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MariettaO [177]
1 year ago
9

What happens to the force between two objects if the mass of one object is doubled.

Physics
1 answer:
Citrus2011 [14]1 year ago
4 0

If the mass of one object is doubled, the force between these objects will also double.

Force refers to an influence on a body which can change its state of rest or motion.

Force F = G(mM/d^2) (where m is the mass of first object, M is the mass of the second object and G is the gravitational pull and d is the distance between the two objects)

The force between two objects (m and M) (according to the universal law of gravitation) is proportional to their mass and reciprocally proportional to the square of their separation (R) between them.

So F' = G(2mM/d^2),

which means F'=2F

Therefore, the when the mass is doubled, the force also doubles.

To learn more about force, click

brainly.com/question/26115859

#SPJ4

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Answer:

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Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

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