The problem involves the concept of acceleration due to gravity. On the moon, the acceleration due to gravity is about 1/6th of that on Earth, or approximately 1.6 m/s². However, the problem states that the marble falls 4.00 m in the first second. This suggests that the acceleration due to gravity in this scenario is actually 4.00 m/s², not 1.6 m/s².

The distance an object falls under constant acceleration is given by the equation d = 0.5*a*t², where d is the distance, a is the acceleration, and t is the time.

In the first second (t = 1.0 s), the marble falls 4.00 m. So we have 4.00 m = 0.5*a*(1.0 s)². Solving for a gives a = 8.00 m/s².

In the next second, the marble continues to fall. However, it's important to note that the marble has already been falling for 1.0 s, so we want to know the distance it falls between t = 1.0 s and t = 2.0 s.

The total distance the marble falls in 2.0 s is given by d = 0.5*a*(2.0 s)² = 0.5*8.00 m/s²*(2.0 s)² = 16.00 m.

So, the additional distance the marble falls in the next 1.0 s is 16.00 m - 4.00 m = 12.00 m.

Question

The problem involves the concept of acceleration due to gravity. On the moon, the acceleration due to gravity is about 1/6th of that on Earth, or approximately 1.6 m/s². However, the problem states that the marble falls 4.00 m in the first second. This suggests that the acceleration due to gravity in this scenario is actually 4.00 m/s², not 1.6 m/s².

The distance an object falls under constant acceleration is given by the equation d = 0.5*a*t², where d is the distance, a is the acceleration, and t is the time.

In the first second (t = 1.0 s), the marble falls 4.00 m. So we have 4.00 m = 0.5*a*(1.0 s)². Solving for a gives a = 8.00 m/s².

In the next second, the marble continues to fall. However, it's important to note that the marble has already been falling for 1.0 s, so we want to know the distance it falls between t = 1.0 s and t = 2.0 s.

The total distance the marble falls in 2.0 s is given by d = 0.5*a*(2.0 s)² = 0.5*8.00 m/s²*(2.0 s)² = 16.00 m.

So, the additional distance the marble falls in the next 1.0 s is 16.00 m - 4.00 m = 12.00 m.

Knowee AI · Accepted Answer

The problem involves the concept of acceleration due to gravity. On the moon, the acceleration due to gravity is about 1/6th of that on Earth, or approximately 1.6 m/s². However, the problem states that the marble falls 4.00 m in the first second. This suggests that the acceleration due to gravity in this scenario is actually 4.00 m/s², not 1.6 m/s².

The distance an object falls under constant acceleration is given by the equation d = 0.5*a*t², where d is the distance, a is the acceleration, and t is the time.

In the first second (t = 1.0 s), the marble falls 4.00 m. So we have 4.00 m = 0.5*a*(1.0 s)². Solving for a gives a = 8.00 m/s².

In the next second, the marble continues to fall. However, it's important to note that the marble has already been falling for 1.0 s, so we want to know the distance it falls between t = 1.0 s and t = 2.0 s.

The total distance the marble falls in 2.0 s is given by d = 0.5*a*(2.0 s)² = 0.5*8.00 m/s²*(2.0 s)² = 16.00 m.

So, the additional distance the marble falls in the next 1.0 s is 16.00 m - 4.00 m = 12.00 m.

17. On the moon’s far east, a marble is released from rest at the top of USSR rover machine. It falls 4.00 𝘮 in the first 1.0 𝘴 of its motion. Through what additional distance does it fall in the next 1.0 𝘴?

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