Ask question

Ask question

All categories

2 years ago

6

An extremely long thin wire carries a uniform linear charge density of 358 nC/m. Find the potential difference between points 5.

0 m and 6.0 m from the wire, provided they are not near either end of the wire. (

1 answer:

Romashka-Z-Leto [24]2 years ago

7 0

Answer:

1.2kV

Explanation:

We are given that

Linear charge density, $\lambda=358nC/m=358\times 10^{-9} C/m$

$1 nC=10^{-9} C$

$r_1=5 m$

$r_2=6 m$

We have to find the potential difference between poinst 5.0 m and 6.0 m from the wire.

We know that potential difference of wire

$V=2k\lambda ln\frac{r_2}{r_1}$

Where

$k=9\times 10^9$

Substitute the values

$V=2\times 9\times 10^9\times 358\times 10^{-9}ln\frac{6}{5}$

$V=1174.8 V$

$V=1.17 kV\approx 1.2 kV$

1kV=1000 V

You might be interested in

If a rock is thrown upward on the planet Mars with a velocity of 18 m/s, its height (in meters) after t seconds is given by H =

lukranit [14]

Answer:

a) The velocity of the rock after 1 s is 14.28 m/s.

b) The velocity of the rock after "a" seconds is 18 m/s - 3.72 m/s² · a

c) The rock will hit the ground after 9.7 s.

d) The velocity with which the rock hits the surface is -18.1 m/s

Explanation:

Hi there!

Let´s write the equation:

H(t) = 18 m/s · t - 1.86 m/s² · t²

The velocity is the variaiton of the height over time, in other words, it is the derivative of the height function with respect to time. Then:

v(t) = dH/dt = 18 m/s - 2 · 1.86 m/s² · t

v(t) = 18 m/s - 3.72 m/s² · t

a) When t = 1 s:

v(1 s) = 18 m/s - 3.72 m/s² · 1 s

v(1 s) = 14.28 m/s

b) When t = a:

v(a) = 18 m/s - 3.72 m/s² · a

c) When the rock hits the surface, the height of the rock is 0:

H(t) = 18 m/s · t - 1.86 m/s² · t²

0 = 18 m/s · t - 1.86 m/s² · t²

0 = t(18 m/s - 1.86 m/s² · t)

t = 0

and

18 m/s - 1.86 m/s² · t = 0

t = -18 m/s /- 1.86 m/s²

t = 9.7 s

d) Let´s use the equation of velocity to find the velocity at t = 9.7 s when the rock hits the ground.

v(t) = 18 m/s - 3.72 m/s² · t

v(9.7 s) = 18 m/s - 3.72 m/s² · 9.7 s

v(9.7 s) = -18.1 m/s

4 0

1 year ago

John is running down the street and hears dogs barking in the distance. How do the sound waves change as John approaches the bar

FrozenT [24]

Answer:

The height of the sound waves increases

Explanation:

3 0

2 years ago

Some of the fastest dragsters (called "top fuel) do not race for more than 300-400m for safety reasons. Consider such a dragster

Masja [62]

Answer:

1.10261 times g

416.17506 mph

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 400=0\times 8.6+\frac{1}{2}\times a\times 8.6^2\\\Rightarrow a=\frac{400\times 2}{8.6^2}\\\Rightarrow a=10.81665\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{10.81665}{9.81}\\\Rightarrow \dfrac{a}{g}=1.10261\\\Rightarrow a=1.10261g$

The acceleration is 1.10261 times g

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 10.81665\times 1.6\times 10^3+0^2}\\\Rightarrow v=186.04644\ m/s$

In mph

$186.04644\times \dfrac{3600}{1609.34}=416.17506\ mph$

The speed of the dragster is 416.17506 mph

5 0

2 years ago

In the system shown above, the pulley is a uniform disk with a mass of .75 kg and a radius of 6.5 cm. The coefficient of frictio

lord [1]

Answer:

i am answering the same question 3rd time

please find the answer in the images attached.

5 0

1 year ago

A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad

maksim [4K]

To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

$\Upsilon = \frac{F/A}{\Delta L/L_0}$

Where,

F = Force/Weight

A = Area

$\Delta L$ = Compression

$L_0$ = Original Length

According to the values given we have to

$\Upsilon_{steel} = 200*10^9Pa$

$\Delta L = 5.6*10^{-7}m$

$L_0 = 3.2m$

$r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2$

Replacing this values at our previous equation we have,

$\Upsilon = \frac{F/A}{\Delta L/L_0}$

$200*10^9 = \frac{F/1.0935}{5.6*10^{-7}/3.2}$

$F = 38272.5N$

Therefore the Weight of the object is 3.82kN

4 0

1 year ago