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3 years ago

Which statement best describes covalent bonding?

Physics

2 answers:

chubhunter [2.5K]3 years ago

6 0

b- electrons are shared between two atoms

slamgirl [31]3 years ago

4 0

Electrons are shared between two atoms, describes covalent bonding.

HOPE THIS HELPS! ^_^

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The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x

Diano4ka-milaya [45]

Answer:

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Explanation:

Given that

At X=0 V=Vo

At X=X1 V=0

As we know that friction force is always try to oppose the motion of an object. It means that it provide acceleration in the negative direction.

We know that

$V^2=U^2-2aS$

$0=V_o^2-2a X_1$

$a=\dfrac{V_o^2}{2X_1}$

So the friction force on the box

Ff= m x a

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Where m is the mass of the box.

4 0

3 years ago

In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21

Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

$z = 20km$

And the velocity can be written as,

$V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$V = 980.55m/s$

From the properties of standard atmosphere at altitude z = 20km temperature is

$T = 216.66K$

$k = 1.4$

$R = 287 J/kg$

Velocity of sound at this altitude is

$a = \sqrt{kRT}$

$a = \sqrt{(1.4)(287)(216.66)}$

$a = 295.049m/s$

Then the Mach number

$Ma = \frac{V}{a}$

$Ma = \frac{980.55}{296.049}$

$Ma = 3.312$

So front stagnation temperature

$T_0 = T(1+\frac{k-1}{2}Ma^2)$

$T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)$

$T_0 = 689.87K$

Therefore the temperature at its front stagnation point is 689.87K

6 0

4 years ago

What is the ideal banking angle for a gentle turn of 1.20-km radius on a highway with a 105 km/h speed limit (about 65 mi/h), as

Mnenie [13.5K]

Answer:

4.14°

Explanation:

given:

r = 1.2 km

v = 105 km/h

1) convert your given

a) r = 1.2 km to m = 1200m

b) v = 105 km/h to m/s = 29.2 m/s

2) plug into your ideal banking angle equation

$tan^-1$ ( $\frac{v^2}{rg}$ ) = $\frac{29.2^2}{(1200)(9.8)}$ = 4.14°

8 0

3 years ago

A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N forc

Setler79 [48]

Answer:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = $\frac{1}{2}ke^{2}$

==> $\frac{1}{2}* 550* 0.02^{2}$ = 0.11 J

3 0

3 years ago

A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge

Kisachek [45]

Answer:

$\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

$\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r$

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

$\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

5 0

3 years ago