A student with mass 50 kg jumps off a high diving

board.Using 6.0×1024 kg for the mass of the earth, what is the acceleration of the earth toward her as she accelerates toward the earth with an acceleration of 9.8 m/s2 ? Assume that the net force on the earth is the force of gravity she exerts on it.

as always, a = F/m

= (50*9.8)N/(6.0*10^24)kg
= 8.2*10^-23 m/s^2

Well, well, well, jumping off a diving board, huh? That's a pretty bold move! Now, let's answer your question about the acceleration of the Earth when a student jumps off.

First things first, we need to calculate the force that the student exerts on the Earth. We can use good old Newton's second law: force equals mass times acceleration. The student's mass is 50 kg, and her acceleration is 9.8 m/s². So, the force exerted on the Earth is 50 kg × 9.8 m/s².

Now, remember that every action (or force) has an equal and opposite reaction. In this case, the Earth experiences a force equal in magnitude but opposite in direction to the force exerted by the student. So, the force exerted by the Earth on the student is also 50 kg × 9.8 m/s².

When it comes to acceleration, we can use Newton's second law in a slightly different way: force equals mass times acceleration. Since we already know the force exerted by the Earth (50 kg × 9.8 m/s²), and we know the mass of the Earth is 6.0 × 10^24 kg, we can calculate the acceleration of the Earth using the equation: acceleration equals force divided by mass.

So, let's plug in the numbers: acceleration = (50 kg × 9.8 m/s²) divided by 6.0 × 10^24 kg.

And the answer is... drumroll, please... a teeny-tiny value. You see, the mass of the Earth is huge compared to the student, so the acceleration of the Earth is practically negligible. In other words, the Earth barely feels a thing when the student makes her great dive! The value is so small, I couldn't even fit it in this reponse! So, let's just say that the Earth's acceleration... *puts on clown wig* ...is clown-sized! Not very noticeable, my friend!

To find the acceleration of the Earth toward the student, we can use Newton's third law of motion, which states that the force exerted by object A on object B is equal in magnitude and opposite in direction to the force exerted by object B on object A.

In this case, as the student accelerates towards the Earth, she exerts a force on the Earth, and the Earth exerts an equal and opposite force on her.

The force exerted by the student on the Earth is given by Newton's second law:

Force = mass × acceleration

The mass of the student is 50 kg, and the acceleration is 9.8 m/s^2. Hence,

Force = 50 kg × 9.8 m/s^2 = 490 N

According to Newton's third law, the force exerted by the Earth on the student is also 490 N but in the opposite direction.

Now we can find the acceleration of the Earth using Newton's second law:

Force = mass × acceleration

The mass of the Earth is given as 6.0 × 10^24 kg, and the force exerted by the Earth on the student is 490 N. Hence,

490 N = 6.0 × 10^24 kg × acceleration

To find the acceleration, we rearrange the equation:

acceleration = 490 N / (6.0 × 10^24 kg)

Using a calculator, we can find the acceleration to be approximately 8.17 × 10^-23 m/s^2.

Therefore, the acceleration of the Earth toward the student is approximately 8.17 × 10^-23 m/s^2.

To find the acceleration of the Earth toward the student, we need to use Newton's third law of motion, which states that every action has an equal and opposite reaction.

Given:
Mass of the student (m1) = 50 kg
Mass of the Earth (m2) = 6.0 × 10^24 kg
Acceleration of the student toward the Earth (a1) = 9.8 m/s²

We know that force (F) is given by the equation:

F = m1 × a1

The force exerted by the student on the Earth is equal in magnitude but opposite in direction, so we have:

F = -m2 × a2

Here, a2 represents the acceleration of the Earth toward the student. The negative sign indicates that the direction of this acceleration is opposite to that of the student.

Now, we can equate the two forces:

m1 × a1 = -m2 × a2

To find a2, we rearrange the equation:

a2 = (-m1 × a1) / m2

Substituting the given values, we get:

a2 = (-50 kg × 9.8 m/s²) / (6.0 × 10^24 kg)

Calculating this expression, we find that the acceleration of the Earth toward the student is approximately -8.1678 × 10^(-23) m/s². The negative sign indicates that the acceleration is in the opposite direction of the student's acceleration.

Please note that this value is extremely small due to the massive mass of the Earth compared to the student.