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4 years ago

Which change will always result in an increase in the gravitational force between two objects?

Physics

1 answer:

Sunny_sXe [5.5K]4 years ago

4 0

Answer:

Increasing the masses of the objects and decreasing the distance between the objects.

Explanation:

Gravitational force always act between two objects that have mass. The gravitational force is a weak force and attractive in nature.

The force of pull depends on the masses of the two objects and the distance between them.

The formula to calculate gravitational force between two objects having masses 'm' and 'M' and separated by a distance 'd' is given as:

$F_g=\frac{GmM}{d^2}$

Where, 'G' is called the universal gravitational constant.

Now, from the above formula, it is clear that, the force of gravitation is inversely proportional to the square of the distance between the two objects and directly proportional to the masses of the two objects.

This means that, if masses increase with other parameters constant, force will increase and vice-versa. If distance decreases with other parameters constant, force will increase and vice-versa.

So, on combining the above two relations,we are certain that if one increases the masses and decrease the distance between, the gravitational force will definitely increase.

So, the third option is correct.

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4 years ago

Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diamet

Alekssandra [29.7K]

Answer:

(a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times(5\times10^{-2})^2$

$A=7.85\times10^{-3}\ m^2$

We need to calculate the capacitor

Using formula of capacitor

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}$

$C=6.94\times10^{-12}\ F$

We need to calculate the resistance of the wire

Using formula of resistivity

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}$

$R=4.27\times10^{-3}\ \Omega$

We need to calculate the voltage

Using formula of charge

$q=CV$

$V=\dfrac{q}{C}$

Put the value into the formula

$V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}$

$V=1.801\times10^{3}\ V$

(a). We need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

$I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}$

$I=421779.85\ A$

$I=4.217\times10^{5}\ A$

(b). We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}$

$E=11.2\times10^{5}\ N/C$

The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

$E=\dfrac{1}{2}CV^2$

Put the value into the formula

$E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2$

$E=1.126\times10^{-5}\ J$

The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Hence, (a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

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3 years ago

Static discharges occur when

vredina [299]

The sudden flow of electricity between two electrically charged objects caused by contact, an electrical short, or dielectric breakdown. A buildup of static electricity can be caused by tribocharging or by electrostatic induction.

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3 years ago

Read 2 more answers

A train is traveling at a velocity of 90 m/s and it has to travel a distance of 2500 meters to reach the train station. How long

Vlada [557]

Answer:

27.7 or 28 seconds

Explanation:

2500/90=27.7

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4 years ago

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