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Yuliya22 [10]
2 years ago
14

About how much of Earth's water is available for humans to use?

Physics
1 answer:
lesya692 [45]2 years ago
5 0

Answer:

0.3 %

Explanation:

Earth cleans and replenishes the water supply through the hydrologic cycle. The earth has an abundance of water, but unfortunately, only a small percentage, is even usable by people.

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A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.
bija089 [108]

a)

i) Potential for r < a: V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

ii) Potential for a < r < b:  V(r)=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

iii) Potential for r > b: V(r)=0

b) Potential difference between the two cylinders: V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c) Electric field between the two cylinders: E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

Explanation:

a)

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

E=\frac{\lambda}{2\pi \epsilon_0 r}

where

\lambda is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

E=0

So the potential where the electric field is zero is constant:

V=const.

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density +\lambda and an equal negative charge density -\lambda. Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)

However, we know that the potential at b is zero, so

V(r)=V(b)=0

ii) The electric field in the region a < r < b instead it is given only by the positive charge +\lambda distributed over the surface of the inner cylinder of radius a, therefore it is

E=\frac{\lambda}{2\pi r \epsilon_0}

And so the potential in this region is given by:

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r} (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

And so, for r<a,

V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

b)

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

- Potential at the surface of the outer cylinder:

V(b)=0

Therefore, the potential difference is simply equal to

V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c)

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

The electric field is just the derivative of the electric potential:

E=-\frac{dV}{dr}

so we can find it by integrating the expression for the electric potential. We find:

E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
Who said the greater the temperature the greater the volume
Serjik [45]
The greater the temperature, the greater the volume - this is Charles's law, said by Jacques Charles, a French inventor, scientist, and mathematician. 
3 0
2 years ago
Read 2 more answers
A +6.33 uC charge q1 is attracted by a force of 0.115 N to a second charge q2 that is 1.44 m away. What is the value of q2?
jarptica [38.1K]

Answer:-4.1858

Explanation: I hate Acellus

7 0
2 years ago
A vertical spring with spring stiffness constant 305 N/m oscillates with an amplitude of 28.0 cm when 0.235 kg hangs from it. Th
den301095 [7]

Answer:

The function that describe the motion in the time

y (t) = 0.28m * sin ( 36.025 * t)

Explanation:

The angular frequency of oscillation of the spring

w = √k/m

w = √305 N/m / 0.235 kg

w = 36.025 rad / s

To determine the function of the motion knowing as a motion oscillation in a amplitude a frequency

y(t) = A * sin (w t )

So

A = 28.0 cm * 1 m / 100 cm = 0.28 m

So replacing to determine the function of the motion in the time

y (t) = A sin (w t)

y (t) = 0.28m * sin ( 36.025 * t)

7 0
3 years ago
A golf club with mass 0.5 kg is swung at 23 m/s toward a stationary golf ball on a tee. The club barely slows down to 17 m/s, bu
mylen [45]

Answer:

I gotchu fam. 0.0435 kg

Explanation:

m1v1+m2v2=m1v1f+m2v2f

(0.5)(23)+(m2)(0)=(0.5)(17)+(m2)(69)

11.5=8.5+69m2

3=69m2

0.0435 kg

5 0
2 years ago
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