Answer:
ΔΦ = -3.39*10^-6
Explanation:
Given:-
- The given magnetic field strength B = 0.50 gauss
- The angle between earth magnetic field and garage floor ∅ = 20 °
- The loop is rotated by 90 degree.
- The radius of the coil r = 19 cm
Find:
calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.
Solution:
- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.
- The strength of magnetic field B and the are of the loop A remains constant. So we have:
Φ = B*A*cos(θ)
ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )
- The initial angle θ_1 between the normal to the coil and B was:
θ_1 = 90° - ∅
θ_1 = 90° - 20° = 70°
The angle θ_2 after rotation between the normal to the coil and B was:
θ_2 = ∅
θ_2 = 20°
- Hence, the change in flux can be calculated:
ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )
ΔΦ = -3.39*10^-6
Answer: The angle of incidence is not always equal to the angle of reflection.
Explanation:
The angle of incidence might not be equal to the angle of reflection. It depends of the type of surface in consideration. If the surface is smooth, the incident ray will reflect out at the same angle the incident ray makes with surface. This is not the same for a rough or irregular surface.
For an irregular surface, the angle of incident is not equal to the angle if reflection because the reflected ray always reflects at different angles to the horizontal.
The escape speed for a spacecraft at the surface of a planet of mass M and radius R is:
where G is the gravitational constant. We can use this formula to solve both parts of the problem, using the data of Jupiter and Mars.
a) Mars:
Mars mass is
, while Mars radius is
, so the escape speed of the spacecraft at Mars surface is
b) Jupiter:
Jupiter mass is
while its radius is
, so the escape speed at its surface is
