Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.
Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m
Now you need the final speed to use it as initial speed of the next part.
Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s
Part B) Free fall
Maximum height, y max ==> Vf = 0
Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s
ymax = yo + Vo*t - g[t^2] / 2
ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m
Answer: ymax = 10084.2m
Answer:
L = 1.15 m
Explanation:
The diffraction phenomenon is described by the equation
a sin θ = m λ
Where a is the width of the slit, λ the wavelength and m is an integer, the order of diffraction is left.
The diffraction measurements are made on a screen that is far from the slit, and the angles in the experiment are very small, let's use trigonometry
tan θ = y / L
tan θ = sint θ / cos θ≈ sin θ
We substitute in the first equation
a (y / L) = m λ
The first maximum occurs for m = 1
The distance is measured from the center point of maximum, which coincides with the center of the slit, in this case the distance is the total width of the central maximum, so the distance (y) measured from the center is
y = 1.15 / 2 = 0.575 cm
y = 0.575 10⁻² m
Let's clear the distance to the screen (L)
L = a y / λ
Let's calculate
L = 115 10⁻⁶ 0.575 10⁻² / 575 10⁻⁹
L = 1.15 m
Answer:
I don't know why you are asking me?
Answer:
The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm
Explanation:
Given;
wavelength of ultraviolet light, λ = 270 nm
work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J
The energy of the ultraviolet light is given by;
The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;
E = φ + K.E
K.E = E - φ
K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )
K.E = 3.677 x 10⁻¹⁹ J
K.E = ¹/₂mv²
mv² = 2K.E
velocity of the electron is given by;
the shortest de Broglie wavelength for the electrons is given by;
Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm
