Answer:
When her hands extends, her momen of inertia is
.
Explanation:
Given that,
Initial angular speed, 
Initial moment of inertia, 
Final angular speed, 
Initially, a skater rotates with her arms crossed and finally she extends her arms. The momentum remains conserved. Using the conservation of momentum as :

is final moment of inertia

So, when her hands extends, her momen of inertia is
. Hence, this is the required solution.
Answer:
Radius of the loop is 0.18 m or 18 cm
Explanation:
Given :
Current flowing through the wire, I = 45 A
Magnetic field at the center of the wire, B = 1.50 x 10⁻⁴ T
Number of turns in circular wire, N = 1
Consider R be the radius of the circular wire.
The magnetic field at the center of the current carrying circular wire is determine by the relation:
Here μ₀ is vacuum permeability constant and its value is 4π x 10⁻⁷ Tm/A.
Substitute the suitable values in the above equation.

R = 0.18 m
Answer:
41.74 m/s
Explanation:
The energy used to draw the bowstring = the kinetic energy of the arrow.
Fd = 1/2mv²................................ Equation 1
Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.
make v the subject of the equation
v = √(2Fd/m)...................... Equation 2
Given: F = 201 N, m = 0.3 kg, d = 1.3 m.
Substitute into equation 2
v = √(2×201×1.3/0.3)
v = √(1742)
v = 41.74 m/s.
Hence the arrow leave the bow with a speed of 41.74 m/s