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2 years ago

Please help!! I'll name you brainliest!!

Physics

1 answer:

kow [346]2 years ago

7 0

Answer:

Explanation:

in this condition the atom is not stable so in order to become stable ( trying to have the last energy level completely filled like the Nobel gases) it tends to lose or gain or share the electrons and that debend on its condition . I hope that's the right answer

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Write an expression for the potential energy UD of a particle when it is at a distance D from the force center, in terms of B an

nadezda [96]

Answer:

E+mc+UD-3

Explanation:

7 0

3 years ago

What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +

Mumz [18]

Complete Question:

Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, $\sigma_n = 28 MPa$

There are shear stresses acting on the surface since $\tau \neq 0$

Explanation:

$\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$

equation of the normal, $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$

$\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

Traction vector on n, $T_n = \sigma \b n$

$T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

$T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]$

$T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z$

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

$\sigma_n = T_n . \b n$

$\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa$

If the shear stress, $\tau$ , is calculated and it is not equal to zero, this means there are shear stresses.

$\tau = T_n - \sigma_n \b n$

$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$

$\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa$

Since $\tau \neq 0$ , there are shear stresses acting on the surface.

3 0

2 years ago

1 kg air in a piston-cylinder assembly is heated at constant pressure, resulting the expansion of the volume. the initial temper

rewona [7]

Answer:

57,42 KJ

Explanation:

By a isobaric proces, the expresion for the works in the jpg adjunt. Then:

W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)

By the ideal gases law: PV=RTn

Then, in (1): (remember Pa = Pb)

W = R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)

Since we have 1 Kg air: How much is this in moles?

From bibliography: 28.96 g/mol

Then, in 1 Kg (1000 g) there are:

n = 34,53 mol

Finally, in (2):

W = (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ

4 0

3 years ago

A 263 g ball of clay falls onto a vertical spring.

Elena-2011 [213]

Answer:

d. According to the previous theorem:

(kh^2)/2-mgh-(m(2v)^2)/2=0,

126h^2-2.577h-8.236=0,

h=0.266 m.

Explanation:

3 0

2 years ago

The legislative branch has the power to (3 points)

maw [93]

Hi there!

The Legislative Branch of government includes the House of Representatives and the Senate. This branch makes laws. Its powers include passing laws, originating spending bills (House), impeaching government officials (Senate), and approving treaties (Senate).

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!

3 0

3 years ago

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