Determine the magnitude of the magnetic flux through the south-facing window of a house in British Columbia, where Earth's B⃗ field has a magnitude of 5.8 × 10−5 T and the direction of B⃗ field is 72∘ below horizontal with the horizontal component of B⃗ field directed to the north. Assume the area of the window is 3 m^2 . My main issue is how to setup the field in relation to the window's direction. I get how to calculate using the magnetic flux=area*B*cos(theta), but in my homework, I keep getting told I'm solving for the angle wrong??

not moving. pos accel,

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