Answer:
B. Newton's second law of motion
Explanation:
Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.
Mathematically, it is given by the formula;
Making acceleration the subject of formula, we have;
In this scenario, the acceleration of a baseball after it is hit by a bat depends on the mass of the ball and the net force on the ball. Thus, this example best illustrates Newton's second law of motion.
Answer:
Explanation:
From the question we are told that
Initial velocity of 60 m/s
Wind speed 
Generally Resolving vector mathematically
Generally the equation Pythagoras theorem is given mathematically by
Therefore Resultant velocity (m/s)
b)Resultant direction
Generally the equation for solving Resultant direction
Therefore
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s
a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49
Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.
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This is incomplete question Complete Question is:
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?
To solve this problem, we use the equation:
<span>d = (v^2 - v0^2) /
2a</span>
where,
d = distance of collapse
v0 = initial velocity = 101 km / h = 28.06 m / s
v = final velocity = 0
a = acceleration = - 300 m / s^2
d = (-28.06 m / s)^2 / (2 * - 300 m / s^2)
<span>d = 1.31 m</span>
Its a thermometer . i hope this helps you
