Answer:
a) 17.8 m/s
b) 28.3 m
Explanation:
Given:
angle A = 53.0°
sinA = 0.8
cosA = 0.6
width of the river,d = 40.0 m,
the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,
The river itself was 100 m below the ramp H = 100 m,
(a) find speed v
vertical displacement
putting values h=15 m, v=0.8
............. (1)
horizontal displacement d = vcosA×t = 0.6×v ×t
so v×t = d/0.6 = 40/0.6
plug it into (1) and get
solving for t we get
t = 3.734 s
also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s
(b) If his speed was only half the value found in (a), where did he land?
v = 17.8/2 = 8.9 m/s
vertical displacement = 
⇒ 
t = 5.30 s
then
d =v×cosA×t = 8.9×0.6×5.30= 28.3 m
Explanation:
The magnetic force acting on a current carrying wire in a uniform magnetic field is given by :
or
Where
is the angle between length and the magnetic field
The magnetic force is perpendicular to both current and magnetic field. It is maximum when it is perpendicular to both current and magnetic field.
So, the correct options are :
- The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.
- .The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.
- The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.
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