This problem can be solved based on the rule of energy conservation, as the energy of the photon covers both the energy needed to overcome the binding energy as well as the energy of ejection.
The rule can be written as follows:
energy of photon = binding energy + kinetic energy of ejectection
(hc) / lambda = E + 0.5 x m x v^2 where:
h is plank's constant = 6.63 x 10^-34 m^2 kg / s
c is the speed of light = 3 x 10^8 m/sec
lambda is the wavelength = 310 nm
E is the required binding energy
m is the mass of photon = 9.11 x 10^-31 kg
v is the velocity = 3.45 x 10^5 m/s
So, as you can see, all the parameters in the equation are given except for E. Substitute to get the required E as follows:
(6.63x10^-34x3x10^8)/(310x10^-9) = E + 0.5(9.11 x 10^-31)(3.45x10^5)^2
E = 6.41 x 10^-16 joule
To get the E in ev, just divide the value in joules by 1.6 x 10^-19
E = 4.009 ev
Answer:
(A) = 3.57 m
Explanation:
from the question we are given the following:
diameter (d) = 3.2 m
mass (m) == 42 kg
angular speed (ω) = 4.27 rad/s
from the conservation of energy
mgh = 0.5 mv^{2} + 0.5Iω^{2} ...equation 1
where
Inertia (I) = 0.5mr^{2}
ω = \frac{v}{r}
equation 1 now becomes
mgh = 0.5 mv^{2} + 0.5(0.5mr^{2})(\frac{v}{r})^{2}
gh = 0.5 v^{2} + 0.5(0.5)(v)^{2}
4gh = 2v^{2} + v^{2}
h = 3v^{2} ÷ 4 g .... equation 2
from ω = \frac{v}{r}
v = ωr = 4.27 x (3.2 ÷ 2)
v = 6.8 m/s
now substituting the value of v into equation 2
h = 3v^{2} ÷ 4 g
h = 3 x (6.8)^{2} ÷ (4 x 9.8)
h = 3.57 m
Answer:
Explanation:
For a charge moving perpendicularly to a magnetic field, the force experienced by the charge is given by:
where
q is the magnitude of the charge
v is the velocity
B is the magnetic field strength
In this problem,
So the force experienced by the electrons is
Answer:
0.3677181864 m
Explanation:
u = Velocity = 1.5 m/s
= Angle = 20°
y = -20 cm
Velocity components
Acceleration components
Time taken is 0.26088 seconds
The distance the beetle travels on the ground is 0.3677181864 m
The volumes are 200cm3 and 0.0002m3
