Explanation:
It is given that,
Diameter of loop, d = 1.4 cm
Radius of loop, r = 0.7 cm = 0.007 m
Magnetic field, 
(A) Magnetic field of a current loop is given by :

I is the current in the loop


I = 27.85 A
(B) Magnetic field at a distance r from a wire is given by :



r = 0.00222 m

Hence, this is the required solution.
Answer:
2.4 m/s
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.08 kg)(0.5 m/s) + (0.05 kg)(0 m/s) = (0.08 kg)(-0.1 m/s) + (0.05 kg) v
0.04 kg m/s = -0.08 kg m/s + (0.05 kg) v
0.12 kg m/s = (0.05 kg) v
v = 2.4 m/s
Answer:
(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).
(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .
Explanation:
m= 3kg
a= 2 i + 5 j = 5 .38 < 68.19 º
F= m * a
F= 3* ( 5.38 < 68.19º )
F= 16.4 N < 68.19º
Fx= F * cos(68.19º)
Fx= 5.99
Fy= F* sin(68.19º)
Fy= 14.98
<span>D. density is your answer</span>
Stopped at the end of the tracks by a spg-damper system, as shown in fig. 1