Drew wants to convert square mils to circular mils in an area calculation. How does he do this?

An electron accelerates through a 12.5 V potential difference, starting from rest, and then collides with a hydrogen atom, excit

Yanka [14]

Answer:

Initial state Final state

3 ⇒ 2

3 ⇒ 1

2 ⇒ 1

Explanation:

For this exercise we must use Bohr's atomic model

E = - 13.606 / n²

where is the value of 13.606 eV is the energy of the ground state and n is the integer.

The energy acquired by the electron in units of electron volt (eV)

E = e V

E = 12.5 eV

all this energy is used to transfer an electron from the ground state to an excited state

ΔE = 13.6060 (1 / n₀² - 1 / n²)

the ground state has n₀ = 1

ΔE = 13.606 (1 - 1/n²)

1 /n² = 1 - ΔE/13,606

1 / n² = 1 - 12.5 / 13.606

1 / n² = 0.08129

n = √(1 / 0.08129)

n = 3.5

since n is an integer, maximun is

n = 3

because it cannot give more energy than the electron has

From this level there can be transition to reach the base state.

Initial state Final state

3 ⇒ 2

3 ⇒ 1

2 ⇒ 1

8 0

2 years ago

Name:

Brums [2.3K]

Answer:

1.a) 1 kJ

1.b) 4 kJ

ratio 1:4

1.c) 4 times as before

2.a) 3.33 m/s2

Explanation:

1.a) bicycle's velocity =Displacement/time

=100/20 m/s

=5 m/s

bicycler's KE =1/2 *mass*(velocity)^2

=1/2*80*5^2

=1000 J = 1 kJ

1.b) bicycle's new velocity =200/20 m/s

=10 m/s

bicycler's new KE =1/2*80*10^2

=4000 J = 4 kJ

Ratio= KE 1 :KE new

= 1 :4

1.c) when bicycler's speed was doubled it increased the KE by 4 times (2^2). because In KE we consider the square of the speed , so the factor we increase the speed , the KE will get increased with the square value of it

ex : speed is triple the prior value , then the KE is as 3^2 times as before. that is 9 times

2.a) car acceleration = (20-0)/6 m/s2

= 3.33 m/s2

4 0

2 years ago

This illustration represents the compoundA)carbon oxide.B)carbon dioxide.C)carbon monoxide.EliminateD)monocarbon oxide.

pentagon [3]

B.) Carbon Dioxide because the carbon is surrounded by oxygen

3 0

2 years ago

Read 2 more answers

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

Known Data

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

Time of the dropped Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

Time of the Thrown Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$