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) xf = 5.1 m
b) u = 0.304
c) x = 10.3 m
Explanation:
we will use the following formula:
u = 0.1 + A*x
Si x = 12.5 m, u = 0.6
Clearing A:
A = 0.5/12.5 = 0.04 m^-1
a) we have to:
W = Ekf - Eki
where Ekf = final kinetic energy
Eki = initial kinetic energy
9.8*(0.1xf + ((0.04*xf^2)/(2))) = (4.5^2)/(2)
Clearing xf, we have:
xf = 5.1 m
b) Replacing values for u:
u = 0.1 + (0.04*5.1) = 0.304
c) Wf = Ekf - Eki
-u*m*x*g = 0 - (m*v^2)/2
Clearing x:
x = v^2/(2*u*g) = (4.5^2)/(2*0.1*9.8) = 10.3 m
Answer:
V(car) = V(truck) at t = Dt/2
acceleration = v(car) = D/t^2
Explanation:
acceleration = v(car) = D/t^2
Since the average velocities must be the same, the car's final velocity must be twice the trunk velocity assuming the car start with zero velocity, since acceleration remain the same throughout the journey velocities at half-time point must be equal.
Let Karen's forward speed be considered as positive.
Therefore, before the headband is tossed backward, the speed of the headband is
V = 9 m/s
The headband is tossed backward relative to Karen at a speed of 20 m/s. Therefore the speed of the headband relative to Karen is
U = -20 m/s
The absolute speed of the headband, relative to a stationary observer is
V - U
= 9 + (-20)
= - 11 m/s
Answer:
The stationary observes the headband traveling (in the opposite direction to Karen) at a speed of 11 m/s backward.
Decomposing the vector b on the x-axis and the y-axis, we get a rectangle triangle where the two sides are bx (x-axis) and by (y-axis), and b is the hypothenuse.
The component in x, bx, is equal to the product between the hypothenuse and the cosine of the angle between b and the x-axis, which is
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