Answer:
force of the breaks is 6650 N, direction opposite to direction of movement
Explanation:
can i get the question so that i can answer your question
The maximum force that the tires can exert on the road before slipping is 16200 N.
From the information in the question;
The coefficient of static friction = 0.9
The mass of the car = 1800 kg
Using the formula;
μ = F/R
μ = coefficient of static friction
F = force on the tires
R = the reaction force
But recall that the reaction is equal in magnitude to the weight of the car.
W=R
Hence; R = 1800 kg × 10 ms-2 = 18000 N
Making F the subject of the formula;
F = μR
Substituting values;
F = 18000 N × 0.9
F = 16200 N
Hence, the maximum force that the tires can exert on the road before slipping is 16200 N.
Learn more: brainly.com/question/18754989
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>
