A uniform line charge that has a linear charge density λ = 3.3 nC/m is on the x axis between x = 0 to x = 5.0 m.

Please help on this one?

diamong [38]

Answer:

ITS C

Explanation:

7 0

3 years ago

Scientists have proven that genes play no role in self-esteem. Please select the best answer from the choices provided. T F

IRINA_888 [86]

The answer is False give thanks for the answer m8 and happy Halloween

6 0

3 years ago

Read 2 more answers

Does anybody know how to answer any of these questions? Any of them without a check mark.

aivan3 [116]

Explanation:

sorry I can't help u right now

3 0

3 years ago

two barges full of salted toad guts have a collision. the red barge has a mass of 150000kg and is traveling northwest at 0.25m/s

solniwko [45]

The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.

Final velocity(v3) of the red barge is calculated by following formula

m1×v1+ m2×v2= (m1+m2)v3

Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s

150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3

37500+ 320000= 1150000×v3

357500= 1150000×v3

v3= 0.311 m/s

<h3>What is elastic collision velocity? </h3>

The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.

For more information on elastic collision velocity kindly visit to

brainly.com/question/29051562

#SPJ9

6 0

1 year ago

Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha

Westkost [7]

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, $A_p = 180 cm^2$

- The charge on each plate, $q = 17 * 10^-^6 C$

- Permittivity of free space, $e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}$

- The radius for the flux region, $r = 3.3 cm$

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

$A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2$

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

σ = $\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\$

σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

$E+ = E- = \frac{sigma}{2*e_o} \\\\$

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

$E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\$

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

Φ = E_net * Ar * cos ( θ )

Φ = (106214689.26553) * (0.00342) * cos ( 5 )

Φ = 361872 N.m^2 / C

5 0

3 years ago