A box of groceries requires five newtons of force to lift it one meter to the counter. How many joules of work is done?

Atomic hydrogen produces a well-known series of spectral lines in several regions of the electromagnetic spectrum. Each series f

navik [9.2K]

Answer:

n₁ = 3

Explanation:

The energy of the states in the hydrogen atom is explained by the Bohr model, the transitions heal when an electron passes from a state of higher energy to another of lower energy,

ΔE = $E_{nf}$ - E₀ = - k²e² / 2m (1 / $n_{f}$ ²2 - 1 / n₀²)

The energy of this transition is given by the Planck equation

E = h f = h c / λ

h c / λ = -k²e² / 2m (1 / no ²- 1 / no²)

1 / λ = Ry (1/ $n_{f}$ ² - 1 / n₀²)

Let's apply these equations to our case

λ = 821 nm = 821 10⁻⁹ m

E = h c / λ

E = 6.63 10⁻³⁴ 3 10⁸/821 10⁻⁹

E = 2.423 10⁻¹⁹ J

Now we can use the Bohr equation

Let's reduce to eV

E = 2,423 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹) = 1,514 eV

$E_{nf}$ - E₀ = -13.606 (1 / $n_{f}$ ² - 1 / n₀²) [eV]

Let's look for the energy of some levels

n $E_{n}$ (eV) $E_{nf}$ - E $E_{ni}$ (eV)

1 -13,606 E₂-E₁ = 10.20

2 -3.4015 E₃-E₂ = 1.89

3 -1.512 E₄- E₃ = 0.662

4 -0.850375

We see the lines of greatest energy for each possible series, the closest to our transition is n₁ = 3 in which a transition from infinity (n = inf) to this level has an energy of 1,512 eV that is very close to the given value

8 0

3 years ago

The index of refraction of n-propyl alcohol is 1.39. Find the angle of refraction of light in that medium if light comes from ai

iogann1982 [59]

Answer:

36.11 degrees

Explanation:

index of refraction n = sin i/sinr

i is the angle of incidence

r is the angle of refraction

Substitute into the expression

1.39 = sin55/sin(r)

1.39 = 0.8191/sin(r)

sin(r) = 0.8191/1.39

sin(r) = 0.5893

r = arcsin(0.5893)

r = 36.11

hence the angle of refraction of light is 36.11 degrees

7 0

3 years ago

I wonder if people that say that there ugly is to get attention (not to be mean...)

alexandr1967 [171]

They do. mostly because they want validation from others

5 0

2 years ago

Read 2 more answers

A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial

Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

$t=\frac{2usin\theta}{g}$

Here θ = 30° and t = 20 s

a) $t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s$

Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

$H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m$

Maximum altitude = 490.5 m

c) Range of projectile is given by

$R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m$

Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

We have equation of motion s = ut + 0.5 at²

Horizontal motion

u = 169.91 m/s

a = 0 m/s²

t = 15 s

Substituting

s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

Vertical motion

u = 98.1 m/s

a = -9.81 m/s²

t = 15 s

Substituting

s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

$\texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m$

Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0

4 years ago

Can a moving car be a frame of reference?

GalinKa [24]

Answer:can a moving car be a frame of reference?

Therefore, a frame of reference can be in motion, and that motion can be of several different types. Anyone who has ever taken a ride in a car has been in a moving frame of reference and has experienced the several different types of motion through which this frame can move.

Explanation:

6 0

3 years ago