<span>A moving electrical charge produces a magnetic field and a moving magnetic field produces an electrical field. An electromagnet works by coiling a bunch of wire and spinning a couple of magnets around that wire at high speeds. When this occurs the magnets induce an electric current in the wire and hence the electricity production. Once the magnets stop spinning, the induced electrical field dissipates and the current stops flowing through the wire.
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Answer:
1500 mph
Explanation:
Take east to be +x and north to be +y.
The x component of the velocity is:
vₓ = 889 cos 0° + 830 cos 59°
vₓ = 1316.5 mph
The y component of the velocity is:
vᵧ = 889 sin 0° + 830 sin 59°
vᵧ = 711.4 mph
The speed is found with Pythagorean theorem:
v² = vₓ² + vᵧ²
v² = (1316.5 mph)² + (711.4 mph)²
v = 1496 mph
Rounded to two significant figures, the jet's speed relative to the ground is 1500 mph.
All of the above, work is a measurement of energy transfer, in Joules.
Potential energy = Joules
Kinetic energy = Joules
The key thing here is that anything having to do with just energy or energy transfer is measured in joules.
The 'Rite of Spring' was composed by Igor Stravinsky.
Answer:
h’ = 1/9 h
Explanation:
This exercise must be solved in parts:
* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy
starting point. Higher
Em₀ = U = m g h
final point. Lower, just before the crash
Em_f = K = ½ m
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v_b =
* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2m 0 + m v_b
final instant. Right after the crash
p_f = (2m + m) v
the moment is preserved
p₀ = p_f
m v_b = 3m v
v = v_b / 3
v = ⅓ 
* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together
Starting point. Lower
Em₀ = K = ½ 3m v²
Final point. Higher
Em_f = U = (3m) g h'
Em₀ = Em_f
½ 3m v² = 3m g h’
we substitute
h’=
h’ =
h’ = 1/9 h