Gaussian surface 1 has twice the area of Gaussian surface 2. Both surfaces enclose the same charge Q. Is the electric flux through surface 1 greater than, less than, or the same as the electric flux through surface 2?

The electric flux through surface 1 is greater than the electric flux through surface 2.

Step-by-step explanation:

Electric Flux is given as:

Φ = E*A

where E = Electric field

A = surface area.

For surface 1,

Φ(1) = E(1)*A(1)

For surface 2,

Φ(2) = E(2)*A(2)

Since both surfaces have the same charge, the electric fields will be the same.

Hence, E(1) = E(2) = E

From the question we are given that:

A(1) = 0.5A(2)

Hence,

Φ(1) = E*A(1)

Φ(2) = E*A(2) = E*0.5A(1)

Φ(2) = 0.5E*A(1)

Comparing Φ(1) and Φ(2), we see that Φ(1) is 2 time greater than Φ(2), hence, the Flux through surface 1 is greater than the Flux through surface 2.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

c.

step-by-step explanation:

because first you have to add the 4x, 9x, and the 7x together and you get 20x and then you add the last numbers up and you get 5and then you put the subtraction sign in there and that's how you get the answer.