Answer:
<h3>1.01 s</h3>
Explanation:
Using the equation of motion S = ut+1/2gt² to solve the problem where;
u is the initial velocity of the chocolate = 0m/s
t is the time taken
g is the acceleration due to gravity = 9.81m/s²
S is the height of fall = 5.0m
Substituting the given parameter into the formula to get the time t we have;
5 = 0(t)+1/2(9.81)t²
5 = 4.905t²
t² = 5/4.905
t² = 1.019
t = √1.019
t = 1.009 secs
<em>Hence it will take 1.01 secs for me to catch the chocolate bar</em>
1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.
2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

Where,
a = Acceleration
A = Amplitude
= Angular velocity
From a reference system in which the downward acceleration is negative due to the force of gravity we will have to



From the definition of frequency and angular velocity we have to




Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz
What are your answer choices?
Answer:
a) It takes her 1.43 s to reach a speed of 2.00 m/s.
b) Her deceleration is - 2.50 m/s²
Explanation:
The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:
v = v0 + a · t
Where:
v = velocty.
v0 = initial velocity.
a = acceleration.
t = time.
a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:
v = v0 + a · t
2.00 m/s = 0 m/s + 1.40 m/s² · t
t = 2.00 m/s / 1.40 m/s²
t = 1.43 s
It takes her 1.43 s to reach a speed of 2.00 m/s
b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:
v = v0 + a · t
0 = 2.00 m/s + a · 0.800 s
-2.00 m/s / 0.800 s = a
a = -2.50 m/s²
Her deceleration is - 2.50 m/s²