Which of the following items best embodies the physical property of conductivity?

Which one of the following is NOT a simple machine?

ivann1987 [24]

It is hammer because hammers are not examples of a simple machine

6 0

3 years ago

A student of mass 40kg takes 10s to run up a flight of 50steps. If each step is 15cm high, calculate the Potential Energy of the

astra-53 [7]

Answer:

2943 J

Explanation:

Potential energy = mass x acceleration due to gravity x height in meters

(P.E. = mgh)

Substitute the given numbers: (I take acceleration due to gravity as 9.81 m s^-2)

PE = 40 x 9.81 x (0.15x50)

PE = 2943 J

3 0

4 years ago

A 10-cm-long thin glass rod uniformly charged to 8.00 nCnC and a 10-cm-long thin plastic rod uniformly charged to - 8.00 nCnC ar

ch4aika [34]

Complete Question

A 10-cm-long thin glass rod uniformly charged to 8.00 nC and a 10-cm

long thin plastic rod uniformly charged to -8.00 nC are placed side by

side, 4.20 cm apart. What are the electric field strengths E_1 to E_3 at

distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line

connecting the midpoints of the two rods

a.) Specify the electric field strength E1

b.) Specify the electric field strength E2

c.) Specify the electric field strength E3

Answer:

$E_1=7.13*10^5 N/C$

$E_2= 2.95*10^{5} N/C$

$E_3= 3.84*10^5 N/C$

Explanation:

From the question we are told that

The length of the thin glass is $L = 10 cm$

The charge on the glass rod is $q_g = 8.00nC = 8* 10^{-9} C$

The length of the plastic rod is $L_p = 10cm$

The charge on the plastic rod is $q_p =- 8.00nC = -8.0*10^{-9}C$

The distance between the materials is $d = 4.20cm = \frac{4.2}{100} =0.042m$

The various distances to obtain electric field of are $r_1 = 1.0cm$

$r_2 = 2.0cm$

$r_3 = 3.0cm$

The objective of the solution is to obtain the electric field $E_1 , E_2 \ and E_3$ at distance $d_1 , d_2 \ and \ d_3$ from the glass rod along the line connecting its mid point

Generally electric field of a charge rod at a distance of r the line dividing the rod into half is mathematically represented as

$E = k \frac{2Q}{r\sqrt{L^2 + 4r^2} }$

For the $r_2 = 1.0cm = \frac{1}{100} = 0.01m$

The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

$E_l = k \frac{2Q }{r \sqrt{L^2 + 4r^2_1} }$

The electric filed by the positively charge glass rod on the right side of the dividing line is mathematically represented as

$E_r = k \frac{2Q }{(0.044 - r_1) \sqrt{L^2 + 4r^2_1} }$

The net electric field is,

$E_{net} =E_1= E_l + E_r$

$= k \frac{2Q}{r_1\sqrt{L^2 + 4 r^2_1 } } + k \frac{2Q}{(0.04-r_1) \sqrt{L^2 + 4 (0.044 -r_1)^2} }$

Where k is know as the coulomb's constant with a constant value of

$k = 9*10^9 \ kgm^3 s^{-4} A^{-2}$

$=(9*10^9) \frac{(2) (8*10^{-9})}{(0.01)\sqrt{(0.01^2 + 4(0.01)^2)} } + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.01)\sqrt{(0.01)^2 + (4) (0.042 - 0.01)^2} }$

$= 6.44*10^5 + 6.9*10^4$

$E_1=7.13*10^5 N/C$

For the $r_2 = 2.0cm = \frac{2}{100} = 0.02m$

The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

$E_l = k \frac{2Q }{r_2 \sqrt{L^2 + 4r^2_2} }$

The electric filed by the positively charge glass rod on the right side of the dividing line is mathematically represented as

$E_r = k \frac{2Q }{(0.044 - r_2) \sqrt{L^2 + 4r^2_2} }$

The net electric field is,

$E_{net} =E_2= E_l + E_r$

$= k \frac{2Q}{r_2\sqrt{L^2 + 4 r^2_2 } } + k \frac{2Q}{(0.04-r_2) \sqrt{L^2 + 4 (0.044 -r_2)^2} }$

Where k is know as the coulomb's constant with a constant value of

$k = 9*10^9 \ kgm^3 s^{-4} A^{-2}$

$=(9*10^9) \frac{(2) (8*10^{-9})}{(0.02)\sqrt{(0.02^2 + 4(0.02)^2)} } + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.02)\sqrt{(0.02)^2 + (4) (0.042 - 0.02)^2} }$

$= 1.6*10^{5}+ 1.3*10^{5}$

$E_2= 2.95*10^{5} N/C$

For the $r_3 = 3.0cm = \frac{3}{100} = 0.03m$

The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

$E_l = k \frac{2Q }{r_3 \sqrt{L^2 + 4r^2_3} }$

The electric filed by the positively charge glass rod on the right side of the dividing line is mathematically represented as

$E_r = k \frac{2Q }{(0.044 - r_3) \sqrt{L^2 + 4r^2_3} }$

The net electric field is,

$E_{net} =E_3= E_l + E_r$

$= k \frac{2Q}{r_3\sqrt{L^2 + 4 r^2_3 } } + k \frac{2Q}{(0.04-r_3) \sqrt{L^2 + 4 (0.044 -r_3)^2} }$

Where k is know as the coulomb's constant with a constant value of

$k = 9*10^9 \ kgm^3 s^{-4} A^{-2}$

$=(9*10^9) \frac{(2) (8*10^{-9})}{(0.03)\sqrt{(0.03^2 + 4(0.03)^2)} } + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.03)\sqrt{(0.03)^2 + (4) (0.042 - 0.03)^2} }$