Answer:
using the lens formula: 1/f = 1/u + 1/v
focal length f = -30 (negative because it is concave lens)
object distance u = 60
image distance v= unknown
1/-30 = 1/60 + 1/v
v = -20
So, the image is 20cm from lens (on the same side along with the object), and it is virtual (because of negative sign) and erect (concave lens must produce erect images).
Shadows blocking part of the light from the star.
A quick warning though this only works on planets either close to the star or planets that are very large.
Also to ensure that the shadows are planets the shadows have to move or orbit around the star. IE The shadow moves
Answer:
t = 4 s
Explanation:
As we know that the particle A starts from Rest with constant acceleration
So the distance moved by the particle in given time "t"
Now we know that B moves with constant speed so in the same time B will move to another distance
now we know that B is already 349 cm down the track
so if A and B will meet after time "t"
then in that case
on solving above kinematics equation we have
Answer:
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.
