Answer:
The amount of current that must flow through the wire for it to be suspended against gravity by magnetic force = 6.125 A
Explanation:
Force on a wire carrying current in an electric field is given by
F = (B)(I)(L) sin θ
For this question,
The magnetic force must match the weight of the wire.
F = mg
mg = (B)(I)(L) sin θ
(m/L)g = (B)(I) sin θ
Mass per unit length = 75 g/m = 0.075 kg/m
B = magnetic field = 0.12 T
I = ?
g = acceleration due to gravity = 9.8 m/s
θ = angle between wire's current direction and magnetic field = 90°
0.075 × 9.8 = 0.12 × I sin 90°
I = 0.075 × 9.8/0.12 = 6.125 A
Answer:
acc. = 4-(-6) /5= 10/5=2 m/s^2
Answer:
96 m
Explanation:
Given,
Initial velocity ( u ) = 4 m/s
Final velocity ( v ) = 20 m/s
Time ( t ) = 8 s
Let Acceleration be " a ".
Formula : -
a = ( v - u ) / t
a = ( 20 - 4 ) / 8
= 16 / 8
a = 2 m/s²
Let displacement be " s ".
Formula : -
s = ut + at² / 2
s = ( 4 ) ( 8 ) + ( 2 ) ( 8² ) / 2
= 32 + ( 2 ) ( 64 ) / 2
= 32 + ( 2 ) ( 32 )
= 32 + 64
s = 96 m
Therefore, it travels 96 m in time 8 s.
