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Basile [38]
3 years ago
14

Draw the Lewis dot Structure for the molecule C2H6​

Physics
1 answer:
zzz [600]3 years ago
6 0

Answer:

This means that the Lewis dot structure for C2H6 must account for 14 valence electrons, either through bonding between atoms, or through lone pairs. So, the two C atoms are placed in the center of the molecule.

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Amount of force exerted on an object due to gravity is called
MAXImum [283]
The gravitational force between a mass and the Earth is the object'sweight. Mass is considered a measure of an object's inertia, and its weight is the force exerted on the object in a gravitational field. On the surface of the Earth, the two forces are related by the acceleration due to gravity: Fg = mg.

Hoped this helped!
6 0
3 years ago
Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du
ELEN [110]

With constant angular acceleration \alpha, the disk achieves an angular velocity \omega at time t according to

\omega=\alpha t

and angular displacement \theta according to

\theta=\dfrac12\alpha t^2

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}

b. Under constant acceleration, the average angular velocity is equivalent to

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2

where \omega_f and \omega_i are the final and initial angular velocities, respectively. Then

\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}

c. After 1.00 s, the disk has instantaneous angular velocity

\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}

d. During the next 1.00 s, the disk will start moving with the angular velocity \omega_0 equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle \theta according to

\theta=\omega_0t+\dfrac12\alpha t^2

which would be equal to

\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}

5 0
3 years ago
The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
Margaret [11]

Let  us consider two bodies having masses m and m' respectively.

Let they are  separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -  F = G\frac{mm'}{r^{2} }   where G is the gravitational force constant.

From the above we see that F ∝ mm' and F\alpha \frac{1}{r^{2} }

Let the orbital radius of planet  A is r_{1}  = r and mass of planet is m_{1}.

Let the mass of central star is m .

Hence the gravitational force for planet A  is f_{1} =G \frac{m_{1}*m }{r^{2} }

For planet B the orbital radius  r_{2} =2r_{1} and mass m_{2} = 3 m_{1}

Hence the gravitational force f_{2} =G\frac{m m_{2} }{r^{2} }

                                                 f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }

                                                 = \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }

Hence the ratio is  \frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2}  }{Gmm_{1}/r_{1} ^2 }

                                      =\frac{3}{4}     [ ans]


                                                 

                           

3 0
3 years ago
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A ball is thrown upward from ground level at a velocity of 55.5 m/s. What is the ball maximum height?
ollegr [7]
I believe the answer is 153.8 m.
5 0
3 years ago
In a house, a circuit often includes multiple electrical outlets which are parallelconnected to a single voltage source. Each of
valkas [14]

Answer:

In parallel circuits, if one element has a problem, it can be eliminated and the rest of the elements continue to work.

Explanation:

There are two ways to connect electrical circuits, in series and in parallel.

Series circuits have the problem that when one element is damaged, the entire circuit is interrupted and runs out of power.

In parallel circuits, if one element has a problem, it can be eliminated and the rest of the elements continue to work.

In Breaker it is connected in series so that when some maintenance is needed, all the energy can be interrupted simultaneously and if some equipment tries to consume a lot of energy, the circuit prevents this action by interrupting the energy of the entire circuit

5 0
3 years ago
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