A flat sheet if in the shape of a rectangle with sides of length 0.400 m and 0.600 m. The sheet is immersed in a uniform electric field of magnitude 95 N/C that is directed 20° from the plane of the sheet. Find the magnitude of the electric flux through the sheet.

Φ= 17 N•m²•C⁻¹

Explanation:

Gauss's Law states that electric flux equals the surface integral of E•dA. But since we are given all the variables as finite values, we can simplify it into EAcosφ.

-E is given as 95N/C

-A is simply (.4)(.6)=.24m²

-φ is the angle between the E field/vector and the normal/perpendicular vector to the surface. We know that E makes a 20° to the surface here, so the angle φ=(90-20)°=70°. So the E vector makes a 70° angle to the normal of the surface. (I can see this portion as being the point of confusion, as it was for me at first.)

With all that we can say that the flux Φ is:

Φ=(95)(0.24)(cos[70°])=17.4384... N•m²•C⁻¹

I'll approximate to 2 sigfigs in my answer, since that'd be the technical answer.

*I believe V/m are also correct units for electric flux.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

part 1: which substance has a higher solubility?

a

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by