The tomato is dropped. What is the velocity, , of the tomato when it hits the ground? Assume 82.7 of the work done in Part A is transferred to kinetic energy, , by the time the tomato hits the ground.

Express your answer numerically in meters per second.

3.119

To determine the velocity of the tomato when it hits the ground, we need to use the principle of conservation of energy.

The given information states that 82.7% of the work done is transferred to kinetic energy by the time the tomato hits the ground.

We can express this as:

Work done = Kinetic energy

Now, let's denote the initial potential energy of the tomato as PE and the final kinetic energy as KE.

Since the tomato is dropped, it starts with only potential energy, which is converted into kinetic energy when it hits the ground.

Therefore, we can represent this as:

PE = KE

Now, since the work done is transferred to kinetic energy:

Work done = KE - Initial Kinetic energy

Since there is no initial kinetic energy:

Work done = KE

Now, using the given information that 82.7% of the work done is transferred to kinetic energy:

0.827 * Work done = KE

Since the work done is equal to the initial potential energy:

0.827 * PE = KE

To calculate the velocity, we need to express the kinetic energy in terms of velocity:

KE = (1/2) * m * v^2

Where m is the mass of the tomato and v is its velocity.

Now, combining the equations:

0.827 * PE = (1/2) * m * v^2

We can rearrange this equation to solve for v:

v^2 = (2 * 0.827 * PE) / m

Taking the square root of both sides:

v = sqrt((2 * 0.827 * PE) / m)

To proceed, we need to know the mass of the tomato and the height from which it was dropped to calculate the initial potential energy (PE).

To find the velocity of the tomato when it hits the ground, we need to use the principle of conservation of energy.

The work done on the tomato can be calculated using the formula: Work = Force × Distance. However, since we are not given the force, we will use another approach.

According to the given information, 82.7% of the work done is transferred to kinetic energy. This means that the work done is equal to 82.7% of the total kinetic energy of the tomato when it hits the ground.

The formula for work done is given by: Work = Change in Kinetic Energy

Therefore, the work done on the tomato is equal to the change in kinetic energy. Let's denote the initial kinetic energy as K_i and the final kinetic energy as K_f.

The work done on the tomato can be expressed as: Work = K_f - K_i

Since 82.7% of the work done is transferred to kinetic energy, we can write the equation as: 0.827 × Work = K_f - K_i

However, since the tomato is dropped, its initial velocity is zero, and therefore its initial kinetic energy is also zero. Hence, the equation simplifies to: 0.827 × Work = K_f

Now, we can use the formula for kinetic energy: K = 0.5 × mass × velocity^2

Let's assume the mass of the tomato is denoted as m and its velocity when it hits the ground is v.

At the point of impact, the tomato is at its maximum height, and all of its gravitational potential energy is converted into kinetic energy. The gravitational potential energy is given by: Potential Energy = mass × acceleration due to gravity × height

Let's assume the height is denoted as h and the acceleration due to gravity is denoted as g.

Therefore, the work done on the tomato can be expressed as: Work = m × g × h

Now, replacing the Work done in the equation 0.827 × Work = K_f, we get: 0.827 × (m × g × h) = 0.5 × m × v^2

Canceling out the mass from both sides, we have: 0.827 × g × h = 0.5 × v^2

Now, we can solve for v. Divide both sides of the equation by 0.5 and take the square root: v = √((0.827 × g × h) / 0.5)

To get the exact numerical value, substitute the known values for g (acceleration due to gravity, approximately 9.8 m/s^2) and h (height from which the tomato was dropped).

v = √((0.827 × 9.8 × h) / 0.5)

Now, calculate the height from which the tomato is dropped and substitute it into the equation. Once you have the height value, you can plug it into the formula and calculate the velocity.

I don't believe this is chemistry.