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irakobra [83]
3 years ago
9

This is a bond in which a single pair of electrons is shared between a pair of atoms.

Physics
1 answer:
Murrr4er [49]3 years ago
8 0

Answer:single bond

Explanation:

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two masses are kept 2 metre apart there is gravitational force of 2 Newton what is the gravitational force when they are kept at
sukhopar [10]

Answer: 0.5N

Explanation:

Gravitational force is calculated using the formula :

F = Gm1m2/r^2

Where G is the gravitational constant (6.67 × 10^-11)

At a distance 'r' of 2metres apart:

Mass of objects are m1 and m2

Gravitational force 'F1' = 2N

Inputting values into the formula :

2 = Gm1m2 / 2^2 - - - - - (1)

At a distance 'r' of 4meters apart:

Mass of objects are m1 and m2

Gravitational force 'F2' = y

Inputting values

F2 = Gm1m2 / 4^2 - - - - - (2)

Dividing equations 1 and 2

2 = Gm1m2 / 2^2 ÷ F2 = Gm1m2 / 4^2

2 / F2 = (Gm1m2 / 4) / (Gm1m2 / 16)

2 / F2 = (Gm1m2 / 4) × (16 / Gm1m2)

2/F2 = 16 / 4

Cross multiply

2 × 4 = 16 × F2

8 = 16F2

F2 = 8/16

F2 = 0.5N

7 0
3 years ago
Preston tossed a red ball upward and it reaches a maximum height of 3.0. What is the final velocity when it returns to prestons
Leokris [45]
That will depend on the units of the 3.0. We need to know if it's 3 feet, 3 yards, 3 meters, or 3 miles. Each one will have a different answer.
5 0
3 years ago
You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac
dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

I_i = Initial moment of inertia = 9.3 kgm²

I_f = Final moment of inertia = 5.1 kgm²

\omega_i = Initial angular speed = 1.8 rev/s

\omega_f = Final angular speed

As the angular momentum of the system is conserved

I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

6 0
3 years ago
Give me the ans of this plsss​
MissTica

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5 0
3 years ago
Propose a hypothesis for how the position of the ball will affect the amount of its gravitational pull energy
Ray Of Light [21]

<em><u>throwing a ball up initially has a lot of kinetic energy because it is moving upwards ( kinetic energy is energy which a body possesses by virtue of being in motion.) this all then get converted to gravitational potential energy, and for a moment it is stationary before it begins to fall again.  by the time it has returned again, all the gravitational potential energy has turned back into kinetic.</u></em>

4 0
3 years ago
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