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 common values. In a thunderstorm, the electric field under a cloud can reach 15,000 N/C, directed upward. For a droplet exposed to this field, how do the magnitude and direction of the electric force compare to those of the weight force

Answers

Answer 1

Answer:

W = 2.3 10² [tex]F_{e}[/tex]

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 = [tex]\frac{4}{3}[/tex] π r³

         V = [tex]\frac{4}{3}[/tex] 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

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

           \frac{W}{F_e} = 2,281 10²

             

             W = 2.3 10² [tex]F_{e}[/tex]

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


Related Questions

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.4 cm. Assuming heat to be transferred uniformly from all surfaces, determine (a) the amount of heat this resistor dissipates during a 24-hour period, (b) the heat flux, and (c) the fraction of heat dissipated from the top and bottom surfaces.

Answers

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ([tex]Q[/tex]), measured in joules, by the cylindrical resistor is the power multiplied by operation time ([tex]\Delta t[/tex]), measured in hours. That is:

[tex]Q = \dot Q \cdot \Delta t[/tex] (1)

If we know that [tex]\dot Q = 1.2\,W[/tex] and [tex]\Delta t = 86400\,s[/tex], then the amount of heat dissipated by the resistor is:

[tex]Q = (1.2\,W)\cdot (86400\,s)[/tex]

[tex]Q = 103680\,J[/tex]

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ([tex]Q'[/tex]), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ([tex]A[/tex]), measured in square meters:

[tex]Q' = \frac{\dot Q}{A}[/tex] (2)

[tex]Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }[/tex] (3)

Where:

[tex]D[/tex] - Diameter, measured in meters.

[tex]h[/tex] - Length, measured in meters.

If we know that [tex]\dot Q = 1.2\,W[/tex], [tex]D = 4\times 10^{-3}\,m[/tex] and [tex]h = 2\times 10^{-2}\,m[/tex], the heat flux of the resistor is:

[tex]Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }[/tex]

[tex]Q' \approx 4340.589\,\frac{W}{m^{2}}[/tex]

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ([tex]r[/tex]), no unit, is the ratio of the top and bottom surfaces to total surface:

[tex]r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}[/tex] (3)

If we know that [tex]A \approx 2.765\times 10^{-4}\,m^{2}[/tex] and [tex]D = 4\times 10^{-3}\,m[/tex], then the fraction is:

[tex]r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}[/tex]

[tex]r = 0.045[/tex]

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

The moon accelerates because it is

A. in a vacuum in space.

B. continuously changing direction.

C. a very large sphere.

D. constantly changing its shape.

Answers

Answer:

the Answer is b

Explanation:

because the moon usually orbits around our solar system

As the moon continuously changing direction, it accelerates.

Option B. is correct.

Define acceleration.

The rate at which an object's velocity changes with respect to time is called acceleration. Accelerations are measured in terms of vectors. The orientation of the net force acting on an object determines the orientation of its acceleration.

The Moon is kept in orbit around us by the gravity of the Earth. It constantly shifting the Moon's velocity direction. This means that, despite its constant speed, gravity causes the Moon to accelerate all the time.

So, as the moon continuously changing direction, it accelerates.

Option B. is correct.

Find out more information about acceleration here:

https://brainly.com/question/2437624?referrer=searchResults

An 80 N rightward force is applied to a 10 kg object to accelerate it to the right.
The object encounters a friction force of 50 N.

Answers

net force = 30 N

mass = 8.16 kg

acceleration = 3.68 m/s²

Further explanation

Given

80 N force applied

mass of object = 10 kg

Friction force = 50 N

Required

Net force

mass

acceleration

Solution

net force

Net force = force applied(to the right) - friction force(to the left)

Net force = 80 - 50 = 30 N

mass

Gravitational force(downward) : F = mg

m = F : g

m = 80 : 9.8

m = 8.16 kg

acceleration

a = F net / m

a = 30 / 8.16

a = 3.68 m/s²

Two balls are thrown against a wall with the same velocity. The first ball is made of rubber and bounces straight back with some non-zero speed. The second ball is made of clay and sticks to the wall after impact. If we assume the collision time was the same for each ball, which ball experienced a greater average acceleration during the collision with the wall? A. the average acceleration was the sameB. the clay ball C. there is not enough information D. the rubber ball

Answers

Answer:

A. the average acceleration was the same

Explanation:

Acceleration is calculated by finding the difference of the initial velocity from the final velocity (on impact, usually 0) and then dividing by the amount of time that took place. If we assume that both balls were thrown at the same initial force, and ended up hitting the wall at the same time then we can say that the average acceleration was the same. If the initial velocity was not the same then we would need the initial velocity of each ball in order to calculate the acceleration of each object and determine which had a greater acceleration.

Although the vocal tract is quite complicated, we can make a simple model of it as an open-closed tube extending from the opening of the mouth to the diaphragm, the large muscle separating the abdomen and the chest cavity. What is the length of this tube if its fundamental frequency equals a typical speech frequency of 230 Hz? Assume a sound speed of 350 m/s.

