The yellow star has much less mass the blue star and so who will live longer

Find the order of magnitude of your age in seconds

Natali5045456 [20]

The order of magnitude of my age in seconds is 10^9. I think you'll find that this is true for anyone who is 32 or older.

6 0

2 years ago

LOL SOMEONE DO THIS FOR ME!!!

olganol [36]

Build a filter.
Dig water from the ground

4 0

2 years ago

PLZZ HELP

Digiron [165]

It would be the temperature of the steel decreases and the temperature of the water increases

7 0

2 years ago

Read 2 more answers

Below is a single strand of DNA, what is the complementary base pair for the strand?

GREYUIT [131]

The complementary base pair is:
TTC, CTG, AGT, CTA.

5 0

2 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

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

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

$= 7.2 *10^{4} + 3.1*10^5$

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

8 0

2 years ago