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3 years ago

Once precipitation has fallen on land what paths are available to it

1 answer:

5 0

When water falls on land, there are several different paths that are available for it. It can evaporate back into the atmosphere, go into rivers, streams or lakes, it can sink into the ground so therefore, downward and laterally.

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A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th

saveliy_v [14]

Answer :

Velocity will be $3.28\times 10^{-11}m/sec$

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere = $=1.6+\frac{1.5}{2}=2.35mm = 0.00235m$

Force on electron is given by $F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}$ , here q is charge on sphere and e is charge on electron

$F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N$

This force work as centripetal force

So $F=\frac{mv^2}{r}$

$4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}$

v = $=0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec$

6 0

3 years ago

What does saturns core look like?

HACTEHA [7]

Answer:

At Saturn's center is a dense core of metals like iron and nickel surrounded by rocky material and other compounds solidified by the intense pressure and heat. It is enveloped by liquid metallic hydrogen inside a layer of liquid hydrogen—similar to Jupiter's core but considerably smaller

Explanation:

7 0

3 years ago

The right eye and right lung are __________.

PilotLPTM [1.2K]

The correct answer is A) Ipsilateral

Explanation:

The term ipsilateral is commonly used to describe objects or structures that are on the same side of a body or structure. This term is correct to describe the right eye and the right lung because these two organs are on the same side of the body (the right side). This can also be used to describe other organs such as the left humerus and the left hand or the right ear and the right feet because these pairs are also on the same side. According to this, the correct answer is A.

3 0

3 years ago

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]

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:

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,

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

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

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

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

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#LearnwithBrainly

7 0

3 years ago

A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r

Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

Explanation:

The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t Formula (1)

θ= ω₀*t + (1/2)*α*t² Formula (2)

at = α*R Formula (3)

v= ω*R Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s : wheel’s initial angular velocity

R = 1.0 m : wheel’s radium

(a) Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ= ω₀*t + (1/2)*α*t²

θ= (2)*(10) + (1/2)*(4)*(10)²

θ= 220 rad

θ= 220 rad

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

6 0

3 years ago