Calculate the number of millimeters in 2.13 kilometer.

Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to

jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x $10^{-8}$ J.

(b) The wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency of the photon is 2.47 x $10^{25}$ Hz.

Explanation:

Let;

$E_{1}$ = -13.60 ev

$E_{2}$ = -3.40 ev

(a) Energy of the emitted photon can be determined as;

$E_{2}$ - $E_{1}$ = -3.40 - (-13.60)

= -3.40 + 13.60

= 10.20 eV

= 10.20(1.6 x $10^{-9}$ )

$E_{2}$ - $E_{1}$ = 1.632 x $10^{-8}$ Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x $10^{-8}$ Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x $10^{-34}$ Js), c is the speed of light (3 x $10^{8}$ m/s) and λ is the wavelength.

10.20(1.6 x $10^{-9}$ ) = (6.6 x $10^{-34}$ * 3 x $10^{8}$ )/ λ

λ = $\frac{1.98*10^{-25} }{1.632*10^{-8} }$

= 1.213 x $10^{-17}$

Wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x $10^{-8}$ = 6.6 x $10^{-34}$ x f

f = $\frac{1.632*10^{-8} }{6.6*10^{-34} }$

= 2.47 x $10^{25}$ Hz

Frequency of the emitted photon is 2.47 x $10^{25}$ Hz.

6 0

3 years ago

A 1.5m wire carries a 6 A current when a potential difference of 61 V is applied. What is the resistance of the wire?

Alisiya [41]

Resistance = (voltage) / (current)

For this piece of wire . . .

Resistance = (61 volts) / (6 Amperes)

Resistance = (61/6) (V/A)

<em>Resistance = (10 and 1/6) ohms</em>

Since you know the voltage and current, the length doesn't matter.

7 0

3 years ago

Kayla is measuring how long it takes a ball to roll down a hill. The ball has a

Anarel [89]

Answer:

اخلع

تم

Explanation:

تتمنننبـ]بؤتر5ي لفق8

14زتععياغغءة

عنف ذي

8 0

3 years ago

Read 2 more answers

A crow is flying horizontally with a constant speed of 2.70m/s when it releases a claim from its beak. The clan lands on the roc

jenyasd209 [6]

Given:

Speed = 2.70 m

Time, t = 2.10 seconds

Let's solve for the following:

• (a) The horizontal component of the velocity.

To find the horizontal component, apply the formula:

$V_{ox}=V_o\cos \theta$

Where:

Vo is the initial speed = 2.70 m

θ = 0 degrees

Hence, we have:

$\begin{gathered} V_{ox}=2.70\cos 0 \\ \\ V_{ox}=2.7\text{ m/s} \end{gathered}$

The horizontal component of the velocity just before it lands is 2.70 m/s.

• (b) The vertical component of the velocity.

To find the vertical component, apply the formula:

$V_{oy}=V_{0y}-gt=\text{V}_{oy}\text{ sin}\Theta-gt$

Where:

g is the acceleration due to gravity = 9.8 m/s²

t is the time = 2.10 s

Hence, we have:

$\begin{gathered} V_{oy}=V_{oy}\sin \theta-gt \\ \\ V_{oy}=2.70\sin 0-9.8(2.10) \\ \\ V_{oy}=0-20.58 \\ \\ V_{oy}=-20.58\text{ m/s} \end{gathered}$

The vertical component of the velocity just before it lands is -20.58 m/s.

(c) Here, the initial speed is equal to the constant horizontal speed.

Therefore, in part (a) the horizontal component will increase in the x-direction if the speed of the crow is increased.

The initial vertical velocity is 0 m/s in both cases.

Therefore, in part (b) the vertical component will remain constant.

ANSWER:

(a) 2.70 m/s

(b) -20.58

(c) In part (a) the horizontal component will increase, while in part (b) the vertical component will remain constant.

6 0

1 year ago

In a loading a lorry a man lifts boxes each weight 200N through a height of 3.5m.

vekshin1

The work done on the boxes and the energy spent is 700J while the power required to lift 4 boxes per minute is 2.8kW

This work done on an object is the force required to move that object from point A to point B.

Data Given;

Weight (mg) = 200N
Distance (s) = 3.5m

This is the work done to move the boxes from point A to point B.

$w = F.s\\F = ma = mg = 200N\\w = 200 * 3.5\\w = 700J$

The work done to move the object is 700J

<h3>Energy</h3>

This is used to calculated the energy used to carry the work done and the formula is given as

$E = force * displacement\\E = F.s\\E = 200 * 3.5\\E = 700J$

The energy transferred when he lifts one box is 700J

<h3>Power</h3>

Power is the rate at which energy is used with respect to time

$P = E/t\\$

The energy required to lift 4 boxes in one minutes is

$700 * 4 = 2800$

Now we can calculate the power used

$P = Energy / time\\P = 2800 / 1 \\P = 2800W = 2.8kW$

The power required to move four boxes in 1 minutes is 2.8kW

From the calculations above, the following was gotten

work done = 700J
Energy = 700J
Power = 2800W or 2.8kW

Learn more on work done, energy and power here;

brainly.com/question/10063455

4 0

3 years ago