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trasher [3.6K]
4 years ago
6

How may stars are in the sky

Physics
2 answers:
Veseljchak [2.6K]4 years ago
7 0
Is impossible say how many stars are in the sky haha
sergeinik [125]4 years ago
4 0
It is impossible of how much stars in the sky
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relationship between the length (l) and ohmic conductor, its thickness (d) and the opposition (R) it offers a current passage​
Free_Kalibri [48]

Answer:

Ohm’s Law states that current is proportional to voltage; circuits are ohmic if they obey the relation V=IR.

Explanation:

LEARNING OBJECTIVES

Contrast shape of current-voltage plots for ohmic and non-ohmic circuits

7 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 <em>E</em> is ionization potential, <em>Z</em> is atomic number and <em>n</em> 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 <em>Z</em> and 1 for <em>n</em> in the equation (1).

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

<em> E = </em>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
4 years ago
Water flows through a 4.50-cm inside diameter pipe with a speed of 12.5 m/s. At a later position, the pipe has a 6.25-cm inside
jek_recluse [69]

Given,

The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m

The initial speed of the water, v₁=12.5 m/s

The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m

From the continuity equation,

\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}

Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.

On substituting the known values,

\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}

Thus, the flow rate of the water at the later position is 5.99 m/s

4 0
1 year ago
Please help me this is a test and it needs to be done in a few​
Amanda [17]

Answer:

I think A because astroids are stronger than a crash. Hope I helped! :)

Explanation:

7 0
3 years ago
Read 2 more answers
A 650 kg car accelerates at 4.0 m/s2 south. what is the net force acting on it?
Lelu [443]
Fnet=mass x acceleration 
       =650 x 4
       =2600N
8 0
4 years ago
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