Answer:
c. 2 MeV.
Explanation:
The computation of the binding energy is shown below
![= [Zm_p + (A - Z)m_n - N]c^2\\\\=[(1) (1.007825u) + (2 - 1 ) ( 1.008665 u) - 2.014102 u]c^2\\\\= (0.002388u)c^2\\\\= (.002388) (931.5 MeV)\\\\=2.22 MeV](https://tex.z-dn.net/?f=%3D%20%5BZm_p%20%2B%20%28A%20-%20Z%29m_n%20-%20N%5Dc%5E2%5C%5C%5C%5C%3D%5B%281%29%20%281.007825u%29%20%2B%20%282%20-%201%20%29%20%28%201.008665%20u%29%20-%202.014102%20u%5Dc%5E2%5C%5C%5C%5C%3D%20%280.002388u%29c%5E2%5C%5C%5C%5C%3D%20%28.002388%29%20%28931.5%20MeV%29%5C%5C%5C%5C%3D2.22%20MeV)
= 2 MeV
As 1 MeV = (1 u) c^2
hence, the binding energy is 2 MeV
Therefore the correct option is c.
We simply applied the above formula so that the correct binding energy could come
And, the same is to be considered
Answer:
1 cm⁻¹ =1.44K 1 ev = 1.16 10⁴ K
Explanation:
The relationship between temperature and thermal energy is
E = K T
The relationship of the speed of light
c =λ f = f / ν 1/λ= ν
The Planck equation is
E = h f
Let's start the transformations
c = f λ = f / ν
f = c ν
E = h f
E = h c ν
E = KT
h c ν = K T
T = h c ν / K =( h c / K) ν
Let's replace the constants
h = 6.63 10⁻³⁴ J s
c = 3 10⁸ m / s
K = 1.38 10⁻²³ J / K
v = 1 cm-1 (100 cm / 1 m) = 10² m-1
T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²
A = h c / K = 1,441 10⁻²
T = 1.44K
ν = 103 cm⁻¹ = 103 10² m
T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 103 10²
T = 148K
1 Rydberg = 1.097 10 7 m
As we saw at the beginning the λ=1 / v
T = (h c / K) 1 /λ
T = 1,441 10⁻² 1 / 1,097 10⁷
T = 1.3 10⁻⁹ K
E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J
E = KT
T = E/K
T = 1.6 10⁻¹⁹ /1.38 10⁻²³
T = 1.16 10⁴ K
Answer:
μk = 0.26885
Explanation:
Conceptual analysis
We apply Newton's second law:
∑Fx = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
a= -0.9 m/s²,
g = 9.81 m/s² : acceleration due to gravity
W= 75 N : Block weight
W= m*g
m = W/g = 75/9.8= 7.65 kg : Block mass
Friction force : Ff
Ff= μk*N
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W-25 = 0
N = 75
+25
N= 100N
∑Fx = m*ax
20-Ff= m*ax
20-μk*100
= 7.65*(-0.90 )
20+7.65*(0.90) = μk*100
μk = ( 20+7.65*(0.90)) / (100)
μk = 0.26885
The work-energy theorem says that the total work done on the block is equal to the difference of its kinetic energies at points B and A. Then the total work done on the block is
Friction acts on the block to oppose its motion, so it does negative work on the block, -4.5 J.
The only other force acting on the block as it moves is the force <em>P</em>. Let 
be the work done by the force <em>P</em>. Then the total work done on the block is
Answer:
only ten meters north from starting point
Explanation:
