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jenyasd209 [6]
3 years ago
13

Please help I need serious help .

Physics
1 answer:
Basile [38]3 years ago
7 0
C , is the answer k????
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INT Raindrops acquire an electric charge as they fall. Suppose a 2.0-mm-diameter drop has a charge of 12 pC; these are both very
horsena [70]

Answer:

W = 2.3 10² F_{e}

Explanation:

The force of the weight is

         W = m g

         

let's use the concept of density

         ρ= m / v

the volume of a sphere is

         V = \frac{4}{3} π r³

         V = \frac{4}{3} π (1.0 10⁻³)³

         V = 4.1887 10⁻⁹ m³

the density of water ρ = 1000 kg / m³

          m = ρ V

          m = 1000 4.1887 10⁻⁹

          m = 4.1887 10⁻⁶ kg

therefore the out of gravity is

          W = 4.1887 10⁻⁶ 9.8

          W = 41.05 10⁻⁶ N

now let's look for the electric force

           F_e = q E

           F_e = 12 10⁻¹² 15000

           F_e = 1.8 10⁻⁷ N

         

the relationship between these two quantities is

          \frac{W}{F_e} = 41.05 10⁻⁶ / 1.8 10⁻⁷

           \frac{W}{F_e} = 2,281 10²

             

             W = 2.3 10² F_{e}

therefore the weight of the drop is much greater than the electric force

3 0
3 years ago
What is government?
Mandarinka [93]

A government is the system or group of people governing an organized community, often a state.

3 0
4 years ago
If it has enough kinetic energy, a molecule at the surface of the Earth can "escape the Earth's gravitation", in the sense that
user100 [1]

Answer:

K_E=mgr_E

Explanation:

By conservation of energy, the sum of the kinetic and gravitational potential energies at the surface of the Earth must be equal than their sum at infinity, so we have:

K_E+U_E=K_\infty+U_\infty

\frac{mv_E^2}{2}-\frac{GM_Em}{r_E}=\frac{mv_\infty^2}{2}-\frac{GM_Em}{r_\infty}

Where G=6.67\times10^{-11}Nm^2/kg^2 is the gravitational constant,M_E=5.97\times10^{24}kg and r_E=6371000m are the mass and radius of the Earth, <em>m </em>is the mass of the particle, v_E its velocity at the surface of the Earth (which would be its escape velocity) and v_\infty and r_\infty are the velocities and distance at infinity, which would be null and infinity respectively, so the right hand side of our equation is 0J, which leaves us with:

\frac{GM_Em}{r_E}=\frac{mv_E^2}{2}=K_E

Also, since the force the molecule experiments is the force of gravity (disregarding drag), we can write its weight in terms of Newton's Law of Gravitation:

F=mg=\frac{GM_Em}{r_E^2}

Which means that:

\frac{GM_Em}{r_E}=mgr_E

So finally putting all together we can write:

K_E=\frac{GM_Em}{r_E}=mgr_E

4 0
3 years ago
How long will it take a car to go from 0 to 45 km/hr if they are accelerating at 5 km/hr/s?
Mandarinka [93]

Answer:

9 seconds

Explanation:

acceleration =  \frac{final \: speed - initial \: speed}{time \: taken}

5km/ hr/ s =  \frac{45 - 0 (km/ h)}{t}

cross multiplying

5t = 45 secs \\ t = 9 \: secs

it will take 9secs for the car to go from 0 to 45 km/ h

4 0
3 years ago
If you want to decrease the current created by a generator what can you do?
Rasek [7]

Answer:

To decrease the current;

1) Use fewer loops or number of turns

2) Use a lower speed of rotation of the coil in the magnetic field

3) Use a weaker magnetic

Explanation:

According to Faraday's Law of induction, which is the basis of the electromagnetism, electromagnetic induction and therefore the basis of the  electric generator, can be written as follows;

\epsilon = -N \cdot \dfrac{\Delta \phi}{\Delta t}

Where;

ε = The induced voltage

N = The number of turns (loops)

ΔФ = The change in the magnetic flux

Δt = The change in the time (the duration)

Given that voltage is directly related to the current, decreasing the voltage, decreases the current

To decrease the voltage, and therefore, the current we can;

1) Reduce the number of loops in the coil

2) Increase the time change per unit change in flux by slowing down the speed of rotation of the generator

3) Decrease the amount of change in the magnetic field per turn, by using a weaker magnetic

3 0
3 years ago
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