Answer:
When there is a change in magnetic flux linkage through a loop of wire, an electromotive force is induced in the loop, according to the Faraday-Newmann-Lenz Law:
where
N is the number of turns in the loop
is the change in magnetic flux through the loop
is the time elapsed
The negative sign in the formula represents Lenz's Law, and tells us about the direction of the electromotive force.
In fact, the negative sign means that the direction of the induced emf is such that to oppose to the change in the magnetic flux that originated the induced emf.
This is a consequence of the law of conservation of energy: no energy can be created out of nowhere. In fact, when the emf is induced in the loop, electrical energy appears in the circuit; however, this electric energy cannot come out of nowhere. Instead, it is just "created" from the transformation of some other form of energy (for instance, the mechanical energy that is used to move the loop in the magnetic field, and changing its magnetic flux).
The negative sign in Lenz's Law tells exactly this: the direction of the induced emf is such that it opposes the initial change in magnetic flux that generated the induced emf, so that overall the total energy is conserved.
Transverse wave as the wave is going up and down no compressions
Momentum is conserved if and only if sum of all forces which are exserted on system equals zero. In our situation there are only internal forces, so by Newton's third law their vector sum is 0.
So 
.
Kinetic energy of system at first: 
. After: 
. The secret is that other energy is in work of deformation forces (they in turn heat a bullet and a block).
Answer is A)
Your answer is C) The speed of sound is higher in solids than in liquids.
Assume no air resistance, and g = 9.8 m/s².
Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity
The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s
With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0
Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°
The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s
If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)
Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s
The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m
Answer: h = 110.4 m
