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Bess [88]
3 years ago
10

In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere

d to be distributed uniformly throughout the cloud. The charge builds up until the electric field at the surface of the cloud reaches the value at which the surrounding air "breaks down."
In general, the term "breakdown" refers to the situation when a dielectric (insulator) such as air becomes a conductor. In this case, it means that, because of a very strong electric field, the air becomes highly ionized, enabling it to conduct the charge from the cloud to the ground or another nearby cloud. The ionized air then emits light as the electrons and ionized atoms recombine to form excited molecules that radiate light. The resulting large current heats up the air, causing its rapid expansion. These two phenomena account for the appearance of lightning and the sound of thunder.

The point of this problem is to estimate the maximum amount of charge that a cloud can contain before breakdown occurs. For the purposes of this problem, take the cloud to be a sphere of diameter 1.00 km. Take the breakdown electric field of air to be Eb=3.00×10^6N/C.


Assuming that the cloud is negatively charged, how many excess electrons are on this cloud?
Physics
1 answer:
muminat3 years ago
5 0

Answer:

2.1\cdot 10^{21} electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

E=\frac{kQ}{r^2}

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a  charged sphere, so we have:

E_b=3.00\cdot 10^6 N/C is the maximum electric field strength tolerated by the air before breakdown occurs

r=1.00 km = 1000 m is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C

Assuming that the cloud is negatively charged, then

Q=-333.3 C

And since the charge of one electron is

e=-1.6\cdot 10^{-19}C

The number of excess electrons on the cloud is

N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}

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A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her orig
atroni [7]

Answer:

The plane is 2353.7 mi from the starting position.

Explanation:

Please, see the attached figure for a graphic representation of the problem.

We have 2 displacement vectors "a" and "b" and a vector "c" that is the sum of vectors "a" plus "b" (c = a + b). The module of "c" will be the distance of the plane from the starting point.

vector a = (xa, ya)

vector b = (xb, yb)

where “xa” and “xb” are the horizontal components of the vectors “a” and “b” respectively and “ya” and “yb” are the vertical components of each vector.

Then, the vector c = a + b will be:

c = (xa + xb, ya + yb)

The module of a vector is calculated using the following expression for a vector “v”:

module of v = \sqrt{x^{2} + y^{2} }

Then, the module of c will be:

module of c = \sqrt{(xa + xb)^{2} + (ya + yb)^{2}} = distance from starting point

Then, we have to find the components of vectors “a” and “b”

The distance traveled during the first 1.5 hours of the trip is the module of the vector “a”. Then:

module of a = \sqrt{xa^{2} + ya^{2} } = distance traveled during the first 1.5 hours.

The distance can be calculated using the equation of the position of an object moving in a straight line at constant speed:

x = x0 + v * t

where

x = position at time t

x0 = initial position

v = speed

t = time

Considering x0 as the starting point (x0 = 0)

x = 675 mi/h * 1.5 h = 1012.5 mi

Then:

module of a = \sqrt{xa^{2} + ya^{2} } = 1012. 5 mi

Since the plane moves only on the horizontal (see figure), the "y" component of the vector, "ya", will be 0.

Then:

(1012.5 mi)² = xa²

xa = 1012. 5mi

a = (1012.5 mi, 0)

In the same way, we have fo find the components of the vector “b”. The module of “b” will be the distance traveled during this part of the flight:

module of b = \sqrt{xb^{2} + yb^{2} } = x = x0 + v * t

Considering x0 as the point at which the plane turns (x0 = 0)

x = 675 mi / h * 2 h = 1350 mi

Using trigonometry, we can calculate xb and yb (see figure):

sin angle = opposite / hypotenuse

cos angle = adjacent / hypotenuse

In this case:

opposite = yb

adjacent = xb

hypotenuse = module of “b”

Then:

sin 10° = yb / module of “b”

sin 10° * module of “b” = yb

In the same way:

cos 10° * module of “b” = xb

Since module of “b” = 1350 mi

xb = 1329.5 mi

yb = 234.4 mi

b = (1329.5 mi, 234.4 mi)

The vector c = a+b can now be calculated:

c = (xa + xb, ya + yb)

c =(1012.5 mi + 1329.5 mi, 0 mi + 234.4 mi) = (2342 mi, 234.4 mi)

The module of c will be:

module of c = \sqrt{(2342 mi)^{2} + (234.4 mi)^{2} } = 2353.7 mi

The plane is 2353.7 mi from the starting position.

4 0
3 years ago
A straight wire of length 0.62 m carries a conventional current of 0.7 amperes. What is the magnitude of the magnetic field made
anyanavicka [17]

Answer:

Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T

Explanation:

Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2}  } }      ......(1)

Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

In this problem,

Current, I = 0.7 A

Length of wire, L = 0.62 m

Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m

Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

Substitute these values in equation (1).

B=\frac{4\pi\times10^{-7}\times  0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2}  } }

B = 6.99 x 10⁻⁶ T

3 0
3 years ago
How much work is done by the force lifting a
Ann [662]

The work done in lifting the hamburger is equal to the increase in gravitational potential energy of the hamburger, given by

W=\Delta U=mg \Delta h

where

m=0.1 kg is the mass of the hamburger

g=9.81 m/s^2 is the gravitational acceleration

\Delta h=0.3 m is the increase in height of the hamburger


Substituting numbers into the equation, we find

W=(0.1 kg)(9.81 m/s^2)(0.3 m)=0.3 J


So, the correct answer is

(3) 0.3 J

3 0
3 years ago
Which of the following statements about the force on a charged particle due to a magnetic field are not valid?
Vinil7 [7]
The correct answer is "None of the above; all of these statements are valid." All the statements namely, it depends on the particle's charge, it depends on the strength of the external magnetic field, it depends on the particle's velocity, and it acts at right angles to the direction of the particle's motion are all valid. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help. 
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3 years ago
What are some ways you can vary your tone of voice to help you communicate effectively with others? A. Speaking aggressively
Leno4ka [110]
The answer is D, talking more loudly or quietly.
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3 years ago
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