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4 years ago

Write two key facts about covalent bondsc

Physics

1 answer:

Lelechka [254]4 years ago

5 0

Explanation:

Covalent bonds:

They are bonds that are formed by atoms that shares their valence electrons.
They are formed by atoms with very low electronegativity difference.
It is usually a bond between non-metals.

Covalent bonds are bonds formed when two atoms jointly shared their valence electrons in order to form a bond.

For the formation of this bond type, each of the atom requires odd or unpaired electrons.

To form a covalent bond, each of the two participating atoms would put down an unpaired electron to be used in forming a shared pair of electrons between them.

Learn more:

Covalent bonds brainly.com/question/6029316

#learnwithBrainly

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3 years ago

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During a chemical reaction, what remains constant? composition or tempature or or energy or bonds

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Energy. Energy cannot be created or destroyed

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3 years ago

A force is described by its __________________ and by the direction in which it acts.

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A force is described by its MAGNITUDE and by the direction in which it acts

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4 years ago

: The shaft is made of L2 tool steel, has a diameter of 40 mm, and is fixed at its ends A and B. If it is subjected to the torqu

Dominik [7]

Answer:

the maximum shear stress in the bar = 9.55 MPa

the twisting angle = $0.146^0$

Explanation:

From the diagram below;

Torque (T) = 2 kN × 5 mm + 2 kN × 5 mm

= $2*10^3 *0.05 + 2*10^3 *0.05$

= 200 N.m

Now; If we divide the shaft into two parts AC & BC

Then :

$T = T_1 + T_2 = 200$ ------------ equation (1)

At the junction :

$\phi _{AC} = \phi _{BC}$

$\phi = \frac{Tl}{GJ}$

⇒ $\frac{T_1 \ l_{AC}}{GJ_1}= \frac{T_2 \ l_{BC}}{GJ_2}$

$T_1l_{AC} = T_2 l_{BC}$

$T_1*400 = T_2 * 600\\T_1 = \frac {600T_2}{400}\\T_1 = 1.5 T_2$

replace $T_1 = 1.5 T_2$ into above equation (1)

$1.5 T_2 +T_2 = 200\\2.5 T_2 = 200\\T_2 = \frac{200}{2.5}\\T_2 = 80 N.m$

Again:

$T_1 +T_2 = 200\\T_1 + 80 = 200\\T_1 = 200 - 80\\T_1 = 120 N.m$

Now ; we can deduce that the maximum shear comes from $\phi_{AC}$ since $T_1 = 120 \ N.m$

So;

$\gamma_{max} = \frac{T_1 *R}{ \frac {\pi D^4}{32}}$

where;

R = 20 mm

D = 40 mm = 0.04 m

Then;

$\gamma_{max} = \frac{120 *20*10^{-3}}{ \frac {\pi (0.04)^4}{32}}$

$\gamma_{max} = 9.55 \ MPa$

Thus ; the maximum shear stress in the bar = 9.55 MPa

Again: $\phi_c = \frac{T_1l_{AC}}{JG} =\frac{T_2l_{AC}}{JG}$

$\phi_c = \frac{120*0.4}{75*10^9*\pi *\frac{(0.04)^2}{32}}$

$\phi_c = 2.55 *10^{-3} \ rads$

$\phi_c = 0.146^0$

Thus, the twisting angle = $0.146^0$

4 0

3 years ago

A 350-g air track cart is traveling at 1.25 m/s and a 280-g cart traveling in the opposite direction at 1.33 m/s. What is the sp

kumpel [21]

Answer:

The speed of the center of mass of the two carts is 0.103 m/s

Explanation:

It is given that,

Mass of the air track cart, m₁ = 350 g = 0.35 kg

Velocity of air track cart, v₁ = 1.25 m/s

Mass of cart, m₂ = 280 g = 0.28 kg

Velocity of cart, v₂ = -1.33 m/s (it is travelling in opposite direction)

We need to find the speed of the center of mass of the two carts. It is given by the following relation as :

$v_{cm}=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}$

$v_{cm}=\dfrac{0.35\ kg\times 1.25\ m/s+0.28\ kg\times (-1.33\ m/s)}{0.35\ kg+0.28\ kg}$

$v_{cm}=0.103\ m/s$

Hence, this is the required solution.

4 0

3 years ago