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4 years ago

10

the gravitational force acting between two objects is 785 newton’s. if the mass of one object doubles, what is the new force act

ing between the objects

1 answer:

Tom [10]4 years ago

6 0

Use formula F=(G Mm)/R^2
from the relation When Mass of one object doubles the new gratification force will be 785*2=1570

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3 years ago

The x coordinate of an electron is measured with an uncertainty of 0.200 mm . What is vx, the x component of the electron's velo

Vikentia [17]

Answer:

Velocity of electron along x direction is 57.9 m/s

Explanation:

The uncertainty in x coordinate of electron, Δx = 0.200 mm = 0.2 x 10⁻³ m

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The uncertainty in x component of electrons momentum is:

Δpₓ = mΔvₓ

Here m is mass of the electron.

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So, the above equation can be written as :

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The minimum uncertainty principle is:

$\Delta x\Delta p_{x} = \frac{h}{2\pi }$ ....(2)

Here h is Planck's constant.

From equation (1) and (2),

$\Delta x\times0.01m v_{x} = \frac{h}{2\pi }$

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3 years ago

An object 1 of mass m1 is separated by some distance d from an object 2 of mass 2m1 . An object 3 of mass m3 is to be placed bet

Harrizon [31]

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

Object 3 should be placed closer to object 1.

Object 3 should be placed on a halfway between object 2 and object 1.

Object 3 should be placed closer to object 2.

Solution

I think that Object 3 should be placed closer to object 2.

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3 years ago

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4 years ago

Given that the electromagnetic force is far stronger than gravity on a per-particle basis, why doesn't the electromagnetic force

Kaylis [27]

Explanation:

The electro magnetic force is given by

F = $\frac{k.q_{1}.q_{2}}{r^{2}}$

where $q_{1}$ and $q_{2}$ are charged particles

k =Coulombs constant

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F = $\frac{G.m_{1}.m_{2}}{r^{2}}$

where $m_{1}$ and $m_{2}$ are masses

G =Garvitation constant

r = distance between two masses

Now since the planets, stars and galaxies are electrically neutral, therefore they have zero electrical charge and so electro magnetic forces have no affect on these planets, stars and heavenly bodies.

Whereas the masses of the heavenly bodies are very large, so they are largely affected by the gravitational force since Gravitational force is directly proportional to the product of the masses of a body.

Therefore, though the electromagnetic force is stronger than the gravitational force, the electromagnetic force does not dominate the forces in the heavenly bodies as they as not electrically charged.

8 0

4 years ago