Which type of bond shows two or more atoms sharing electrons?

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Yanka [14]

Answer:

yes yes so do I

6 0

2 years ago

Help with this question.....

mariarad [96]

Answer:

$g_{moon}=1.67 [m/s^{2} ]$

Explanation:

The weight of some mass is defined as the product of mass by gravitational acceleration. In this way using the following formula we can find the weight.

$w =m*g\\$

where:

w = weight [N]

m = mass = 0.06 [kg]

g = gravity acceleration = 10 [N/kg]

Therefore:

$w=0.06*10\\w=0.6[N]$

By Hooke's law we know that the force in a spring can be calculated by means of the following expression.

$F=W\\F = k*x$

where:

k = spring constant [N/m]

x = deformed distance = 6 [cm] = 0.06 [m]

We can find the spring constant.

$k= F/x\\k=0.6/0.06\\k=10 [N/m]$

Since we use the same spring on the moon and the same mass, the constant of the spring does not change, the same goes for the mass.

$F_{moon}=k*0.01\\F = 10*0.01\\F=0.1[N]$

Since this force is equal to the weight, we can now determine the gravitational acceleration.

$F=m*g_{moon}\\g=F/m\\g = 0.1/0.06\\g_{moon} = 1.67[m/s^{2} ]$

6 0

2 years ago

The temperature scale which starts at absolute zero is the _____.

oksian1 [2.3K]

The temperature scale which starts at absolute zero is the Kelvin scale. The correct option in respect to the given question is the last option. William Thompson was the British scientist and inventor that invented the Kelvin scale. William Thompson was also popularly known as Lord Kelvin.His discovery of the Kelvin scale is considered one among the three best scales in use for measuring temperatures.Each measuring unit of this scale is never called a degree but a Kelvin. This specialized scale gives the option of measuring temperature in both centigrade and Fahrenheit.

6 0

3 years ago

Read 2 more answers

A proton and an electron are both accelerated to the same final speed. If λp s the de Broglie wavelength of the proton and e is

meriva

Answer:

Explanation:

The relation between the de Broglie wavelength and the momentum of the particle is given by

$\lambda =\frac{h}{m\times v}$

where, m is the mas of the particle and v be the velocity of the particle and h be the Plank's constant.

So, the de broglie wavelength of proton is given by

$\lambda _{p}=\frac{h}{m_{p}\times v}$ .... (1)

The de broglie wavelength of electron is given by

$\lambda _{e}=\frac{h}{m_{e}\times v}$ .... (2)

Divide equation (2) by equation (1), we get

$\frac{\lambda _{e}}{\lambda _{p}}=\frac{m_{p}}{m_{e}}$

As the mass of proton is much more than the mass of electron, so the de broglie wavelength of electron is more than the de Broglie wavelength of proton.

5 0

3 years ago

You are traveling along a freeway at 65 mi/h. You suddenly skid to a stop because of congestion in traffic. Where is the energy

sergey [27]

The work and energy theorem allows finding the result for where the kinetic energy of the car is before stopping is:

The energy becomes:

An important part in work on discs.

A part in non-conservative work due to friction.

Work is defined by the scalar product of force and displacement.

W = F . d

Where the bold indicate vectors, W is work, F is force and d is displacement.

The work energy theorem relates work and kinetic energy.

W = ΔK = $K_f - K_o$

In this case the vehicle stops therefore its final kinetic energy is zero, consequently the work is:

W = - K₀

Therefore, the initial kinetic energy that the car has is converted into work in its brakes. In reality, if assuming that there is friction, an important part is transformed into non-conservative work of the friction force, this work can be seen in a significant increase in the temperature of the discs on which the work is carried out.

In conclusion, using the work-energy theorem we can find the result for where the kinetic energy of the car is before stopping is:

The energy becomes:

An important part in work on the discs.

A part in non-conservative work due to friction.

Learn more here: brainly.com/question/17056946

8 0

2 years ago