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WITCHER [35]
3 years ago
11

When a car drives over a speed bump and oscillates up and down in simple harmonic motion, at which position during the motion is

the acceleration of the car the greatest? when a car drives over a speed bump and oscillates up and down in simple harmonic motion, at which position during the motion is the acceleration of the car the greatest? at the equilibrium position, x = 0 at half the maximum amplitude, x = a/2 at the maximum amplitude, x = a none; the acceleration is constant?
Physics
1 answer:
Oksana_A [137]3 years ago
3 0
Using the second Law of Newton, F = m * a, you know that acceleration is maximum when the force is maximum.


Using Hooke's Law, F = K Δx, you know that the force is maximum when the displacement from the equilibrium (Δx) is maximum.


So the answer is that the acceleration is maximum at the maximum amplitude x = a.
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You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.
Andru [333]

Answer:

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Explanation:

Hi there!

The equations of height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity of the ball at time t.

Placing the origin at the throwing point, y0 = 0.

Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.

v = v0 + g · t

6.00 m/s = 12.0 m/s -9.81 m/s² · t

(6.00 - 12.0)m/s / -9.81 m/s² = t

t = 0.612 s

Now, let´s calculate the height of the baseball at that time:

y = y0 + v0 · t + 1/2 · g · t²     (y0 = 0)

y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²

y = 5.51 m

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Have a nice day!

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3 years ago
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