where
P=presure
α=9.810KN/m^3 specific weight for water
V=speed
g=gravity
solving for P1
the pressure at exit is -37.5kPa