Explanation:
Below is an attachment containing the solution.
If the angle between the vector and the y-axis is Θ ,
then the y-component of the vector is
(length of the vector) times cosine( Θ ),
and the x-component of the vector is
(length of the vector) times sine( Θ ) .
The force of the throw is an applied force
moving through the air is drag
and the downward pull would be due to gravity
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.
The ant's resultant velocity is 0.51 m/s at 78.7°
Since the ant is moving with a velocity of 0.1 m/s in the +x direction relative to the treadmill, and the treadmill is moving 0.5 m/s in the +y direction, the magnitude of the resultant velocity of the ant relative to the ground (since both directions are perpendicular) is thus
V = √(x² + y²)
= √[(0.1 m/s)² + (0.5 m/s)²]
= √[0.01 m²/s² + 0.25 m²/s²]
= √(0.26 m²/s²)
= 0.51 m/s.
The direction is Ф = tan⁻¹(y/x)
= tan⁻¹(0.5 m/s ÷ 0.1 m/s)
= tan⁻¹(5)
= 78.69°
≅ 78.7°
The ant's resultant velocity is 0.51 m/s at 78.7°
Learn more about resultant velocity here:
brainly.com/question/15093278