Answers

Answer:

0.76m

Explanation:

Given data

Frequency= 230Hz

speed= 350m/s

Since we are told that the frequency is the fundamental frequency n= 1

For a standing wave

Fn= nv/2L

n= 1

230= 1*350/2*L

230= 350/2L

cross multiply

2L= 350/230

2L=1.521

L=1.521/2

L=0.76m

Hence the length is 0.76m

At what latitude are there almost no differences between the seasons? Explain
why this occurs?

Answers

Answer:

The four-season year is typical only in the mid-latitudes. The mid-latitudes are places that are neither near the poles nor near the Equator. The farther north you go, the bigger the differences in the seasons.

Explanation:

hope this helps have a good day :) ❤

At a latitude equal to zero degrees there is little seasonal variation. This phenomenon is due each day the Sun's rays strike the Earth's surface at approximately the same angle near the Equator.

The Equator is the line of (zero degrees) latitude around the middle of the Earth.

Moroever, the intensity of solar radiation and therefore also the temperature at the Earth's surface largely depends on the angle of incidence of the Sun's rays.

At 0° latitude, there is a very little seasonal variation because all days the Sun's rays strike the Earth's surface at approximately the same angle. At the Equator, the Sun's rays strike the Earth's surface at an angle of 90°, causing warmer temperatures compared to higher latitudes.

In additon, at 0° latitude, all days also have the same number of hours of light and dark (i.e., approximately 12 hours of sunlight).

In conclusion, at a latitude equal to zero degrees (i.e., at the Equator) there is little seasonal variation. This phenomenon is due each day the Sun's rays strike the Earth's surface at approximately the same angle near the Equator.

Learn more in:

https://brainly.com/question/744653

What is the period, in seconds, that corresponds to each of
the following frequencies: (a) 10 Hz, (b) 0.2 Hz, (c) 60 Hz?

Answers

Answer:

0.1s,5s,0.017s

Explanation:

T=1÷frequency

Answer:

a =

✔ 6

The period is

✔ 2 seconds.

b =

✔ pi

Explanation:

Graph the function using the graphing calculator. Find the least positive value of t at which the pendulum is in the center.

t =

✔0.5 sec

To the nearest thousandth, find the position of the pendulum when t = 4.25 sec.

d =

✔ 4.243 in.

A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s. Assuming the damping to be negligible, calculate the motor inertia in Nm·s2.

Answers

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ([tex]\tau[/tex]), measured in Newton-meters, is:

[tex]\tau = I\cdot \alpha[/tex] (1)

Where:

[tex]I[/tex] - Moment of inertia, measured in Newton-meter-square seconds.

[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

[tex]\alpha = \frac{\omega - \omega_{o}}{t}[/tex] (2)

Where:

[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.

[tex]\omega[/tex] - Final angular speed, measured in radians per second.

[tex]t[/tex] - Time, measured in seconds.

If we know that [tex]\tau = 3\,N\cdot m[/tex], [tex]\omega_{o} = 0\,\frac{rad}{s }[/tex], [tex]\omega = 145.875\,\frac{rad}{s}[/tex] and [tex]t = 4\,s[/tex], then the moment of inertia of the motor is:

[tex]\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}[/tex]

[tex]\alpha = 36.469\,\frac{rad}{s^{2}}[/tex]

[tex]I = \frac{\tau}{\alpha}[/tex]

[tex]I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }[/tex]

[tex]I = 0.0823\,N\cdot m\cdot s^{2}[/tex]

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

A wave in which the movement of the wave is perpendicular to the movement of the wave traveling through the medium-compression wave (longitudinal wave)

True or false?

Answers

Answer:

I'm rusty sorry if I'm wrong but true?

A 0.75 kg golf ball is being launched from the ground. What is its potential energy​

Answers

Explanation:

the question it's not complete as I don't know the height as the formula for potential energy is : PE = mgh

(m) - mass acceleration due to gravity

(g) - acceleration due to gravity (9.8 m/s2)

(h) - height

Identify the independent variable(s) in Asch's original experiment.

Answers

Answer:

How do I answer this I don't understand the question

Explanation:

Surface Features on earth combine to form the(blank) or area ,Will mark brainliest if someone give’s me the word that fit’s in there.

Answers

i believe it’s topography

The plates of a parallel plate capacitor each have an area of 0.40 m2 and are separated by a distance of 0.02 m. They are charged until the potential difference between the plates is 3000 V. The charged capacitor is then isolated. Determine the magnitude of the electric field between the capacitor plates.

Answers

Answer:

 E = -1.5 10⁵ N / C

Explanation:

In a capacitor the electric field is uniform between the blades, therefore we use the expression

                   V = - E s

                   E = - V / s

let's calculate

                   E = - 3000 / 0.02

                   E = -1.5 10⁵ N / C

the sign indicates that the field and the potential are opposite, when one increases the gold decreases

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