KE = 1/2mv^2
KE = 1/2 (5kg)(3m/s)
KE = 22.5 J
Answer:
a) The magnitude of the magnetic field inside the toroid at the inner radius is
0.000447 T = 0.447 mT
b) The magnitude of the magnetic field inside the toroid at the outer radius is
0.000344 T = 0.344 mT
Explanation:
With a logical assumption that magnetic permeability of vacuum would be used,
the magnetic field at a distance r, from the centre of the loop is given as
B = μ₀I (N/2πr)
B = ?
μ₀ = (4π × 10⁻⁷) H/m
I = 0.739 A
N = 590 turns
For the inner radius,
r = 19.5 cm = 0.195 m
a) B = μ₀I (N/2πr)
B = (4π × 10⁻⁷ × 0.739 × 590) ÷ (2π × 0.195)
B = 0.0004471897 T = 0.000447 T = 0.447 mT
b) Magnetic field at the outer radius
r(outer) = r(inner) + length of the square
r(outer) = 19.5 cm + 5.83 cm = 25.33 cm = 0.2533 m
B = μ₀I (N/2πr)
B = (4π × 10⁻⁷ × 0.739 × 590) ÷ (2π × 0.2533)
B = 0.0003442637 T = 0.000344 T = 0.344 mT
Hope this Helps!!!
Answer: vf1/vf2= 1/ sqrt(2)
Explanation :on the moon no drag force so we have only the force of gravity. aceleration is g(moon)= 1.62m/s2.the rest is basic kinematics
if the rock travels H to the bottom we can calculate velocity:
vo=0m/s (drops the rock) , yo=0
vf*vf= vo*vo+2g(y-yo)
when the rock is halfway y = H/2 so:
vf1*vf1=2*g*H/2 so vf1 = sqrt(gH)
when the rock reach the bottom y=H so:
vf2*vf2=2*g*H so vf2 = sqrt(2gH)
so vf1/vf2= 1/ sqrt(2)
good luck from colombia
8. The answer is C
9. Sorry I don't know