Paleontologists estimate that if a Tyrannosaurus rex were to trip and fall, it would have experienced a force of approximately 260,000 N acting on its torso when it hit the ground.

Question 1
Assuming the torso has a mass of 3800 kg , find the magnitude of the torso's upward acceleration as it comes to rest. (For comparison, humans lose consciousness with an acceleration of about 7g .)

Question 2
Assuming the torso is in free fall for a distance of 1.21 m as it falls to the ground, how much time is required for the torso to come to rest once it contacts the ground?

How do you do this......formula's anything.

1st one is 59

See:

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To solve these questions, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = ma). We can rearrange this formula to solve for acceleration (a = F/m) and use kinematic equations to answer the questions.

Question 1:
Given:
Force (F) = 260,000 N
Mass (m) = 3800 kg

Using Newton's second law, we can find the magnitude of the torso's upward acceleration (a):
a = F/m
a = 260,000 N / 3800 kg
a ≈ 68.42 m/s²

Keep in mind that acceleration due to gravity on Earth is approximately 9.8 m/s², so the torso would experience an acceleration much greater than that of humans losing consciousness.

Question 2:
Given:
Distance (d) = 1.21 m

We need to find the time taken for the torso to come to rest once it contacts the ground. To do this, we'll use the kinematic equation that relates displacement, initial velocity, final velocity, acceleration, and time:
v^2 = u^2 + 2ad

In this case, the initial velocity (u) is 0 m/s, the final velocity (v) is also 0 m/s because the torso comes to rest, and the acceleration (a) is approximately 68.42 m/s².

Plugging in the values, we get:
0^2 = 0^2 + 2 * 68.42 * 1.21
0 = 0 + 2 * 82.8982 * t
0 = 165.7964 * t
t ≈ 0 seconds

Therefore, it would take approximately 0 seconds for the torso to come to rest once it contacts the ground.

To answer these questions, we need to use Newton's second law of motion, which states that the net force exerted on an object is equal to the product of its mass and acceleration (F = ma). We can rearrange this equation to solve for acceleration (a = F/m) and solve for the two questions separately.

For Question 1, we need to find the magnitude of the torso's upward acceleration as it comes to rest. We know the force acting on the torso (F = 260,000 N) and its mass (m = 3800 kg). Plugging these values into the equation, we get:

a = F/m
= 260,000 N / 3800 kg
≈ 68.42 m/s²

Therefore, the magnitude of the torso's upward acceleration is approximately 68.42 m/s².

As for Question 2, we need to determine the time it takes for the torso to come to rest once it contacts the ground when it falls a distance of 1.21 m. In this case, we'll be using the kinematic equation:

vf² = vi² + 2ad

where
vf = final velocity (0 m/s since the torso comes to rest)
vi = initial velocity (unknown)
a = acceleration (-68.42 m/s²)
d = distance (1.21 m)

Since the torso is in free fall, its initial velocity vi is 0 m/s. Plugging in the values into the equation, we have:

0² = 0 + 2(-68.42)(1.21)
0 = -166.28d
d ≈ 0.007 m/s²

Now, we can use the equation for velocity:

v = vi + at

Since the final velocity vf is 0 m/s and the initial velocity vi is 0 m/s, we can simplify the equation to:

0 = 0 + (-68.42)t
t = 0 / -68.42
t ≈ 0 seconds

The torso comes to rest instantaneously upon hitting the ground.

To summarize:
Question 1: The torso's upward acceleration is approximately 68.42 m/s².
Question 2: The time required for the torso to come to rest once it contacts the ground is approximately 0 seconds.