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3 years ago

Have you ever heard about chemical bond

Physics

2 answers:

Alex Ar [27]3 years ago

6 0

Answer:

well yea I did So hmmmmmm

Harman [31]3 years ago

3 0

Yes, I have heard about chemical bond.

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Which statement is true regarding coppers ability to conduct electricity

Alecsey [184]

Copper is the second best known electrical conducting substance. The first best one is silver. We almost always use copper because silver costs too much.

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3 years ago

Two identical bowling balls are moving down a bowling alley so that their centers of mass have the same velocity, but one just s

tamaranim1 [39]

Answer:

Rolling Ball

Explanation

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3 years ago

Help me with this please

pochemuha

Answer:

1. 31.5

3. 3.5

Explanation:

4 0

3 years ago

Calculate the ionization potential for C+5 ( 5 electrons removed for the C atom) and in addition compute the wavelength of the t

sveta [45]

Answer:

Ionization potential of C⁺⁵ is 489.6 eV.

Wavelength of the transition from n=3 to n=2 is 1.83 x 10⁻⁸ m.

Explanation:

The ionization potential of hydrogen like atoms is given by the relation :

$E = \frac{13.6Z^{2} }{n^{2} } eV$ .....(1)

Here E is ionization potential, Z is atomic number and n is the principal quantum number which represents the state of the atom.

In this problem, the ionization potential of Carbon atom is to determine.

So, substitute 6 for Z and 1 for n in the equation (1).

$E = \frac{13.6\times(6)^{2} }{1^{2} }$

 E = 489.6 eV

The wavelength (λ) of the photon due to the transition of electrons in Hydrogen like atom is given by the relation :

$\frac{1}{\lambda} =RZ^{2}[\frac{1}{n_{1} ^{2}}-\frac{1}{n_{2} ^{2} }]$ ......(2)

R is Rydberg constant, n₁ and n₂ are the transition states of the atom.

Substitute 6 for Z, 2 for n₁, 3 for n₂ and 1.09 x 10⁷ m⁻¹ for R in equation (2).

$\frac{1}{\lambda} =1.09\times10^{7} \times6^{2}[\frac{1}{2 ^{2}}-\frac{1}{3 ^{2} }]$

$\frac{1}{\lambda}$ = 5.45 x 10⁷

λ = 1.83 x 10⁻⁸ m

7 0

3 years ago

A rocket with a total mass of 330,000 kg when fully loaded burns all 280,000 kg of fuel in 250 s. The engines generate 4.1 MN of

lutik1710 [3]

Answer:

6900 m/s

Explanation:

The mass of the rocket is:

m = 330000 − 280000 (t / 250)

m = 330000 − 1120 t

Force is mass times acceleration:

F = ma

a = F / m

a = F / (330000 − 1120 t)

Acceleration is the derivative of velocity:

dv/dt = F / (330000 − 1120 t)

dv = F dt / (330000 − 1120 t)

Multiply both sides by -1120:

-1120 dv = -1120 F dt / (330000 − 1120 t)

Integrate both sides. Assuming the rocket starts at rest:

-1120 (v − 0) = F [ ln(330000 − 1120 t) − ln(330000 − 0) ]

-1120 v = F [ ln(330000 − 1120 t) − ln(330000) ]

1120 v = F [ ln(330000) − ln(330000 − 1120 t) ]

1120 v = F ln(330000 / (330000 − 1120 t))

v = (F / 1120) ln(330000 / (330000 − 1120 t))

Given t = 250 s and F = 4.1×10⁶ N:

v = (4.1×10⁶ / 1120) ln(330000 / (330000 − 1120×250))

v = 6900 m/s

4 0

3 years ago