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3 years ago

The number of energy levels to which an electron can jump depends on the

1 answer:

8 0

C. The amount of energy it absorbs.

Electrons can only transition between energy levels when they absorb or release certain quanta of energy.

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Review please help.

iragen [17]

Answer:

1 and 3

Explanation:

because they are going up from 0

7 0

3 years ago

Unpolarized light with an intensity of 655 W / m2 is incident on a polarizer with an unknown axis. The light then passes through

Norma-Jean [14]

Answer:

1. $\theta=29.84^{0}$

2. $\theta=60.15^{0}$

Explanation:

Polarizes axis can create two possible angles with the vertical.

first we have to find the intensity of first polarizer

which is given as

$I=\frac{I_{0} }{2}$

$I= \frac{655\frac{W}{M^{2} } }{2}$

$I=327.5\frac{W}{m^{2} }$

For a smaller angle for the first polarizer:

According to Malus Law

$I_{2} =I_{1} Cos^{2}(90^{0} - \theta)$

$I_{2} =I_{1} sin^{2}\theta$

$\frac{I_{2} }{I_{1} }=Sin^{2}\theta$

taking square root on both sides

$\sqrt{\frac{163}{327.5} } = sin\theta$

$\theta=Sin^{-1}(0.4977)$

$\theta=29.84^{0}$

For a larger angle for the first polarizer:

According to Malus Law

$I_{2} =I_{1} cos^{2}\theta$

$\frac{I_{2} }{I_{1} }=Cos^{2}\theta$

taking square root on both sides

$\sqrt{\frac{163}{327.5} } = cos\theta$

$\theta=Cos^{-1}(0.4977)$

$\theta=60.15^{0}$

7 0

3 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago

A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab

shutvik [7]

Answer:

a) θ₁ = 0.487º , b) t = 0.400 s , x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

y = y₀ + $v_{oy}$ t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

x = v₀ₓ t

t = x / v₀ₓ

We replace

y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

v_{oy} = v₀ sin θ

v₀ₓ = vo cos θ

y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

Sec² θ = 1 - tan² θ

1 = 120 tan θ - 0.0213 (1 –tan²θ)

0.0213 tan²θ + 120 tanθ -1.0213 = 0

We change variables

u = tan θ

u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

u = [- 5633.8 ± 5633.82] / 2

u₁ = 0.0085

u₂= -5633.81

u = tan θ

θ = tan⁻¹ u

For u₁

θ₁ = tan⁻¹ 0.0085

θ₁ = 0.487º

For u₂

θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

x = v₀ₓ t

t = x / v₀ cos θ

t = 120 / (300 cos 0.487)

t = 0.400 s

The deer must be at a distance of

v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

x = v t

x = 29.33 0.4

x = 11.73 ft

3 0

3 years ago

An object that looks white when exposed to sunlight reflects all colors of light. What

Ierofanga [76]

It looks blue as it is only reflecting blue light

6 0

2 years ago