Im not exactly sure but I think the answer is techtonic plates collide
We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
where
is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
where
is the final speed of the sled
is the initial speed
is the distance covered
By rearranging the equation, we find d:
Answer:
el plomo será el más largo
Explanation:
Dado que;
longitud inicial (l1) = 4m
Longitud final l2
aumento de temperatura (θ) = 10 ° C
Coeficiente de expansión lineal α
Ahora para el hierro;
α = 11,7 x 10-6
Desde;
l2-l / l1θ = α
l2 = α l1θ + l1
l2 = l1 (αθ + 1)
l2 = 4 ((11,7 x 10-6 * 10) + 1)
l2 = 4.00044 m
Para el plomo
l2 = 4 ((27,3 x 10-6 * 10) + 1)
l2 = 4,00109 m
Para cobre
l2 = 4 ((16,7 x 10-6 * 10) + 1)
l2 = 4.000668 m
Por lo tanto, el plomo será el más largo
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>
