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hoa [83]
3 years ago
5

The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c

enter of the earth. (a) What magnitude and sign of charge would a 60-kg human have to acquire to overcome his or her weight by the force exerted by the earth’s electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 m? Is use of the earth’s electric field a feasible means of flight? Why or why not?
Physics
1 answer:
Bezzdna [24]3 years ago
5 0

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is 13.851\times 10^{6} newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

\Sigma F = F_{E}-W = 0 (Eq. 1)

Where:

F_{E} - Electrostatic force exerted on human, measured in Newton.

W - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

q\cdot E-m\cdot g = 0

q = \frac{m\cdot g}{E} (Eq. 2)

E - Electric field, measured in Newtons per Coloumb.

m - Mass, measured in kilograms.

g - Gravity acceleration, measured in meters per square second.

q - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that m = 60\,kg, g = 9.807\,\frac{m}{s^{2}} and E = -150\,\frac{N}{C}, the charge that a 60-kg human must have to overcome weight is:

q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }

q = -3.923\,C

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

F = \kappa \cdot \frac{q^{2}}{r^{2}} (Eq. 3)

Where:

\kappa - Electrostatic constant, measured in Newton-square meter per square Coulomb.

q - Electric charge, measured in Coulomb.

r - Distance between two people, measured in meters.

If we know that \kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}, q = -3.923\,C and r = 100\,m, then the force of repulsion between two people is:

F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right]

F = 13.851\times 10^{6}\,N

The force of repulsion between two people is 13.851\times 10^{6} newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

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Answer:

7.36 × 10^22 kg

Explanation:

Mass of the man = 90kg

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radius of the moon =1.74×10^6

Weight =mg

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g = GM/r^2

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1)

In this circuit (see attachment #1), we have:

- A voltage source: in this case, we choose a battery. A voltage source is a device producing an electromotive force (in a battery, this is done by means of a chemical reaction), which is responsible for "pushing" the electrons along the circuit and creating a current. The electromotive force (emf) of the battery is also called voltage, and it is indicated with the letter V.

- Four resistors: a resistor is a device which opposes to the flow of current. The property that describes by "how much" the resistor "opposes" to the flow of current is called "resistance", and it is indicated with the letter R.

- In this circuit, the 4 resistors are in series. Resistors are said to be in series when they are connected along the same branch of the circuit, so that the same current flow across each of them.

- For resistors in series, the equivalent resistance of the circuit is given by the sum of the individual resistances:

R=R_1+R_2+...+R_n

2)

In this circuit (see attachment #2), we have:

- A voltage source: as before, we have chosen a battery, providing an electromotive  force to the circuit

- Three resistors wired in parallel. Resistors are said to be connected in parallel when they are connected along different branches, but with their terminals connected to the same point, so that each of them has the same potential difference across it.

- For resistors in parallel, the equivalent resistance of the circuit is calculated using the formula:

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}

3)

In this circuit (see attachment #3), we have:

- A voltage source (again, we have choosen a battery)

- Three resistors, of which:

-- 2 of them are connected in parallel with each other

-- the 3rd one it is in series with the first two

If we call R_1,R_2 the resistances of the first 2 resistors in parallel, their equivalent resistance is:

\frac{1}{R_{12}}=\frac{1}{R_1}+\frac{1}{R_2}\\\rightarrow R_{12}=\frac{R_1 R_2}{R_1+R_2}

Then, these two resistors are connected in series with resistor R_3; and so, the total resistance of this circuit will be:

R=R_{12}+R_3=\frac{R_1R_2}{R_1+R_2}+R_3=\frac{R_1R_2+R_3(R_1+R_2)}{R_1+R_2}

4)

In this circuit (see attachment #4), we have:

- A voltage source (again, a battery)

- We have 6 resistors, which are arranged as follows:

-- Two branches each containing 3 resistors

-- The two branches are in parallel with each other

So, the total resistance of the two branches are:

R_{123}=R_1+R_2+R_3

R_{456}=R_4+R_5+R_6

And since the two branches are in parallel, their total resistance will be:

\frac{1}{R}=\frac{1}{R_{123}}+\frac{1}{R_{456}}\\\rightarrow R=\frac{R_{123}R_{456}}{R_{123}+R_{456}}=\frac{(R_1+R_2+R_3)(R_4+R_5+R_6)}{R_1+R_2+R_3+R_4+R_5+R_6}

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