Answer:
<h2><em>
12.45eV</em></h2>
Explanation:
Before calculating the work function, we must know the formula for calculating the kinetic energy of an electron. The kinetic energy of an electron is the taken as the difference between incident photon energy and work function of a metal.
Mathematically, KE = hf - Ф where;
h is the Planck constant
f is the frequency = c/λ
c is the speed of light
λ is the wavelength
Ф is the work function
The formula will become KE = hc/λ - Ф. Making the work function the subject of the formula we have;
Ф = hc/λ - KE
Ф = hc/λ - 1/2mv²
Given parameters
c = 3*10⁸m/s
λ = 97*10⁻⁹m
velocity of the electron v = 3.48*10⁵m/s
h = 6.62607015 × 10⁻³⁴
m is the mass of the electron = 9.10938356 × 10⁻³¹kg
Substituting the given parameters into the formula Ф = hc/λ - 1/2mv²
Ф = 6.63 × 10⁻³⁴*3*10⁸/97*10⁻⁹ - 1/2*9.11*10⁻³¹(3.48*10⁵)²
Ф = 0.205*10⁻¹⁷ - 4.555*10⁻³¹*12.1104*10¹⁰
Ф = 0.205*10⁻¹⁷ - 55.163*10⁻²¹
Ф = 0.205*10⁻¹⁷ - 0.0055.163*10⁻¹⁷
Ф = 0.1995*10⁻¹⁷Joules
Since 1eV = 1.60218*10⁻¹⁹J
x = 0.1995*10⁻¹⁷Joules
cross multiply
x = 0.1995*10⁻¹⁷/1.60218*10⁻¹⁹
x = 0.1245*10²
x = 12.45eV
<em>Hence the work function of the metal in eV is 12.45eV</em>
The rate at which the height is changing is ( 5 / x ) m / hr
We know that,
Area of an equilateral triangle A =
/ 4
h =
x / 2
Where,
x = Side
h = Height
Given that,
dA / dt = 5
/ hr
h =
x / 2
Differentiate both sides with respect to t
dh / dt = (
/ 2 ) ( dx / dt )
dx / dt = ( 2 /
) ( dh / dt )
A =
/ 4
Differentiate both sides with respect to t
dA / dt = (
/ 4 ) ( 2x ) ( dx / dt )
5 = (
/ 4 ) ( 2x ) ( 2 /
) ( dh / dt )
dh / dt = ( 5 / x ) m / hr
Rate of change of height is defined as the rate at which height of an object changes with respect to time. It is represented as dh / dt
Therefore, the rate at which the height is changing is ( 5 / x ) m / hr
To know more about Rate of change of height
brainly.com/question/13283964
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Answer:
![\large\boxed{\large\boxed{10ft/s}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%5Clarge%5Cboxed%7B10ft%2Fs%7D%7D)
Explanation:
The question, translated, is:
- <em>A steel ball rolls and falls off the edge of a table from 4ft above the floor. If you hit the ground 5ft from the base of the table, what was your initial horizontal velocity?</em>
<em />
<h2>Solution</h2>
<em />
This is a projectile motion, for which, the equations that you will need are:
![V_x_0= V_x=x\cdot t](https://tex.z-dn.net/?f=V_x_0%3D%20V_x%3Dx%5Ccdot%20t)
![y=y_0+Vy_o\cdot t-g\cdot t^2/2](https://tex.z-dn.net/?f=y%3Dy_0%2BVy_o%5Ccdot%20t-g%5Ccdot%20t%5E2%2F2)
<u />
<u>1. Calculate the time that it takes the ball to fall 4ft</u>
![0=4ft-g\cdot t^2/2\\\\t^2=2\times 4ft/( 32.174ft/s^2)=0.24865s^2\\\\t=0.4986s](https://tex.z-dn.net/?f=0%3D4ft-g%5Ccdot%20t%5E2%2F2%5C%5C%5C%5Ct%5E2%3D2%5Ctimes%204ft%2F%28%2032.174ft%2Fs%5E2%29%3D0.24865s%5E2%5C%5C%5C%5Ct%3D0.4986s)
<u />
<u>2. Calculate the horizontal velocity:</u>
![V_x_0= V_x=x\cdot t\\\\V_x_0=5ft/0.4986s=10.027ft/s\approx 10ft/s](https://tex.z-dn.net/?f=V_x_0%3D%20V_x%3Dx%5Ccdot%20t%5C%5C%5C%5CV_x_0%3D5ft%2F0.4986s%3D10.027ft%2Fs%5Capprox%2010ft%2Fs)
Answer:
245.25 J
Explanation:
Potential Energy = m g h
= 10 * 9.81 * 2.5 = 245.25 J
Answer:
The value is ![E = 8.9 *10^{-5} \ J](https://tex.z-dn.net/?f=E%20%3D%20%208.9%20%2A10%5E%7B-5%7D%20%5C%20%20J)
Explanation:
From the question we are told that
The area is ![A = 0.700 \ m^2](https://tex.z-dn.net/?f=A%20%3D%20%200.700%20%5C%20%20m%5E2)
The root mean square value is ![E_{rms} = 0.0400 \ V/m](https://tex.z-dn.net/?f=E_%7Brms%7D%20%3D%20%200.0400%20%5C%20%20V%2Fm)
The time taken is ![t = 30.0 \ s](https://tex.z-dn.net/?f=t%20%3D%20%2030.0%20%5C%20%20s)
Generally the energy is mathematically represented as
![E = c * \sepsilon_o * A * t * E_{rms}^2](https://tex.z-dn.net/?f=E%20%3D%20%20c%20%2A%20%20%5Csepsilon_o%20%2A%20%20A%20%20%2A%20%20t%20%20%2A%20E_%7Brms%7D%5E2)
=> ![E = 3.0*10^{8} * 8.85*10^{-12} * 0.700 * 30 * (0.04)^2](https://tex.z-dn.net/?f=E%20%3D%20%203.0%2A10%5E%7B8%7D%20%2A%20%208.85%2A10%5E%7B-12%7D%20%2A%20%200.700%20%2A%20%2030%20%2A%20%280.04%29%5E2)
=> ![E = 8.9 *10^{-5} \ J](https://tex.z-dn.net/?f=E%20%3D%20%208.9%20%2A10%5E%7B-5%7D%20%5C%20%20J)