Something to do with how the suns magnetic field interacts with the surface plasmas I think.
Answer:
Explanation:
A plane flies due north (90° from east) with a velocity of 100 km/h for 2 hours.
With no wind, it will be 100*2 = 200 km north of its starting point.
But a steady wind blows southeast at 30 km/h at an angle of 315° from due east.
So the wind itself will blow the plane 30*2 = 60km at an angle of 315° from due east.
That is the same as 60*cos315° = 42.43km due east and 60*sin315° = -42.43km north.
Combining, the plane is at 42.43km due east and 200-42.43 = 157.57km due north from its starting point.
Answer: 1.55 x 10⁴ Nm²c^-1
Explanation: The electric flux, electric field intensity and area are related by the formulae below.
Φ= EAcosθ,
Where Φ= electric flux (Nm²c^-1)
E =electric field intensity (N/m²)
A = Area (m²)
θ= this is angle between the planar area and the magnetic flux
For our question E=3.80KN/c= 3800 N/c
A= 0.700 x 0.350= 0.245m²
θ= 0° ( this is because the electric field was applied along the x axis, thus the electric flux will be parallel to the area).
Hence Φ= 3800 x 0.245 x cos(0)
= 3800 x 0.245 x 1 (value of cos 0° =1)
= 1.55 x 10⁴ Nm²c^-1
Thus the electric field is 1.55 x 10⁴ Nm²c^-1
Answer:
6.5 m/s
Explanation:
We are given that
Distance, s=100 m
Initial speed, u=1.4 m/s
Acceleration, 
We have to find the final velocity at the end of the 100.0 m.
We know that
Using the formula
Hence, her final velocity at the end of the 100.0 m=6.5 m/s
