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kondaur [170]
2 years ago
7

Anyone who wants to shw booooobs than u can see my body/cqg-fhfh-ptx​

Physics
1 answer:
zlopas [31]2 years ago
7 0
The answer yes, she gladly will
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Four identical particles of mass 0.980 kg each are placed at the vertices of a 4.14 m x 4.14 m square and held there by four mas
Zielflug [23.3K]

Answer:

a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²

b) I = 50.39 kg m²

c) I = 16.8 kg m²

Explanation:

a) Given data:

m = 0.98 kg

a = 4.14 * 4.14

The moment of inertia is:

I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}

For 4 particles:

I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}

c)

I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}

8 0
2 years ago
The Balmer series in hydrogen atom produces only infrared emission. O True O False
lorasvet [3.4K]

Answer:

False

Explanation:

As we know that, the Balmer series gives the n values as,

n_{i}=2.

[tex]n_{f}=3,4,5,.....\infty.

Now the value of wavelength can be calculated as,

\frac{1}{\lambda}=R(\frac{1}{n_{i} }-\frac{1}{n_{f} } )z^{2}.

Here, R=109677 cm^{-1}.

And n_{f}=3.

Now,

\frac{1}{\lambda}=109677 cm^{-1}(\frac{1}{2}-\frac{1}{3}).

Therefore,

\lambda=\frac{6}{109677} cm\\\lambda=547\times 10^{-9} m\\ \lambda=547 nm

Therefore, the wavelength of Balmer series lies in visible region which is 547 nm.

8 0
3 years ago
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. If force (F), work (W) and velocity (v) are taken as fundamental quantities.
alex41 [277]

Answer:

∴ [T]=[WF−1V−1]

Hope this answer is right!!

7 0
2 years ago
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3. Si usted duplicara la amplitud de un M.A.S. ¿cómo cambiaría la frecuencia, velocidad máxima, la aceleración máxima y la energ
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Answer:

Explanation:

la frecuencia = ω/2π, nada cambio

v(max) = ωA → ω2Α = 2ωA  duplicara velocidad máxima

a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima

la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía

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2 years ago
Which statement is TRUE concerning the particle arrangement of a solid
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It's packed together
8 0
3 years ago
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