When an object's atoms move faster, its thermal energy increases and the object becomes warmer.
The west constituent of their sequence needs to cancel out 58 mph crosswind. Subsequently a northwest direction is a 45-degree angle up to even with the destination. That is the third point out of the triangle and the right angle is at the destination. The top side is the west constituent of their flight the vertical side is their resultant travel and the hypotenuse is their definite distance flown. Since the 58 mph crosswind was negated by flying northwest, the distance from the beginning to the destination must be the same distance as the west component of their travel. The hypotenuse is square root of twice the side since it has 2 identical sides.
c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02
Alternative solution:
c = sqrt (2) * 58 = 1.414 * 58 = 82.02
Therefore, they have to fly 82.02 mph
By compressing the spring a distance <em>x</em> (in m), you are storing 1/2 <em>k</em> <em>x</em> ² (in J) of potential energy, which is converted completely into kinetic energy 1/2 <em>m v</em> ², where
• <em>k</em> = 40 N/m = spring constant
• <em>m</em> = 10 kg = mass of the ball
• <em>v</em> = 2 m/s = ball's speed (at the moment the spring returns to its equilibrium point)
So we have
1/2 <em>k</em> <em>x</em> ² = 1/2 <em>m</em> <em>v</em> ²
<em>x</em> = √(<em>m</em>/<em>k</em> <em>v</em> ²) = √((10 kg) / (40 N/m) (2 m/s)²) = 1 m
Calderas are what sometimes surround it
De-celleration because speed is lower as time goes on
