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3 years ago

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When two point charges are a distance d apart, the magnitude of the electrostatic force between them is F. If the distance betwe

en the point charges is increased to 3d, the magnitude of the electrostatic force between the two charges will be

2 answers:

dedylja [7]3 years ago

8 0

Decreased by a factor of 9

EastWind [94]3 years ago

3 0

Answer:

The magnitude of the electrostatic force decreases by a factor 9

Explanation:

The electrostatic force between two charges is given by:

$F=k\frac{q_1 q_2}{d^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

d is the distance between the two charges

We see that the magnitude of the force is inversely proportional to the square of the distance. If the distance is increased to 3d: d' = 3d, the new electrostatic force would be:

$F'=k\frac{q_1 q_2}{(d')^2}=k\frac{q_1 q_2}{(3d)^2}=\frac{1}{9} k\frac{q_1 q_2}{d^2}=\frac{F}{9}$

So, the electrostatic force decreases by a factor 9.

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Suppose an individual is lying on his stomach with sheets of paper stacked on his back. If each sheet of paper has a mass of 0.0

AURORKA [14]

Answer:

N = 177843 sheets

Explanation:

We are given;

Mass;m = 0.0035 kg

Pressure; p = 101325 pa = 101325 N/m²

L = 0.279m

W = 0.216m

The weight of N sheets is N(mg)

Where;

m is the mass of one sheet

N is number of sheets

g is the acceleration due to gravity.

The pressure equals weight divided by the area on which the weight presses:

Thus,

p= F/A = Nmg/(L•W)

Therefore, making N the subject;

N = pLW/(mg)

N = 101325 x 0.279 x 0.216/ (0.0035 x 9.81)

N = 177843

5 0

3 years ago

How much work does the electric field do in moving a proton from a point with a potential of +130 V

In-s [12.5K]

Explanation:

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6 0

3 years ago

If you put a drop of food coloring in water and watch the drop disperse, is entropy increasing or decreasing.

den301095 [7]

Answer:

Entropy is increasing. Entropy is decreasing.

Explanation:

The Entropy doesn't change.

4 0

3 years ago

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

3 years ago

When do net force is applied to a moving object it still comes to rest because of its inertia

Paraphin [41]

That is false for that.

6 0

3 years ago