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ale4655 [162]
3 years ago
14

A significant drawback to wind power is: A. oil companies discourage it. B. the cost of the turbines requires significant capita

l investment. C. the novelty of windmill technology raises the risk. D. the turbines obstructing the landscape. E. there are only a few areas on the globe that would benefit.
Physics
1 answer:
sweet [91]3 years ago
6 0
The most significant drawback is B
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When two point charges are a distance dd part, the electric force that each one feels from the other has magnitude F.F . In orde
Harman [31]

Answer:

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r' = r/2.

Explanation :

The electric force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between charges. It is given by :

F=\dfrac{kq_1q_2}{r^2}

r is the separation between charges  

F\propto \dfrac{1}{r^2}

r=\sqrt{\dfrac{1}{F}}

If F'= 2F

r'=\dfrac{1}{\sqrt{2F} }

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r'=\dfrac{1}{\sqrt{2F} }. Hence, this is the required solution.                                                                                    

6 0
3 years ago
What angle is formed by the sun, the earth, and the moon during an eclipse?.
Andrew [12]

Answer:

The Sun-Earth-Moon system happens to exhibit a striking geometric coincidence, which we examine in the first problem. PROBLEM 1. To an observer on Earth, the Sun and the Moon subtend almost the same angle in the sky. The average angle is 0.52 degrees for the Moon and 0.53 degrees for the Sun.

4 0
3 years ago
Explain what you think is happening in this picture
Dima020 [189]

Answer:

sorry, there is no picture, try re-uploading it with the pic

Explanation:

5 0
3 years ago
*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude
kvv77 [185]

Answer:

Approximately 3.86\; {\rm N} (given that the magnitude of this charge is -7.35\; {\rm \mu C}.)

Explanation:

If a charge of magnitude q is placed in an electric field of magnitude E, the magnitude of the electrostatic force on that charge would be F = E\, q.

The magnitude of this charge is q = 7.35\; {\rm \mu C}. Apply the unit conversion 1\; {\rm \mu C} = 10^{-6}\; {\rm C}:

\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}.

An electric field of magnitude E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}} would exert on this charge a force with a magnitude of:

\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}.

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0
2 years ago
A battery is connected to the two ends of a conducting rod and the current through it is measured.a) Another rod, made of the sa
IceJOKER [234]

Answer:

a) - the resistance is proportional to the length of the metal

   - a thicker rod, since resistance decreases with increasing area

b)  The current is proportional to the area so this is the factor that is influenced by the electric field  

Explanation:

a) When the rod is connected to the rod an electrical circuit described by the expression is formed

          V = I R

          I = V / R

The resistance of the rod is

         R = ρ L / A

Where ρ is the resistivity of metal, L Longitud and A cross section.

If a higher current is measured when the other rod is connected, it implies that the resistance has decreased. To have a decrease in resistance you have two possibilities

- use a shorter rod, since the resistance is proportional to the length of the metal

- Use a thicker rod, since resistance decreases with increasing area, remember that the area of ​​a circle is

                  A = π R²

Therefore, small increases in the diameter of the rod could give great changes in resistance.

b) The current is defined with the charge that crosses a surface per unit of time, when the electric field increases the electrons feel a greater force, therefore the current increases.

The charge number on a wire is

              Q = q n v_{d} Δt A

So up to date is

               I = q n When the rod is connected to the rod an electrical circuit described by the expression is formed

          V = I R

          I = V / R

The resistance of the rod is

         R = rho L / A

Where rho is the resistivity of metal, L e; Long and A cross section.

If a higher current is measured when the other rod is connected, it implies that the resistance has decreased. To have a decrease in resistance you have two possibilities

.- use a shorter rod, since the resistance is proportional to the length of the metal

. - Use a thicker rod, since resistance decreases with increasing area, remember that the area of ​​a circle is

                  A = pi R2

Therefore, small increases in the diameter of the rod could give great changes in resistance.

.b) The current is defined with the charge that crosses a surface per unit of time, when the electric field increases the electrons feel a greater force, therefore the current increases.

The load number on a wire is

              Q = q n v_{d} Dt A

So  current  is

               I = q n v_{d} A

The drag speed (v_{d}) is constant in the materials, it depends on the purity and imperfections, so a change in the length does not significantly change the current.

  The current is proportional to the area so this is the factor that is influenced by the electric field  

The drag speed (vd) is constant in the materials, it depends on the purity and imperfections, so a change in the length does not significantly change the current.

  The current is proportional to the area so this is the factor that is influenced by the electric field

8 0
3 years ago
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