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

A hydraulic lift raises a 2000 kg automobile when a 500 N force is applied to the smaller piston. If the smaller piston has an a

Inga [223]

Answer:

The cross-sectional area of the larger piston is 392cm ^{2}[/tex]

Explanation:

To solve this problem we apply the following formula:

Pascal principle: F=P*A Formula (1)

F=Force applied to the piston

P: Pressure

A= Piston area

Nomenclature:

Fp= Force on the primary piston= 500N

W= car weight =m*g=2000kg*9.8m/s2= 19600N

Fs= Force on the secondary piston= W = 19600N

$Ap= Primary piston area=10cm^{2} =10*10^{-4}m^{2}$

As= Secondary piston area=?

The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.

In the equation (1)

P=F/A

Pp=Ps

$\frac{Fp}{Ap} = \frac{Fs}{As}$

$As= \frac{Fs*Ap}{Fp}$

$As=\frac{19600*10*10^{-4} }{500}$

$As=0.0392m^{2} =0.0392*10^{4}cm^{2}$

$As=392cm ^{2}$

5 0

2 years ago

Two wires are stretched between two fixed supports and have the same length. One wire A there is a second-harmonic standing wave

lina2011 [118]

(a) Greater

The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, $f_1$ :

$f_n = n f_1$

So we have:

- On wire A, the second-harmonic has frequency of $f_2 = 660 Hz$ , so the fundamental frequency is:

$f_1 = \frac{f_2}{2}=\frac{660 Hz}{2}=330 Hz$

- On wire B, the third-harmonic has frequency of $f_3 = 660 Hz$ , so the fundamental frequency is

$f_1 = \frac{f_3}{3}=\frac{660 Hz}{3}=220 Hz$

So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

(b) $f_1 = \frac{v}{2L}$

For standing waves on a string, the fundamental frequency is given by the formula:

$f_1 = \frac{v}{2L}$

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

$v=2Lf_1$

We know that the two wires have same length L. For wire A, $f_1 = 330 Hz$ , while for wave B, $f_B = 220 Hz$ , so we can write the ratio between the speeds of the waves in the two wires:

$\frac{v_A}{v_B}=\frac{2L(330 Hz)}{2L(220 Hz)}=\frac{3}{2}$

So, the waves travel faster on wire A.

7 0

2 years ago

Sunitha can type 1800 words in half an hour. What is her typing speed in words per minute?

Andre45 [30]

Answer:

60words/minute

Explanation:

If Sunitha can type 1800 words in half an hour, this can be expressed as;

1800 words = 30 minutes

To get her typing speed per minute, we will use the formula

Speed = Number of words/Time used

Typing speed = 1800/30

Typing speed = 60words/minute

Hence her typing speed in words per minute is 60words/minute

6 0

1 year ago

A 1000.–kilogram car traveling 20.0 meters per second east experiences an impulse of 2000. newton • seconds west. What is the fi

emmainna [20.7K]

Impulse is equal to change in momentum. So if impulse is 2000 then to solve for new velocity we just set it equal to equation for momentum.

First find original momentum by p=mv
p=1000*20=20000

So then taking that value minus the impulse since it was in opposite direction of original momentum it will slow it down some. To find new velocity we just take

20000-2000=18000=mv

v=18000/1000 =18m/s

Hope this helps!! Any questions please ask!!
Thank you!

5 0

2 years ago

A 10.0 cm3 sample of copper has a mass of 89.6

Romashka-Z-Leto [24]

Density is mass divides by volume, so
89.6g / 10cm^3 =8.96g /cm^3

*cm^3 is a standard unit of volume*

4 0

2 years ago