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solmaris [256]
3 years ago
6

Suppose the electrons and protons in 1g of hydrogen could be separated and placed on the earth and the moon, respectively. Compa

re the electrostatic attraction with the gravitational force between the earth and the moon. ( the number of atoms in 1g of hydrogen is Avogadro's number Na. There is one electron and one proton in a hydrogen atom. ) Please explain step by step
Physics
1 answer:
MAXImum [283]3 years ago
7 0

Answer:

The gravitational force is 3.509*10^17 times larger than the electrostatic force.

Explanation:

The Newton's law of universal gravitation and Coulombs law are:

F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}

Where:

G= 6.674×10^−11 N · (m/kg)2

k =  8.987×10^9 N·m2/C2

We can obtain the ratio of these forces dividing them:

\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}   --- (1)

The mass of the moon is 7.347 × 10^22 kilograms

The mass of the earth is  5.972 × 10^24 kg

And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C

Replacing these values in eq1:

\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}

Therefore

\frac{F_{N}}{F_{C}}}}=3.509\times10^{17}

This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field

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A car travels from Boston to Hartford in 4 hours. The two cities are 240 kilometers apart. What was the average speed of the car
Y_Kistochka [10]

Answer:

Average speed is 60 km/hour

Explanation:

When we need to calculate average speed, we use this equation:

V = \frac{x_{f} - x_{o}}{t_{f} - t_{o}}

Where:   x_{o} = 0 km   position at the beginning

              x_{f} = 240 km   at the end

              t_{o} = 0 hours

              t_{f} = 4 hours

Then:     V = \frac{240 km - 0 km}{4 hours - 0 hours}

              V = \frac{240 km}{4 hours}

Finally    V = 60 km/hour

6 0
3 years ago
Dr. Kirwan is preparing a slide show that he will present to the executive board at tonight's committee meeting. He places a 3.5
lisov135 [29]

Answer:

A) d_o = 20.7 cm

B) h_i = 1.014 m

Explanation:

A) To solve this, we will use the lens equation formula;

1/f = 1/d_o + 1/d_i

Where;

f is focal Length = 20 cm = 0.2

d_o is object distance

d_i is image distance = 6m

1/0.2 = 1/d_o + 1/6

1/d_o = 1/0.2 - 1/6

1/d_o = 4.8333

d_o = 1/4.8333

d_o = 0.207 m

d_o = 20.7 cm

B) to solve this, we will use the magnification equation;

M = h_i/h_o = d_i/d_o

Where;

h_o = 3.5 cm = 0.035 m

d_i = 6 m

d_o = 20.7 cm = 0.207 m

Thus;

h_i = (6/0.207) × 0.035

h_i = 1.014 m

8 0
3 years ago
In a parallel circuit, if bulb #2 were to blow out, bulb #1 would stay lit or go out?
valkas [14]
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8 0
3 years ago
Two friends are playing a version of proton golf where the hole is marked by a single proton. The first friend reads his meter,
yarga [219]

Answer:

the electric field strength on the second one is 2.67 N/C.

Explanation:

the electric fiel on the first one is:

E1 = k×q/(r^2)

r^2 = k×q/(E1)

     = (9×10^9)×(q)/(24.0)

     = 375000000q

then the electric field on the second one is:

E2 = k×q/(R^2)

we know that R = 3r

                       R^2 = 9×r^2

E2 = k×q/(9×r^2)

     = k×q/(9×375000000q)

     = k/(9×375000000)

     = (9×10^9)/(9×375000000)

     = 2.67 N/C

Therefore, the electric field strength on the second one is 2.67 N/C.

5 0
3 years ago
Which projectile will be visibly affected
jonny [76]

Answer:

leaf and a balloon is the correct answer

6 0
3 years ago
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